Page 1

13.1. Model: The crankshaft is a rotating rigid body.

Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s, maintains a constant angular velocity for 2 s until t = 4 s, and then speeds up to 200 rad/s from t = 4 s to t = 7 s. The angular acceleration (ι) graph is based on the fact that ι is the slope of the ω -versus-t graph.


13.2. Solve:

Model: The turntable is a rotating rigid body. The angular velocity is the area under the α-versus-t graph:

α=

dω ⇒ ω = ∫ α ( x ) dt = ω 0 + area under the α graph dt

The values of ω at selected values of time (t) are: t (s) 0

ω (rad/s) 0

0.5

(5 + 3.75)(0.5)/2 = 2.18

1.0

(5 + 2.5)(1)/2 = 3.75

1.5

(5 + 1.25)(1.5)/2 = 4.68

2.0

(5 + 0)(2)/2 = 5.0

2.5

5.0

3.0

5.0


13.3. Solve:

Model: (a)

The wheel is a rotating rigid body.

The angular acceleration (ι) is the slope of the ω-versus-t graph. (b) The car is at rest at t = 0 s. It gradually speeds up for 4 s and then slows down for 4 s. The car is at rest from t = 8 s to t = 12 s, and then speeds up in the opposite direction for 4 s.


13.4.

Model:

The angular velocity and angular acceleration graphs correspond to a rotating rigid body.

Solve:

(a) The α-versus-t graph has a positive slope of 5 rad/s2 from t = 0 s to t = 2 s and a negative slope of −5 rad/s2 from t = 2 s to t = 4 s. (b) The angular velocity is the area under the α-versus-t graph:

α=

dω ⇒ ω = ∫ α ( x ) dt = ω 0 + area under α graph. dt


13.5.

Model: Visualize:

Spinning skater, whose arms are outstretched, is a rigid rotating body.

Solve: The speed v = rω, where r = 140 cm/2 = 0.70 m. Also, 180 rpm = (180)2π /60 rad/s = 6π rad/s. Thus, v = (0.70 m)(6π rad/s) = 13.2 m/s. Assess: A speed of 13.2 m/s ≈ 26 mph for the hands is a little high, but reasonable.


13.6.

Model: Visualize:

Solve:

The magnetic computer disk is a rigid rotating body.

Using the rotational kinematic equation ω f = ω i + α∆t, we get

ω1 = 0 rad + (600 rad/s2)(0.5 s − 0 s) = 300 rad/s ω2 = (300 rad/s) + (0 rad/s2)(1.0 s – 0.5 s) = 300 rad/s The speed of the painted dot v2 = rω2 = (0.04 m)(300 rad/s) = 12 m/s. The number of revolutions during the time interval t0 to t2 is 1 1 θ1 = θ 0 + ω 0 (t1 − t 0 ) + α 0 (t1 − t 0 ) 2 = 0 rad + 0 rad + (600 rad /s 2 )(0.5 s − 0 s) 2 = 75 rad 2 2 1 2 θ 2 = θ1 + ω 1 (t2 − t1 ) + α 1 (t2 − t1 ) 2  1 rev  = 75 rad + (300 rad /s)(1.0 s − 0.5 s) + 0 rad = 225 rad = (225 rad )  = 35.8 rev  2π rad 


13.7.

Model: Visualize:

The drill is a rigid rotating body.

The figure shows the drill’s motion from the top. Solve: (a) The kinematic equation ωf = ωi + α(tf – ti) becomes, after using ωi = 2400 rpm = (2400)(2π)/60 = 251.3 rad/s, tf – ti = 2.5 s – 0 s = 2.5 s, and ωf = 0 rad/s, 0 rad = 251.3 rad/s + α(2.5 s) ⇒ α = −100.5 rad /s 2 (b) Applying the kinematic equation for angular position yields: 1 θ f = θ i + ω i (t f − t i ) + α (t f − t i ) 2 2 1 = 0 rad + (251.3 rad /s)(2.5 s − 0 s) + ( −100.5 rad /s 2 )(2.5 s − 0 s) 2 2 = 314.2 rad = 50.0 rev


13.8.

Model: The turbine is a rigid rotating body. Solve: The known values are ωi = 3600 rpm = (3600)(2π)/60 = 120π rad/s, ti = 0 s, tf = 10 min = 600 s, ωf = 0 rad/s, and θi = 0 rad. Using the rotational kinematic equation ωf = ωi + α(tf – ti), we get 0 rad = (120π rad/s) + α(600 s – 0 s). Thus, α = −0.628 rad/s2. Now, 1 θ f = θ i + ω i (t f − t i ) + α (t f − t i ) 2 2 1 = 0 rad + (120π rad /s)(600 s − 0 s) + ( −0.628 rad /s 2 )(600 s − 0 s) 2 2 = 113,100 rad = 18,000 rev Assess: 18,000 revolutions during 10 minutes when the starting angular velocity is 3600 rpm is reasonable.


13.9.

Model: The triangle is a rigid body rotating about its center of mass perpendicular to the plane of the triangle. The center of mass of any symmetrical object of uniform density is at the physical center of the object. Visualize:

The distance to one tip of the triangle from the center of mass is given by r cos 30° = (5/2) cm, which yields r = (2.5 cm)/cos 30° = 2.89 cm. Solve: The speed of the tip is v = rω = (2.89 cm )(120 rpm ) = (2.89 cm )( 4π rad /s) = 36.3 cm/s


13.10. Visualize: Please refer to Figure Ex13.10. The coordinates of the three masses mA, mB, and mC are (0 cm, 0 cm), (10 cm, 10 cm), and (10 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are m x + mB x B + mC x C (100 g)(0 cm ) + (200 g)(10 cm ) + (300 g)(10 cm ) x cm = A A = = 8.33 cm m A + m B + mC (100 g + 200 g + 300 g) ycm =

mA yA + mB yB + mC yC (100 g)(0 cm ) + (200 g)(10 cm ) + (300 g)(0 cm ) = 3.33 cm = (100 g + 200 g + 300 g) m A + m B + mC


13.11. Visualize:

r Solve: Torque by a force is defined as τ = Frsinφ where φ is measured counterclockwise from the r vector to the r F vector. The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 20 N force: (30 N)r1 sinφ1 + (20 N)r2 sinφ 2 = (30 N)(0.02 m) sin (−90°) + (20 N)(0.02 m) sin (90°) = ( −0.60 N m) + (0.40 N m) = −0.20 N m Assess: A negative torque causes a clockwise motion of the pulley.


13.12. Visualize:

The two equal but opposite 50 N forces, one acting at point P and the other at point Q, make a couple that causes a net torque. Solve: The distance between the lines of action is l = d cos 30°. The net torque is given by τ = lF = ( d cos 30°) F = (0.10 m)(0.866)(50 N)

= 4.33 N m


13.13. Visualize:

Solve:

The net torque on the spark plug is τ = Fr sin φ = −38 N m = F(0.25 m )sin( −120°) ⇒ F = 175.5 N

That is, you must pull with a force of 175.5 N to tighten the spark plug. Assess: The force applied on the wrench leads to its clockwise motion. That is why we have used a negative sign for the net torque.


13.14.

Model: Visualize:

The disk is a rotating rigid body.

The radius of the disk is 10 cm and the disk rotates on an axle through its center. Solve: The net torque on the axle is τ = FArA sin φA + FBrB sin φB + FCrC sin φC + FDrD sin φD = (30 N)(0.10 m) sin (−90°) + (20 N)(0.05 m) sin 90° + (30 N)(0.05 m) sin 135° + (20 N)(0.10 m) sin 0° = −3 N m + 1 N m + 1.0607 N m = −0.939 N m Assess: A negative torque means a clockwise rotation of the disk.


13.15.

Model: Visualize:

The beam is a solid rigid body.

r r The steel beam experiences a torque due to the construction worker’s weight wC and the beam’s weight wB . The normal force exerts no torque since the net torque is calculated about the point where the beam is bolted into place. r r Solve: The net torque on the steel beam about point O is the sum of the torque due to wC and the torque due to wB . The weight of the beam acts at the center of mass. τ = (wC)(4.0 m) sin (−90°) + (wB)(2.0 m) sin (−90°) = −(70 kg)(9.80 m/s2)(4.0 m) – (500 kg)(9.80 m/s2)(2.0 m) = −12,500 N m The negative torque means these forces would cause the beam to rotate clockwise. The magnitude of the torque is 12,500 N m.


13.16.

Model: Visualize:

Solve:

Model the arm as a uniform rigid rod. Its mass acts at the center of mass.

(a) The torque is due both to the weight of the ball and the weight of the arm: τ = τ ball + τ arm = ( mb g)rb sin 90° + ( ma g)ra sin 90°

= (3.0 kg)(9.8 m / s 2 ) (0.70 m) + ( 4.0 kg)(9.8 m / s 2 ) (0.35 m) = 34.4 N m (b) The torque is reduced because the moment arms are reduced. Both forces act at φ = 45° from the radial line, so τ = τ ball + τ arm = ( mb g)rb sin 45° + ( ma g)ra sin 45° = (3.0 kg)(9.8 m / s 2 ) (0.70 m)(0.707) + ( 4.0 kg)(9.8 m / s 2 ) (0.35 m)(0.707) = 24.3 N m


13.17.

Model: The moment of inertia of any object depends on the axis of rotation. In the present case, the rotation axis passes through mass A and is perpendicular to the page. Visualize: Please refer to Figure Ex13.17. ∑ mi x i = m A x A + m B x B + m C x C + m D x D Solve: (a) x cm = m A + m B + mC + m D ∑ mi

(100 g)(0 m ) + (200 g)(0 m ) + (200 g)(0.10 m ) + (200 g)(0.10 m ) = 0.0571 m 100 g + 200 g + 200 g + 200 g m y + m B y B + mC yC + m D y D = A A m A + m B + mC + m D =

ycm

=

(100 g)(0 m ) + (200 g)(0.10 m ) + (200 g)(0.10 cm ) + (200 g)(0 m ) = 0.0571 m 700 g

(b) The distance from the axis to mass C is 14.14 cm. The moment of inertia through A and perpendicular to the page is

I A = ∑ mi ri2 = mA rA2 + mB rB2 + mC rC2 + mD rD2 i

= (0.100 kg)(0 m)2 + (0.200 kg)(0.10 m)2 + (0.200 kg)(0.1414 m)2 + (0.200 kg)(0.10 m)2 = 0.0080 kg m2


13.18.

Model: Visualize:

Solve:

(a)

The moment of inertia of any object depends on the axis of rotation.

x cm =

∑m x ∑m i

i

ycm

i

=

m A x A + m B x B + mC x C + m D x D m A + m B + mC + m D

(100 g)(0 m ) + (200 g)(0 m ) + (200 g)(0.10 m ) + (200 g)(0.10 m ) = = 0.0571 m 100 g + 200 g + 200 g + 200 g m y + m B y B + mC yC + m D y D = A A m A + m B + mC + m D =

(100 g)(0 m ) + (200 g)(0.10 m ) + (200 g)(0.10 cm ) + (200 g)(0 m ) = 0.0571 m 700 g

(b) The moment of inertia about a diagonal that passes through B and D is

I BD = mA rA2 + mC rC2 where rA = rC = (0.10 m) cos 45° = 7.07 cm and are the distances from the diagonal. Thus,

I BD = (0.100 kg)rA2 + (0.200 kg)rC2 = 0.0015 kg m 2 Assess: Note that the masses B and D, being on the axis of rotation, do not contribute to the moment of inertia.


13.19.

Model: The three masses connected by massless rigid rods is a rigid body. Visualize: Please refer to Figure Ex13.19. ∑ mi xi = (0.100 kg)(0 m) + (0.200 kg)(0.06 m) + (0.100 kg)(0.12 m) = 0.060 m x cm = Solve: (a) 0.100 kg + 0.200 kg + 0.100 kg ∑ mi

ycm =

∑m y ∑m

i i

=

(

)

(0.100 kg)(0 m ) + (0.200 kg) (0.10 m ) 2 − (0.06 m ) 2 + (0.100 kg)(0 m ) 0.100 kg + 0.200 kg + 0.100 kg

i

= 0.040 m

(b) The moment of inertia about an axis through A and perpendicular to the page is

I A = ∑ mi ri2 = mB (0.10 m ) + mC (0.10 m ) 2 = (0.100 kg)[(0.10 m ) 2 + (0.10 m ) 2 ] = 0.0020 kg m 2 2

(c) The moment of inertia about an axis that passes through B and C is

IBC = mA

(

(0.10 m )2 − (0.06 m )2

) = 0.00128 kg m 2

2

Assess: Note that mass mA does not contribute to IA, and the masses mB and mC do not contribute to IBC.


13.20. Solve: τ = Iα is the rotational analog of Newton’s second law F = ma. We have τ = (2.0 kg m2)(4.0 rad/s2) =

8.0 kg m2/s2 = 8.0 N m.


13.21. Visualize: Since α = τ/I, a graph of the angular acceleration looks just like the torque graph with the numerical values divided by I = 4.0 kg m2.

Solve: From Equation 13.6

ωf = ωi + area under the angular acceleration α curve between ti and tf. The area under the curve between t = 0 s and t = 3 s is 0.75 rad/s. With ωi = 0 rad/s, we have ωf = 0 rad/s + 0.75 rad/s = 0.75 rad/s


13.22.

Model: Two balls connected by a rigid, massless rod are a rigid body rotating about an axis through the center of mass. Assume that the size of the balls is small compared to 1 m. Visualize:

We placed the origin of the coordinate system on the 1.0 kg ball. Solve: The center of mass and the moment of inertia are

x cm =

(1.0 kg)(0 m ) + (2.0 kg)(1.0 m ) = 0.667 m and ycm = 0 m (1.0 kg + 2.0 kg)

Iabout cm = ∑ mi ri2 = (1.0 kg)(0.667 m) 2 + (2.0 kg)(0.333 m) 2 = 0.667 kg m 2 We have ωf = 0 rad/s, tf – ti = 5.0 s, and ω i = 20 rpm = 20(2π rad /60 s) = 23 π rad /s, so ωf = ωi + α(tf – ti) becomes

2π 2π 0 rad /s =  rad /s + α (5.0 s) ⇒ α = − rad /s 2  3  15 Having found I and α, we can now find the torque τ that will bring the balls to a halt in 5.0 s:

2 2π 4π τ = Iabout cmα =  kg m 2   − rad /s 2  = − N m = −0.279 N m 3   15  45 The magnitude of the torque is 0.279 N m.


13.23.

Model: A circular plastic disk rotating on an axle through its center is a rigid body. Assume axis is perpendicular to the disk. Solve: To determine the torque (τ) needed to take the plastic disk from ωi = 0 rad/s to ωf = 1800 rpm = (1800)(2π)/ 60 rad/s = 60π rad/s in tf – ti = 4.0 s, we need to determine the angular acceleration (α) and the disk’s moment of inertia (I) about the axle in its center. The radius of the disk is R = 10.0 cm. We have 1 1 I = MR 2 = (0.200 kg)(0.10 m ) 2 = 1.0 × 10 −3 kg m 2 2 2 ω − ω i 60π rad /s − 0 rad /s ω f = ω i + α (t f − t i ) ⇒ α = f = = 15π rad /s 2 tf − ti 4.0 s Thus, τ = Iα = (1.0 × 10−3 kg m2)(15π rad/s2) = 0.0471 N m.


13.24.

Model: The rocket attached to the end of a rigid rod is a rotating rigid body. Assume the rocket is small compared to 60 cm. Visualize: Please refer to Figure Ex13.24. Solve: We can determine the rocket’s angular acceleration from the relationship τ = Iα. The torque τ can be found from the thrust (F) using τ = Frsinφ. The moment of inertia (I) can be calculated from equations given in Table 13.3. Specifically, I = Irod about one end + Irocket becomes 1 1 M rod L2 + ML2 = (0.100 kg)(0.60 m ) 2 + (0.200 kg)(0.60 m ) 2 3 3 = 0.012 kg m 2 + 0.072 kg m 2 = 0.0840 kg m 2

⇒α =

τ Fr sin φ ( 4.0 N )(0.60 m )sin( 45°) = = = 20.2 rads/s 2 I I 0.0840 kg m 2

Assess: The rocket will accelerate counterclockwise since α is positive.


13.25.

Model: Visualize:

The rod is in rotational equilibrium, which means that τnet = 0.

As the weight of the rod and the hanging mass pull down (the rotation of the rod is exaggerated in the figure), the rod touches the pin at two points. The piece of the pin at the very end pushes down on the rod; the right end of the pin pushes up on the rod. To understand this, hold a pen or pencil between your thumb and forefinger, with your thumb on top (pushing down) and your forefinger underneath (pushing up). Solve: Calculate the torque about the left end of the rod. The downward force exerted by the pin acts through this r point, so it exerts no torque. To prevent rotation, the pin’s normal force npin exerts a positive torque (ccw about the left end) to balance the negative torques (cw) of the weight of the mass and rod. The weight of the rod acts at the center of mass, so

τ net = 0 N m = τ pin − (0.40 m)(2.0 kg)(9.8 m / s 2 ) − (0.80 m)(0.50 kg)(9.8 m / s 2 ) ⇒ τ pin = 11.8 N m


13.26.

Model: Visualize:

Solve:

The massless rod is a rigid body.

r To be in equilibrium, the object must be in both translational equilibrium ( Fnet = 0 N ) and rotational

equilibrium (τnet = 0 Nm). We have ( Fnet ) y = ( 40 N ) − (100 N ) + (60 N ) = 0 N, so the object is in translational equilibrium. Measuring τnet about the left end,

τnet = (60 N)(3.0 m) sin (+90°) + (100 N)(2.0 m) sin (−90°) = −20 N m The object is not in equilibrium.


13.27.

Model: Visualize:

The object balanced on the pivot is a rigid body.

Since the object is balanced on the pivot, it is in both translational equilibrium and rotational equilibrium. r Solve: There are three forces acting on the object: the weight w1 acting through the center of mass of the long r r rod, the weight w2 acting through the center of mass of the short rod, and the normal force P on the object applied

by the pivot. The translational equilibrium equation ( Fnet ) y = 0 N is

− w1 − w2 + P = 0 N ⇒ P = w1 + w2 = (1.0 kg)(9.8 m /s 2 ) + ( 4.0 kg)(9.8 m /s 2 ) = 49.0 N Measuring torques about the left end, the equation for rotational equilibrium τnet = 0 Nm is

Pd − w1 (1.0 m ) − w2 (1.5 m ) = 0 Nm ⇒ (49.0 N)d − (1.0 kg)(9.8 m/s 2 )(1.0 m) − (4.0 kg)(9.8 m/s 2 )(1.5 m) = 0 N ⇒ d = 1.40 m Thus, the pivot is 1.40 m from the left end.


13.28.

Model: The earth is a rigid, spherical rotating body. Solve: The rotational kinetic energy of the earth is K rot = 12 Iω 2 . The moment of inertia of a sphere about its diameter (see Table 13.3) is I = 25 Mearth R 2 and the angular velocity of the earth is

ω=

2π rad = 7.27 × 10 −5 rad /s 24 × 3600 s

Thus, the rotational kinetic energy is

12 Mearth R 2  ω 2  25 1 = (5.98 × 10 24 kg)(6.37 × 10 6 m ) 2 (7.27 × 10 −5 rad /s) 2 = 2.57 × 10 29 J 5

K rot =


13.29.

Model: The triangle is a rigid body rotating about an axis through the center. Visualize: Please refer to Figure Ex13.28. Each 200 g mass is a distance r away from the axis of rotation, where r is given by

0.20 m 0.20 m = cos 30° ⇒ r = = 0.2309 m r cos 30° Solve: The moment of inertia of the triangle is I = 3 × mr2 = 3(0.200 kg)(0.2309 m)2 = 0.0320 kg m2 . The frequency of rotation is given as 5.0 revolution per s or 10π rad/s. The rotational kinetic energy is 1 1 K rot. = Iω 2 = (0.0320 kg m 2 )(10.0π rad /s) 2 = 15.8 J 2 2


13.30.

Model: Visualize:

The disk is a rigid body rotating about an axis through its center.

Solve: The speed of the point on the rim is given by vrim = Rω. The angular velocity ω of the disk can be determined from its rotational kinetic energy which is K = 12 Iω 2 = 0.15 J . The moment of inertia I of the disk about its center and perpendicular to the plane of the disk is given by 1 1 I = MR 2 = (0.10 kg)(0.040 m ) 2 = 8.0 × 10 −5 kg m 2 2 2 0.30 J 2(0.15 J ) 2 ⇒ω = = ⇒ ω = 61.237 rad /s I 8.0 × 10 −5 kg m 2 Now, we can go back to the first equation to find vrim. We get vrim = Rω = (0.040 m)(61.237 rad/s) = 2.45 m/s.


13.31.

Model: Visualize:

The structure is a rigid body rotating about its center of mass.

We placed the origin of the coordinate system on the 300 g ball. Solve: First, we calculate the center of mass: (300 g)(0 cm ) + (600 g)( 40 cm ) x cm = = 26.67 cm 300 g + 600 g Next, we will calculate the moment of inertia about the structure’s center of mass: I = (300 g)(xcm)2 + (600 g)(40 cm – xcm)2 = (0.300 kg)(0.2667 m)2 + (0.600 kg)(0.1333 m)2 = 0.032 kg m2 Finally, we calculate the rotational kinetic energy:

1 2 1 100 × 2π Iω = (0.032 kg m 2 ) rad /s = 1.75 J  60  2 2 2

K rot =


13.32.

Solve:

(a) According to Equation 13.37, the speed of the center of mass of the tire is v 20 m /s 60 vcm = Rω = 20 m /s ⇒ ω = cm = = 66.67 rad /s = (66.7)  rpm = 637 rpm  0.30 m 2π  R (b) The speed at the top edge of the tire relative to the ground is vtop = 2vcm = 2(20 m/s) = 40 m/s. (c) The speed at the bottom edge of the tire relative to ground is vbottom = 0 m/s.


13.33.

Model: The can is a rigid body rolling across the floor. Assume that the can has uniform mass distribution. Solve: The rolling motion of the can is a translation of its center of mass plus a rotation about the center of mass. The moment of inertia of the can about the center of mass is 12 MR 2, where R is the radius of the can. Also vcm = Rω, where ω is the angular velocity of the can. The total kinetic energy of the can is

K = K cm + K rot = =

1 1 1 1 1 v  2 2 Mvcm + Icmω 2 = Mvcm +  MR 2  cm  R  2 2 2 22

3 3 2 Mvcm = (0.50 kg)(1.0 m /s) 2 = 0.375 J 4 4

2


13.34.

Model: Visualize:

The sphere is a rigid body rolling down the incline without slipping.

The initial gravitational potential energy of the sphere is transformed into kinetic energy as it rolls down. Solve: (a) If we choose the bottom of the incline as the zero of potential energy, the energy conservation equation will be Kf = Ui. The kinetic energy consists of both translational and rotational energy. This means 1 1 1 2 1 2 K f = Icmω 2 + Mvcm = Mgh ⇒  MR 2  ω 2 + M ( Rω ) 2 = Mgh  2 2 25 2 7 MR 2ω 2 = Mg(2.1 m )sin 25° ⇒ 10

⇒ω =

10 7

g(2.1 m )(sin 25°) = R2

10 7

g(2.1 m )(sin 25°) = 88.1 rad /s (0.04 m ) 2

(b) From part (a)

K total =

1 1 7 1 1 2 1 2 Icmω 2 + Mvcm = MR 2ω 2 and K rot = Icmω 2 =  MR 2  ω 2 = MR 2ω 2  2 2 10 2 25 5 ⇒

K rot = K total

1 5 7 10

MR 2ω 2 1 10 = × = 0.286 MR 2ω 2 5 7


13.35.

Visualize: Please refer r torFigure Ex13.35. r r (a) The magnitude of A × B is ABsin α = (6)( 4)sin 120° = 20.78 . The direction of A × B is given by the r r r r right hand rule. To curl our fingers from A to B, we have to point our thumb out of the page. Thus, A × B = (20.78, out of the page). r r (b) C × D = ((6)( 4)sin 90°, into the page) = (24, into the page). Solve:


Visualize: Please refer to Figure Ex13.36. To determine angle α, put the tails of the vectors together. r r r r (a) The magnitude of A × B is AB sin α = (6)( 4)sin 45° = 16.97. The direction of A × B, using the right r r hand rule, is out of the page. Thus, A × B = (16.97, out of the page). r r r r r (b) The magnitude of C × D is CD sin α = (6)( 4)sin 180° = 0. Thus C × D = 0.

13.36. Solve:


Solve: (a) (iˆ × jˆ ) × iˆ = kˆ × iˆ = jˆ iˆ × ( jˆ × iˆ ) = iˆ × ( − kˆ ) = −iˆ × kˆ = −( − jˆ ) = jˆ

13.37. (b)


Solve: (a) iˆ × (iˆ × jˆ ) = iˆ × kˆ = − jˆ r (iˆ × jˆ ) × kˆ = kˆ × kˆ = 0

13.38. (b)


13.39.

(b)

Solve:

(a)

r r A × B = (3iˆ − jˆ ) × (2iˆ + 3 jˆ − kˆ ) = 6iˆ × iˆ + 9iˆ × jˆ − 3iˆ × kˆ − 2 jˆ × iˆ − 3 jˆ × jˆ + jˆ × kˆ = 0 + 9kˆ − 3( − jˆ ) − 2( − kˆ ) − 0 + iˆ = iˆ + 3 jˆ + 11kˆ


r

r

r

r

13.40. Solve: (a) C ×r D = 0 implies that D must also be in the same or opposite direction as the C vector or zero, because iˆ × iˆ = 0. Thus D = niˆ, where n could be any real number. r r r r (b) C × E = 6 kˆ implies that E must be along the jˆ vector, because iˆ × jˆ = kˆ. Thus E = 2 jˆ. r r r r (c) C × F = −3 jˆ implies that F must be along the kˆ vector, because iˆ × kˆ = − jˆ. Thus F = 1kˆ.


13.41.

Solve:

r r r τ = r × F = (5iˆ + 5 jˆ ) × ( −10 jˆ ) N m r = [ −50(iˆ × jˆ ) − 50( jˆ × jˆ )] N m = [ −50( + kˆ ) − 0] N m = −50 kˆ N m


13.42.

Solve:

r r r τ = r × F = (5 jˆ ) × ( −10iˆ + 10 jˆ ) N m = ( −5 jˆ × iˆ + 50 jˆ × jˆ ) N m r = [ −50( − kˆ ) + 0] N m = 50 kˆ N m


13.43.

Model: The block attached to a massless rod is a rigid rotating body. Assume the block is a particle. The moment of inertia of the block about the pivot is I = ML2 = (0.200 kg)(0.40 m)2 = 0.032 kg m2 Since τ = Iα = 0.050 Nm, 0.050 Nm 0.050 Nm α= = = 1.56 rad /s 2 I 0.032 kg m 2

Solve:

(a) Substituting θ0 = 0 rad, t0 = 0 s, and ω0 = 0 rad/s into the rotational kinematic equation, we get 1 1 2 θ = θ 0 + ω 0 (t − t 0 ) + α (t − t 0 ) 2 10(2π )rad = 0 rad + 0 rad + (1.56 rad /s 2 )(t − 0 s) ⇒ t = 8.97 s 2 2 (b) The angular momentum is calculated as follows:

L = Iω = I[ω 0 + α (t − t 0 )] = I[0 rad /s + α (t − 0 s)] = Iα t = (0.032 kg m 2 )(1.56 rad /s 2 )(8.97 s) = 0.448 kg m 2 s (c)

∆L 0.448 kg m 2 /s − 0 kg m 2 /s = = 0.050 N m = τ ∆t 8.97 s − 0 s


13.44. Solve:

Visualize: Please refer to Figure Ex.13.44. r r r L = r × mv = (1.0iˆ + 2.0 jˆ ) m × (0.2 kg)(2.0iˆ ) m/s = (0.4iˆ × iˆ + 0.8 jˆ × iˆ ) kg m 2 / s = −(0.80 kˆ ) kg m 2 / s or (0.80 kg m 2 /s, into page)


13.45. Solve:

Visualize: Please refer to Figure Ex13.45. r r r L = r × mv = (3.0iˆ + 2.0 jˆ ) m × (0.1 kg)( 4.0 jˆ ) m /s

= 1.20(iˆ × jˆ ) kg m 2 / s + 0.8( jˆ × jˆ ) kg m 2 / s = 1.20 kˆ kg m 2 / s + 0 kg m 2 / s = 1.20 kˆ kg m 2 / s or (1.20 kg m 2 /s, out of page)


13.46.

Model: The bar is a rotating rigid body. Assume that the bar is thin. Visualize: Please refer to Figure Ex13.46. Solve: The angular velocity ω = 120 rpm = (120)(2π)/60 rad/s = 4π rad/s. From Table 13.3, the moment of inertial of a rod about its center is I = 121 ML2 . The angular momentum is

1 L = Iω =   (0.50 kg)(2.0 m ) 2 ( 4π rad /s) = 2.09 kg m 2 /s  12  If we wrap our fingers in the direction of the rod’s rotation, our thumb will point in the z direction or out of the page. Consequently, r L = (2.09 kg m 2 / s, out of the page)


13.47.

Model: The disk is a rotating rigid body. Visualize: Please refer to Figure Ex.13.47. Solve: From Table 13.3, the moment of inertial of the disk about its center is

I=

1 1 MR 2 = (2.0 kg)(0.020 m ) 2 = 4.0 × 10 −4 kg m 2 2 2

The angular velocity ω is 600 rpm = 600 × 2π/60 rad/s = 20π rad/s. Thus, L = Iω = (4.0 × 10−4 kg m2)(20π rad/s) = 0.0251 kg m2/s. If we wrap our right fingers in the direction of the disk’s rotation, our thumb will point in the –x direction. Consequently, r L = −0.0251 iˆ kg m 2 / s = (0.0251 kg m 2 / s, into page)


13.48.

Model: Visualize:

The earth is a rigid, rotating, and spherical body.

Solve: At a latitude of θ degrees, the radius is r = Re cos θ with Re = 6.37 × 106 m. (a) In Miami θ = 26°, and we have r = (6.37 × 106 m)(cos 26°) = 5.725 × 106 m. The angular velocity of the earth is

ω=

2π 2π = = 7.272 × 10 −5 rad /s 24 × 3600 s T

Thus, vstudent = rω = (5.725 × 106 m)(7.272 × 10−5 rad/s) = 416 m/s. (b) In Fairbanks θ = 65°, so r = (6.37 × 106 m)cos 65° = 2.692 × 106 m and vstudent = rω = (2.692 × 106 m)(7.272 × 10−5 rad/s) = 196 m/s.


13.49.

Model: Visualize:

The wheel is a rigid rolling body.

Solve: The front of the disk is moving forward at velocity vcm. Also, because of rotation the point is moving downward at velocity vrel = Rω = vcm. So, this point has a speed 2 2 v = vcm + vcm = 2 vcm = 2 (20 m /s) = 28.3 m /s

Assess: The speed v is independent of the radius of the wheel.


13.50. Visualize:

Solve: We will consider a vertical strip of width dx and of mass dm at a position x from the origin. The formula for the x component of the center of mass is 1 x cm = x dm M∫ The area of the steel plate is A = 12 (0.2 m )(0.3 m ) = 0.030 m 2 . Mass dm in the strip is the same fraction of M as dA is of A. Thus dm dA M 0.800 kg  = ⇒ dm = dA =  dA = (26.67 kg/ m 2 )ldx  0.030 m 2  M A A The relationship between l and x is

l x 2 = ⇒l= x 0.20 m 0.30 m 3 Therefore,

x cm =

(17.78 kg/ m 2 ) x 3 1 2 (26.67 kg/ m 2 )  x 2 dx = ∫  3 M M 3

Due to symmetry ycm = 0 cm.

0.3 m

= 0m

(17.78 kg/ m 2 ) (0.3 m ) 2 = 20 cm. 0.8 kg 3


13.51.

Model: Visualize:

Solve:

The disk is a rigid rotating body. The axis is perpendicular to the plane of the disk.

(a) From Table 13.3, the moment of inertia of a disk about its center is 1 1 I = MR 2 = (2.0 kg)(0.10 m ) 2 = 0.010 kg m 2 2 2 (b) To find the moment of inertia of the disk through the edge, we can make use of the parallel axis theorem: I = Icenter + Mh2 = (0.010 kg m2) + (2.0 kg)(0.10 m)2 = 0.030 kg m2 Assess: The larger moment of inertia about the edge means there is more inertia to rotational motion about the edge than about the center.


13.52. Visualize:

We chose the origin of the coordinate system to be on the axis of rotation, that is, at a distance d from one end of the rod. Solve: The moment of inertia can be calculated as follows: x2

I=

∫x

2

dm and

x1

⇒I=

M L

L−d

−d

M x3 x 2 dx =    L 3

L−d

= −d

dm dx M = ⇒ dm = dx M L L

M 1 M [( L − d ) 3 − ( − d ) 3 ] = [( L − d ) 3 + d 3 ] 3 L  3L

For d = 0 m, I = 13 ML2 , and for d = 12 L,

I=

3 3 M  L   L   1 ML2   +    = 3L  2 2  12

Assess: The special cases d = 0 m and d = L/2 of the general formula give the same results that are found in Table 13.3.


13.53. Visualize:

Solve: We solve this problem by dividing the disk between radii r1 and r2 into narrow rings of mass dm. Let dA = 2πrdr be the area of a ring of radius r. The mass dm in this ring is the same fraction of the total mass M as dA is of the total area A. (a) The moment of inertia can be calculated as follows:

Idisk = ∫ r 2 dm and dm =

M M dA = (2πr )dr 2 A π (r2 − r12 )

r

⇒ Idisk =

r

2 M 2M 2 3 2M r 4 r 2 (2πr ) dr = 2 r dr = 2 2 2 ∫ 2 ∫ π (r2 − r1 ) r1 (r2 − r1 ) r1 (r2 − r12 ) 4

r2

r1

2M M r 4 − r14 ) = (r22 + r12 ) = 2 2 ( 2 2 4(r2 − r1 ) Replacing r1 with r and r2 with R, the moment of inertia of the disk through its center is Idisk = 12 M ( R 2 + r 2 ). (b) For r = 0 m, Idisk = 12 MR 2 . This is the moment of inertia for a solid disk or cylinder about the center. Additionally, for r ≅ R, we have I = MR2. This is the expression for the moment of inertia of a cylindrical hoop or ring about the center. (c) The initial gravitational potential energy of the disk is transformed into kinetic energy as it rolls down. If we choose the bottom of the incline as the zero of potential energy, and use vcm = ωR, the energy conservation equation Kf = Ui is

v2 1 2 1 1 M 1 2 2 Iω + Mvcm = Mgh ⇒   ( R 2 + r 2 ) cm2 + Mvcm = Mgyi = Mg(0.50 m )sin 20° 2 2 2 2  2 R 2 2 r2  1 2 1 2  R +r  2 1 2 2 ⇒ vcm + = + + v v     = 1.6759 m /s cm cm  4 R2  2  2 4 4 R2 

(0.015 m ) 2  2 3 2 2 vcm  = 1.6759 m /s ⇒ vcm = 1.37 m /s  +  4 4(0.020 m ) 2  For a sliding particle on a frictionless surface K f = U i , so

v 1 2 mvf = mgyi ⇒ vf = 2 gyi = 2 g(0.50 m )sin 20° = 1.83 m /s ⇒ cm = 0.748 2 vf That is, vcm is 74.8% of the speed of a particle sliding down a frictionless ramp.


13.54. Model: The object is a rigid rotating body. Assume the masses m1 and m2 are small and the rod is thin. Visualize: Please refer to P13.54. Solve: The moment of inertia of the object is the sum of the moment of inertia of the rod, mass m1, and mass m2. Using Table 13.3 for the moment of inertia of the rod, we get 2

I rod = I rod about center + I m1 + I m 2 = =

1 L L ML2 + m1   + m2      12 2 4

m  1 1 1 L2  M ML2 + m1 L2 + m2 L2 = + m1 + 2  12 4 16 4 3 4 

Assess: With m1 = m2 = 0 kg, I rod =

1 12

ML2 , as expected.

2


13.55. Visualize:

Solve: We can divide the face of the square rod into small areas dA = dx dy with mass dm. Because the mass of the rod is uniformly distributed, dm is the same fraction of M as dA is of A = L2. That is

dA dm dx dy = ⇒ dm = M  2  2  L M L  The moment of inertia is

I = ∫ r 2 dm = =

L/2 L/2

L/2

L/2

L/2

L/2

M M M ( x 2 + y 2 ) dx dy = 2 ∫ x 2 dx ∫ dy + 2 ∫ dx ∫ y 2 dy L2 − L∫/ 2 − L∫/ 2 L − L/2 L − L/2 − L/2 − L/2 M x3 L2 3

L/2

L/2

y − L/2

− L/2

+

M L/2 y3 x L2 − L / 2 3

L/2

= − L/2

1 ML2 6


13.56.

Model: Visualize:

The beam is a rigid body of length 3.0 m and the student is a particle.

r r Solve: To stay in place, the beam must be in both translational equilibrium ( Fnet = 0 N ) and rotational equilibrium (τ net = 0 Nm ) . The first condition is

∑F

y

= − wbeam − wstudent + F1 + F2 = 0 N

⇒ F1 + F2 = wbeam + wstudent = (100 kg + 80 kg)(9.80 m/s2) = 1764 N Taking the torques about the left end of the beam, the second condition is −wbeam (1.5 m) – wstudent (2.0 m) + F2 (3.0 m) = 0 N m − (100 kg)(9.8 m/s2)(1.5 m) – (80 kg)(9.8 m/s2)(2.0 m) + F2 (3.0 m) = 0 N m ⇒ F2 = 1013 N From F1 + F2 = 1764 N, we get F1 = 1764 N – 1013 N = 751 N. Assess: To establish rotational equilibrium, the choice for the pivot is arbitrary. We can take torques about any point on the body of interest.


13.57.

Model: Visualize:

The structure is a rigid body.

Solve: We pick the left end of the beam as our pivot point. We don’t need to know the forces Fh and Fv because the pivot point passes through the line of application of Fh and Fv and therefore these forces do not exert a torque. For the beam to stay in equilibrium, the net torque about this point is zero. We can write τabout left end = – wB (3.0 m) – wW (4.0 m) + (T sin 150°)(6.0 m) = 0 N m Using wB = (1450 kg)(9.8 m/s2) and wW = (80 kg)(9.8 m/s2), the torque equation can be solved to yield T = 15300 N. The tension in the cable is slightly more than the cable rating. The worker should be worried.


13.58.

Model: Model the forearm as a rigid body. For the forearm parallel to the floor, we require both translational equilibrium ( Fnet = 0 N ) and rotational equilibrium (τ net = 0 N m ) . Visualize:

Solve:

τnet = Fbicep(0.025 m) – wF(0.16 m) – wB(0.32 m) = 0 N m Fbicep (0.025 m) – (1.2 kg)(9.8 m/s2)(0.16 m) – (0.500 kg)(9.8 m/s2)(0.32 m) = 0 N m

⇒ Fbicep = 138 N Assess: 138 N is much larger than the 4.9 N weight of the ball because the moment arm is so much shorter.


13.59. Model: r Model the beam as a rigid body. For the beam not to fall over, it must be both in translational equir librium ( Fnet = 0 N ) and rotational equilibrium (τ net = 0 N m ). Visualize:

The boy walks along the beam a distance x, measured from the left end of the beam. There are four forces acting on r r the beam. F1 and F2 are from the two supports, wb is the weight of the beam, and wB is the weight of the boy. Solve: We pick our pivot point on the left end through the first support. The equation for rotational equilibrium is –wb (2.5 m) + F2 (3.0 m) – wBx = 0 N m –(40 kg)(9.80 m/s2)(2.5 m) + F2 (3.0 m) – (20 kg)(9.80 m/s2)x = 0 N m The equation for translation equilibrium is

∑F

y

= 0 N = F1 + F2 − wb − wB

⇒ F1 + F2 = wb + wB = ( 40 kg + 20 kg)(9.8 m /s 2 ) = 588 N Just when the boy is at the point where the beam tips, F1 = 0 N. Thus F2 = 588 N. With this value of F2, we can simplify the torque equation to: –(40 kg)(9.80 m/s2)(2.5 m) + (588 N)(3.0 m) – (20 kg)(9.80 m/s2)x = 0 N m

⇒ x = 4.0 m Thus, the distance from the right end is 5.0 m – 4.0 m = 1.0 m.


13.60. r Model: The ladder is a rigid rod of length L. To not slip, it must be in both translational equilibrium r ( Fnet = 0 N ) and rotational equilibrium (τ net = 0 N m ). We also apply the model of static friction.

Visualize:

r Since the wall is frictionless, the only force from the wall on the ladder is the normal force n2 . On the other hand, r r r the floor exerts both the normal force n1 and the static frictional force fs . The weight w of the ladder acts through the center of mass of the ladder. r r Solve: The x- and y-components of Fnet = 0 N are

∑F

x

= n 2 − fs = 0 N ⇒ fs = n 2

∑F

y

= n1 − w = 0 N ⇒ n1 = w

The minimum angle occurs when the static friction is at its maximum value fs max = µsn1 . Thus we have n2 = fs = µsn1 = µsw. We choose the bottom corner of the ladder as a pivot point to obtain τnet, because two forces pass through this point and have no torque about it. The net torque about the bottom corner is

τnet = d1w – d2n2 = (0.5L cos θmin)mg – (L sin θmin)µsmg = 0 N m ⇒ 0.5 cosθ min = µ s sin θ min ⇒ tan θ min =

0.5 0.5 = = 1.25 ⇒ θ min = 51.3° 0.4 µs


13.61.

Model: Visualize:

Solve:

(a)

The two masses connected by a massless rod is a rigid body rotating about its center of mass.

x cm =

m1 x1 + m2 x 2 (1.0 kg)(20 cm ) + (3.0 kg)(20 cm ) = = 20.0 cm m1 + m2 1.0 kg + 3.0 kg

ycm =

(1.0 kg)(20 cm ) + (3.0 kg)(100 cm ) = 80.0 cm 1.0 kg + 3.0 kg

(b) Relative to the center of mass, the mass m1 is at a distance of r1 = 80 cm – 20 cm = 60 cm and m2 is at a distance of r2 = 100 cm – 80 cm = 20 cm. The moment of inertia is

Icm = ∑ mr 2 = m1r12 + m2 r22 = (1.0 kg)(0.60 m ) 2 + (3.0 kg)(0.20 m ) 2 = 0.48 kg m 2 (c) Using the rotational kinematic equation ω1 = ω0 + α(t1 – t0), we get

α=

ω 1 − ω 0 6.25 rad /s − 0 rad /s = = 2.083 rad /s 2 t1 − t 0 3.0 s − 0 s

Thus, the torque τ = Iα = (0.48 kg m2)(2.083 rad/s2) = 1.0 N m. (d) To get a torque of 1.0 Nm with forces of 1.5 N on each mass, (1.5 N)(0.2 m) sin β + (1.5 N)(0.6 m) sin β = 1.0 N m 1.0 ⇒ (1.2 N m) sinφ = 1.0 N m ⇒ φ = sin −1   = 56.4°  1.2 


13.62.

Model: Visualize:

Solve:

The compact disk is a rigid body rotating about its center.

(a) The rotational kinematic equation ω 1 = ω 0 + α (t1 − t 0 ) gives

2π 200π (2000 rpm )  rad /s = 0 rad + α (3.0 s − 0 s) ⇒ α = rad /s 2  60  9 The torque needed to obtain this operating angular velocity is

200π τ = Iα = (2.5 × 10 −5 kg m 2 ) rad /s 2  = 1.75 × 10 −3 N m  9  (b) From the rotational kinematic equation, 1 1 200π 2 rad /s 2  (3.0 s − 0 s) θ1 = θ 0 + ω 0 (t1 − t 0 ) + α (t1 − t 0 ) 2 = 0 rad + 0 rad +   2 2 9 100π = 100π rad = revolutions = 50 rev 2π Assess: Fifty revolutions in 3 seconds is a reasonable value.


13.63.

Model: Visualize:

The stars are spherical masses. They each rotate about the system’s center of mass.

Solve: (a) The stars rotate about the system’s center of mass with the same period: T1 = T2 = T. We can locate the center of mass by lettering the origin be at the smaller-mass star. Then

r1 = rcm =

(2.0 × 10 30 kg)(0 m) + (6.0 × 10 30 kg)(2.0 × 1012 m) = 1.5 × 1012 m 2.0 × 10 30 kg + 6.0 × 10 30 kg

Mass m2 undergoes uniform circular motion with radius r2 = 0.5 × 1012 m due to the gravitational force of mass m1 at distance R = 2.0 × 1012 m. The gravitational force is responsible for the centripetal acceleration, so

Gm1m2 m2 v22 m2  2πr2  4π 2 m2 r2 m a = = = = 2 centripetal R2 r2 r2  T  T2 2

Fgrav = 1/ 2

 4π 2  r2 R2  ⇒ T=  Gm1 

1/ 2

 4π 2 (0.5 × 1012 m)(0.5 × 1012 m)2  =  30 2 2 −11  (6.67 × 10 N m / kg ) (2.0 × 10 kg) 

= 7.693 × 108 s = 24.4 years

(b) The speed of each star is v = (2πr)/T. Thus

2πr1 2π (1.5 × 1012 m) = 12, 250 m /s = T 7.693 × 10 8 s 2πr2 2π (0.5 × 1012 m) = = 4080 m /s v2 = 7.693 × 10 8 s T v1 =


13.64.

Model: Visualize:

The bar is a solid body rotating through its center.

Solve:

(a) The two forces form a couple. The net torque on the bar about its center is Iα τ net = LF = Iα ⇒ F = L where F is the force produced by one of the air jets. We can find I and α as follows: 1 1 I= ML2 = (0.50 kg)(0.60 m ) 2 = 0.015 kg m 2 12 12 ω 1 = ω 0 + α (t1 − t 0 ) ⇒ 150 rpm = 5.0π rad /s = 0 rad + α (10 s − 0 s) ⇒ α = 0.50π rad /s 2

(0.015 kg m 2 )(0.5π rad /s 2 ) = 0.0393 N (0.60 m ) (b) The torque of a couple is the same about any point. It is still τ net = LF. However, the moment of inertia has changed. LF 1 1 τ net = LF = Iα ⇒ α = where I = ML2 = (0.500 kg)(0.6 m ) 2 = 0.060 kg m 2 I 3 3 (0.0393 N ) × (0.60 m ) ⇒α = = 0.393 rad /s 2 0.060 kg m 2 ⇒F=

Finally,

ω 1 = ω 0 + α (t1 − t 0 ) = 0 rad /s + (0.393 rad /s 2 )(10 s − 0 s) (3.93)(60) rpm = 37.5 rpm 2π Assess: Note that ω ∝ α and α ∝ 1/ I . Thus, ω ∝ 1/ I . I about the center of the rod is 4 times smaller than I about one end of the rod. Consequently, ω is 4 times larger. = 3.93 rad /s =


13.65.

Model: Visualize:

The flywheel is a rigid body rotating about its central axis.

Solve: (a) The radius of the flywheel is R = 0.75 m and its mass is M = 250 kg. The moment of inertia about the axis of rotation is that of a disk: 1 1 I = MR 2 = (250 kg)(0.75 m ) 2 = 70.31 kg m 2 2 2 The angular acceleration is calculated as follows: τ net = Iα ⇒ α = τ net / I = (50 N m )/(70.31 kg m 2 ) = 0.711 rad /s 2 Using the kinematic equation for angular velocity gives ω 1 = ω 0 + α (t1 − t 0 ) = 1200 rpm = 40 π rad /s = 0 rad /s + 0.711 rad /s 2 (t1 − 0 s)

⇒ t1 = 177 s (b) The energy stored in the flywheel is rotational kinetic energy: 1 1 K rot = Iω 12 = (70.31 kg m 2 )( 40π rad /s) 2 = 5.55 × 10 5 J 2 2 energy delivered (5.55 × 10 5 J )/ 2 (c) Average power delivered = = = 1.39 × 10 5 W = 139 kW time interval 2s (d) Because τ = Iα, ⇒ τ = I

∆ω  ω full energy − ω half energy  = I  . ω full energy = ω1 (from part (a)) = 40π rad /s. ω half energy   ∆t ∆t

can be obtained as:

1 2 1 Iω half energy = K rot ⇒ ω half energy = 2 2

K rot = I

5.55 × 10 5 J = 88.85 rad /s 70.31 kg m 2

Thus

40 π rad /s − 88.85 rad /s  τ = (70.31 kg m 2 ) = 1300 N m   2s


13.66.

Model: Visualize:

The hollow cylinder is a rigid rotating body.

We placed the origin of the coordinate system on the ground. Solve: (a) Newton’s second law for the block is −wB + T = mBay, where T is the tension in the string, wB = mBg is the weight of the block, and ay is the acceleration of block. The string tension exerts a negative (cw) torque on the cylinder, so the rotational form of Newton’s second law for the hollow cylinder is a Ia τ = − TR = Iα = I y ⇒ T = − 2y R R where we used the acceleration constraint a y = αR. With this expression for T, Newton’s second law for the block becomes − mB g Ia − mB g − 2 = mB a y ⇒ a y = ( mB + I / R 2 ) R The moment of inertia of a hollow cylinder is I = mC R 2 , so the equation for ay simplifies to

ay =

− mB g −(3.0 kg)(9.8 m /s 2 ) = = −5.88 m /s 2 3.0 kg + 2.0 kg m B + mC

The speed of the block just before it hits the ground can now be found using kinematics: v12 = v02 + 2 a y ( y1 − y0 ) = 0 m 2 / s 2 + 2( −5.88 m / s 2 )(0 m − 1.0 m ) ⇒ v1 = 3.43 m / s (b) The conservation of energy equation K1 + Ug1 = K0 + Ug0 for the system (block + cylinder + earth) is 1 1 1 1 mB v12 + Iω 12 + mB gy1 = mB v02 + Iω 02 + mB gy0 2 2 2 2 2 m  v 1 1 m mB v12 + mC R 2 12 + 0 J = 0 J + 0 J + mB gy0 ⇒ v12 B + C = mB gy0  2 2 2 2  R

(

⇒ v12 =

)

2 mB gy0 9.8 m /s 2 )(1.0 m ) 2(3.0 kg)(9.8 = = 11.76 m 2 / s 2 ⇒ v1 = 3.43 m / s (3.0 kg) + (2.0 kg) m B + mC

Assess: Newton’s second law and the conservation of energy method give the same result for the block’s final velocity.


13.67.

Model: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution of a disk and that the string does not slip. Visualize:

Because the pulley is not massless and frictionless, tension in the rope on both sides of the pulley is not the same. Solve: Applying Newton’s second law to m1, m2, and the pulley yields the three equations:

T1 − w1 = m1a1

− w2 + T2 = m2 a2

T2 R − T1 R − 0.50 N m = Iα

Noting that –a2 = a1 = a, I = mp R , and α = a/R, the above equations simplify to 1 2

T1 − m1 g = m1a

2

m2 g − T2 = m2 a

a 1 0.50 N m 1 1 0.50 N m T2 − T1 =  mp R 2    + = mp a + 2  R R R 2 0.060 m

Adding these three equations,

1 ( m2 − m1 )g = a m1 + m2 + mp  + 8.333 N  2  ⇒a=

( m2 − m1 )g − 8.333 N ( 4.0 kg − 2.0 kg)(9.8 m /s 2 ) − 8.333 N = = 1.610 m /s 2 2.0 kg + 4.0 kg + (2.0 kg/2) m1 + m2 + 12 mp

We can now use kinematics to find the time taken by the 4.0 kg block to reach the floor:

y1 = y0 + v0 (t1 − t 0 ) +

1 1 a2 (t1 − t 0 ) 2 ⇒ 0 = 1.0 m + 0 + ( −1.610 m /s 2 )(t1 − 0 s) 2 2 2 ⇒ t1 =

2(1.0 m ) = 1.11 s (1.610 m /s 2 )


13.68.

Model: Visualize:

Assume the string does not slip on the pulley.

The free-body diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward force on the block m2, but a downward force on the outer edge of the pulley. Similarly the string exerts a force on block m1 to the right, but a leftward force on the outer edge of the pulley. Solve: (a) Newton’s second law for m1 and m2 is T = m1a1 and T – m2g = m2a2. Using the constraint –a2 = +a1 = a, we have T = m1a and −T + m2g = m2a. Adding these equations, we get m2g = (m1 + m2)a, or m2 g mm g a= ⇒ T = m1a = 1 2 m1 + m2 m1 + m2 (b) When the pulley has mass m, the tensions (T1 and T2) in the upper and lower portions of the string are different. Newton’s second law for m1 and the pulley are: T1 = m1a and T1R – T2R = − Iα We are using the minus sign with α because the pulley accelerates clockwise. Also, a = Rα. Thus, T1 = m1a and I a aI T2 − T1 = = R R R2 Adding these two equations gives I T2 = a m1 + 2   R  Newton’s second law for m2 is T2 – m2g = m2a2 = − m2a. Using the above expression for T2,

m2 g I a m1 + 2  + m2 a = m2 g ⇒ a =  R  m1 + m2 + I / R 2 Since I = 12 mp R 2 for a disk about its center,

a=

m2 g m1 + m2 + 12 mp

With this value for a we can now find T1 and T2:

T1 = m1a =

m1 m2 g m1 + m2 + 12 mp

T2 = a( m1 + I / R 2 ) =

(

)

Assess: For m = 0 kg, the equations for a, T1 and T2 of part (b) simplify to m2 g mm g mm g a= and T1 = 1 2 and T2 = 1 2 m1 + m2 m1 + m2 m1 + m2 These agree with the results of part (a).

(

)

1  m + 1 m  = m2 m1 + 2 mp g p 1 2  m1 + m2 + 12 mp m1 + m2 + 12 mp 

m2 g


13.69.

Model: The disk is a rigid spinning body. Visualize: Please refer to Figure P13.69. The initial angular velocity is 300 rpm or (300)(2π)/60 = 10π rad/s. After 3.0 s the disk stops. Solve: Using the kinematic equation for angular velocity, ω − ω 0 (0 rad /s − 10π rad /s) −10π ω 1 = ω 0 + α (t1 − t 0 ) ⇒ α = 1 = = rad /s 2 3 t1 − t 0 (3.0 s − 0 s) Thus, the torque due to the force of friction that brings the disk to rest is

( 12 mR2 )α = − 1 (mR)α = − 1 (2.0 kg)(0.15 m) −10 π rad /s2  = 1.57 N Iα =−   R R 2 2 3 The minus sign with τ = − fR indicates that the torque due to friction acts clockwise. τ = Iα = − fR ⇒ f = −


13.70.

Model: The entire structure is a rigid rotating body. The two thrust forces are a couple that exerts a torque on the structure about its center of mass. We will assume the thrust forces are perpendicular to the connecting tunnel. Visualize: Please refer to Figure 13.26. We chose a coordinate system in which m1 and m2 are on the x-axis and m1 = 100,000 kg is at the origin. m3 is the mass of the tunnel, whose center is at x3 = 45 m. Solve: (a) Assuming the center of mass of the tunnel is at the center of the tunnel, we get

x cm =

m1 x1 + m2 x 2 + m3 x3 m1 + m2 + m3

x cm =

(1.0 × 10 5 kg)(0 m ) + (2.0 × 10 5 kg)(90 m ) + (5.0 × 10 4 kg)( 45 m ) = 57.9 m 1.0 × 10 5 kg + 2.0 × 10 5 kg + 5.0 × 10 4 kg

(b) Initially, the angular velocity is zero. The structure’s angular velocity after 30 s is ω 1 = ω 0 + α (t1 − t 0 ) = 0 rad /s + α (30 s − 0 s) = α (30 s) The angular acceleration α can be found from τ = Iα , where τ is the net torque on the structure and I is its moment of inertia. The two thrusts form a couple with torque τ = lF = (90 m)(50,000 N) = 4.50 × 106 N m 1 I = I m1 + I m 2 + I tunnel = m1 x12 + m2 x 22 + m3 (90 m ) 2 12 1 5 2 = (1.0 × 10 kg)(57.9 m ) + (2.0 × 10 5 kg)(32.1 m ) 2 + (5 × 10 4 kg)(90 m ) 2 12 = 5.75 × 10 8 kg m 2 τ 4.5 × 10 6 N m ⇒α = = = 7.83 × 10 −3 rad /s 2 I 5.75 × 10 8 kg m 2

⇒ ω 1 = (30 s)(7.83 × 10 −3 rad /s 2 ) = 0.235 rad /s


13.71.

Model: Assume that the hollow sphere is a rigid rolling body and that the sphere rolls up the incline without slipping. We also assume that the coefficient of rolling friction is zero. Visualize:

The initial kinetic energy, which is a combination of rotational and translational energy, is transformed in gravitational potential energy. We chose the bottom of the incline as the zero of the gravitational potential energy. Solve: The conservation of energy equation Kf + Ugf = Ki + Ugi is 1 1 1 1 M (v1 ) 2cm + Icm (ω 1 ) 2 + Mgy1 = M (v0 ) 2cm + Icm (ω 0 ) 2 + Mgy0 2 2 2 2 (v ) 2 1 1 2 1 1 0 J + 0 J + Mgy1 = M (v0 ) 2cm +  MR 2  (ω 0 ) 2cm + 0 J ⇒ Mgy1 = M (v0 ) 2cm + MR 2 0 2cm  2 23 2 3 R

⇒ gy1 =

5 (v ) 2 5 5 (5.0 m /s) 2 = 2.126 m (v0 ) 2cm ⇒ y1 = 6 0 cm = 6 6 9.8 m /s 2 g

The distance traveled along the incline is

y1 2.126 m = = 4.25 m sin 30° 0.5 Assess: This is a reasonable stopping distance for an object rolling up an incline when its speed at the bottom of the incline is approximately 10 mph. s=


13.72. Model: The disk is a rigid body rotating on an axle passing through one edge. The gravitational potential energy is transformed into rotational kinetic energy as the disk is released. Visualize:

We placed the origin of the coordinate system at a distance R just below the axle. In the initial position, the center of mass of the disk is at the same level as the axle. The center of mass of the disk in the final position is coincident with the origin of the coordinate system. Solve: (a) The torque is due to the disk’s weight acting at the center of mass. Thus

τ = ( mg) R = (5.0 kg)(9.8 m /s 2 )(0.30 m ) = 14.7 N m The moment of inertia about the disk’s edge is obtained using the parallel-axis theorem:

I = Icm + mR 2 =

1 3 3 mR 2 + mR 2 = mR 2 =   (5.0 kg)(0.3 m ) 2 = 0.675 kg m 2  2 2 2 τ 14.70 N m ⇒α = = = 21.8 rad /s 2 I 0.675 kg m 2

(b) The energy conservation equation Kf + Ugf = Ki + Ugi is 1 2 1 1 Iω 1 + mgy1 = Iω 02 + mgy0 ⇒ Iω 12 + 0 J = 0 J + mgR 2 2 2

ω1 =

2 mgR = I

2(5.0 kg)(9.8 m /s 2 )(0.30 m ) = 6.60 rad /s 0.675 kg m 2

Assess: An angular velocity of 6.60 rad/s (or 1.05 revolutions/s) as the center of mass of the disk reaches below the axle is reasonable.


13.73.

Model: The hoop is a rigid body rotating about an axle at the edge of the hoop. The gravitational torque on the hoop causes it to rotate, transforming the gravitational potential energy of the hoop’s center of mass into rotational kinetic energy. Visualize:

We placed the origin of the coordinate system at the hoop’s edge on the axle. In the initial position, the center of mass is a distance R above the origin, but it is a distance R below the origin in the final position. Solve: (a) Applying the parallel-axis theorem, Iedge = Icm + mR2 = mR2 + mR2 = 2mR2. Using this expression in the energy conservation equation Kf + Ugf = Ki + Ugi yields:

1 1 I edgeω 12 + mgy1 = I edgeω 02 + mgy0 2 2 (b) The speed of the lowest point on the hoop is

1 (2 mR 2 )ω 12 − mgR = 0 J + mgR ⇒ ω 1 = 2

2g R

2g (2 R) = 8gR R Assess: Note that the speed of the lowest point on the loop involves a distance of 2R instead of R. v = (ω 1 )(2 R) =


13.74.

Model: The long, thin rod is a rigid body rotating about a frictionless pivot on the end of the rod. The gravitational torque on the rod causes it to rotate, transforming the gravitational potential energy of the rod’s center of mass into rotational kinetic energy. Visualize:

We placed the origin of the coordinate system at the pivot point. In the initial position, the center of mass is a distance 1 2 L above the origin. In the final position, the center of mass is at y = 0 m and thus has zero gravitational potential energy. Solve: (a) The energy conservation equation for the rod Kf + Ugf = Ki + Ugi is

1 2 1 Iω 1 + mgy1 = Iω 02 + mgy0 2 2

1  1 2 2 mL ω 1 + 0 J = 0 J + mg( L / 2) ⇒ ω 1 = 3g / L  23

(b) The speed at the tip of the rod is vtip = (ω 1 ) L = 3gL .


13.75.

Model: The sphere attached to a thin rod is a rigid body rotating about the rod. Assume the rod is vertical and the sphere solid. Visualize: Please refer to Figure P13.75. The sphere rotates because the string wrapped around the rod exerts a torque τ. Solve: From the parallel-axis theorem and Newton’s second law, 2

R 2 MR 2 13 Ioff center = Icm + M   = MR 2 + = MR 2  2 5 4 20

α=

τ τ 20τ = = 2 I (13 MR / 20) 13 MR 2


13.76.

Model: Visualize:

The pulley is a rigid rotating body.

We placed the origin of the coordinate system on the floor. The pulley rotates about its center. Solve: Using kinematics for the physics book (mass = m1), 1 1 y1 = y0 + v0 (t1 − t 0 ) + a(t1 − t 0 ) 2 0 m = 1.0 m + 0 m + a(0.71 s − 0 s) 2 ⇒ a = −3.967 m /s 2 2 2 Since the torque acting on the pulley is τ = –TR = Iα , we have TR TR TR 2 I=− =− =− a /R a α We can compare the measured value of I for the pulley with the theoretical value. We first must find the tension T. From the free-body diagram, Newton’s second law of motion is T – m1g = m1a. This means T = m1(g + a) = (1.0 kg)(9.80 m/s2 – 3.967 m/s2) = 5.833 N With these values of T and a, we can now find I as: TR 2 (5.833 N )(0.06 m ) 2 I=− =− = 5.3 × 10 −3 kg m 2 a ( −3.967 m /s 2 ) Let us now calculate the theoretical values of I: hoop about center: I = MR 2 = (2.0 kg)(0.06 m ) 2 = 7.2 × 10 −3 kg m 2

1 1 MR 2 = (2.0 kg)(0.06 m ) 2 = 3.6 × 10 −3 kg m 2 2 2 Since Idisk < I < Ihoop, the mass of the disk is not uniformly distributed. The mass is concentrated near the rim. disk about center: Idisk =


13.77.

Model: The mechanical energy of both the hoop (h) and the sphere (s) is conserved. The initial gravitational potential energy is transformed into kinetic energy as the objects roll down the slope. The kinetic energy is a combination of translational and rotational kinetic energy. We also assume no slipping of the hoop or of the sphere. Visualize:

The zero of gravitational potential energy is chosen at the bottom of the slope. Solve: (a) The energy conservation equation for the sphere or hoop Kf + Ugf = Ki + Ugi is 1 1 1 1 I (ω 1 ) 2 + m(v1 ) 2 + mgy1 = I (ω 0 ) 2 + m(v0 ) 2 + mgy0 2 2 2 2 For the sphere, this becomes

(v ) 2 1 12 mR 2  1 2 s + m(v1 )s2 + 0 J = 0 J + 0 J + mghs  R 25 2 7 (v1 )s2 = gh ⇒ (v1 )s = 10 gh / 7 = 10(9.8 m /s 2 )(0.30 m )/ 7 = 2.05 m /s 10 For the hoop, this becomes (v ) 2 1 1 ( mR 2 ) 1 2 h + m(v1 ) 2h + 0 J = 0 J + 0 J + mghhoop 2 2 R (v ) 2 ⇒ hhoop = 1 h g For the hoop to have the same velocity as that of the sphere, (v ) 2 (2.05 m/s) 2 hhoop = 1 s = = 42.9 cm 9.8 m/s 2 g ⇒

(b) As we see in part (a), the speed of a hoop at the bottom depends only on the starting height and not on the mass or radius. So the answer is No.


13.78.

Model: Model the turntable as a rigid disk rotating on frictionless bearings. As the blocks fall from above and stick on the turntable, the turntable slows down due to increased rotational inertia of the (turntable + blocks) system. Any torques between the turntable and the blocks are internal to the system, so angular momentum of the system is conserved. Visualize: The initial moment of inertia is I1 and the final moment of inertia is I2. Solve: The initial moment of inertia is I1 = Idisk = 12 mR 2 = 12 (2.0 kg)(0.10 m ) 2 = 0.010 kg m 2 and the final moment of inertia is I2 = I1 + 2mR2 = 0.010 kg m2 + 2(0.500 kg) × (0.10 m)2 = 0.010 kg m2 + 0.010 kg m2 = 0.020 kg m2 Let ω1 and ω2 be the initial and final angular velocities. Then Iω (0.010 kg m 2 )(100 rpm ) Lf = Li ⇒ ω 2 I2 = ω 1 I1 ⇒ ω 2 = 1 1 = = 50 rpm 0.020 kg m 2 I2


Model: Model the turntable as a rigid disk rotating on frictionless bearings. For the (turntable + block) system, no external torques act as the block moves outward towards the outer edge. Angular momentum is thus conserved. Visualize: The initial moment of inertia of the turntable is I1 and the final moment of inertia is I2. Solve: The initial moment of inertia is I1 = Idisk = 12 mR 2 = 12 (0.2 kg)(0.2 m ) 2 = 0.0040 kg m 2 . As the block reaches the outer edge, the final moment of inertia is I2 = I1 + mBR2 = 0.0040 kg m2 + (0.020 kg)(0.20 m)2 = 0.0040 kg m2 + 0.0008 kg m2 = 0.0048 kg m2 Let ω1 and ω2 be the initial and final angular velocities, then the conservation of angular momentum equation is Iω (0.0040 kg m 2 )(60 rpm ) Lf = Li ⇒ ω 2 I2 = ω 1 I1 ⇒ ω 2 = 1 1 = = 50 rpm (0.0048 kg m 2 ) I2

13.79.

Assess: A change of angular velocity from 60 rpm to 50 rpm with an increase in the value of the moment of inertia is reasonable.


13.80.

Model: Model the merry-go-round as a rigid disk rotating on frictionless bearings about an axle in the center and John as a particle. For the (merry-go-round + John) system, no external torques act as John jumps on the merry-go-round. Angular momentum is thus conserved. Visualize: The initial angular momentum is the sum of the angular momentum of the merry-go-round and the angular momentum of John. The final angular momentum as John jumps on the merry-go-round is equal to Ifinalωfinal. Solve: John’s initial angular momentum is that of a particle: LJ = mJ vJ R sin β = mJ vJ R. The angle β = 0 since John runs tangent to the disk. The conservation of angular momentum equation Lf = Li is

1 I finalω final = Ldisk + LJ =  MR 2  ω i + mJ vJ R 2  1 2π  rad  2 =   (250 kg)(1.5 m ) 2 (20 rpm )   + (30 kg)(5.0 m /s)(1.5 m ) = 814 kg m / s  2 60  rpm 

⇒ ω final = I final = Idisk + I J =

814 kg m 2 / s I final

1 1 MR 2 + mJ R 2 = (250 kg)(1.5 m ) 2 + (30 kg)(1.5 m ) 2 = 349 kg m 2 2 2 814 kg m 2 / s ω final = = 2.33 rad /s = 22.3 rpm 349 kg m 2


13.81.

Model: Visualize:

The angular momentum of the satellite in the elliptical orbit is a constant.

Solve: (a) Because the r gravitational force is always along the same direction as the direction of the moment arm vector, the torque τ = r × Fg is zero at all points on the orbit. (b) The angular momentum of the satellite at any point on the elliptical trajectory is conserved. The velocity is r perpendicular to r at points a and b, so β = 90° and L = mvr. Thus

r  Lb = La ⇒ mvb rb = mva ra ⇒ vb =  a  va  rb  ra =

30, 000 km − 9, 000 km = 6.00 × 10 6 m 2

and rb =

30, 000 km + 9, 000 km = 2.4 × 10 7 m 2

 6.00 × 10 6 m  ⇒ vb =   (8,000 m /s) = 2000 m /s  2.4 × 10 7 m  (c) Using the conservation of angular momentum Lc = La , we get r  mvc rc sin β c = mva ra ⇒ vc =  a  va / sin β c rc = (9000 km ) 2 + (12,000 km ) 2 = 1.5 × 10 7 m  rc  From the figure, we see that sin β c = 12,000 15,000 = 0.80. Thus

 6.0 × 10 6 m  (8000 m /s) = 4000 m /s vc =   0.80  1.5 × 10 7 m 


13.82.

Model: Model the skater as a cylindrical torso with two rod-like arms that are perpendicular to the axis of the torso in the initial position and collapse into the torso in the final position. Visualize:

Solve: For the initial position, the moment of inertia is I1 = I Torso + 2 I Arm . The moment of inertia of each arm is that of a rod rotating about its end. In the final position, the moment of inertia is I2 = 12 MR 2 . The equation for the conservation of angular momentum Lf = Li can be written I2ω 2 = I1ω 1 ⇒ ω 2 = ( I1 / I2 )ω 1 . Calculating I1 and I2 ,

1 1 MT R 2 + 2  MA L2A  2 3  1 1 = ( 40 kg)(0.10 m ) 2 + 2  (2.5 kg)(0.78 m ) 2  = 1.214 kg m 2 2  3

I1 =

I2 =

(1.214 kg m 2 ) 1 1 (1.5 rev/s) = 8.1 rev/s MR 2 = ( 45 kg)(0.10 m ) 2 = 0.225 kg m 2 ⇒ ω 2 = (0.225 kg m 2 ) 2 2


Model: For the (billiard ball + rod) system the angular momentum is conserved. The ball rolls without slipping and the rod rotates about a frictionless pivot. Visualize:

13.83.

Solve: (a) The initial angular momentum about the pivot due to the ball moving at r = 12 d is Li = mv0 ( 12 d ) and the final angular momentum about the pivot is 1 d Lf = I rodω rod + m(v0 / 2)( d / 2) =   (2 m)( d 2 )ω rod + mv0  12  4 Using Lf = Li, we have

d d 1 (2 m)( d 2 )ω rod + mv0 = mv0 ⇒ ω rod = 12 4 2

( 14 mv0 d ) = 3v0

(

1 6

md 2 )

2d

(b) The initial kinetic energy K i = mv and the final kinetic energy is 1 2

2 0

2

1 1 1 1 1 1 3 5  3v0  2 m(v0 / 2) 2 + I rodω rod = mv02 +  (2 m)d 2  = mv02 + mv02 = mv02     2 2 8 2 12 2d 8 16 16 Because Kf ≠ Ki, the mechanical energy of the system is not conserved. Kf =


13.84.

Model: Assume that the marble does not slip as it rolls down the track and around a loop-the-loop. The mechanical energy of the marble is conserved. Visualize:

Solve: The ball’s center of mass moves in a circle of radius R – r. The free-body diagram on the marble at its highest position shows that Newton’s second law for the marble is mv12 mg + n = R−r The minimum height (h) that the track must have for the marble to make it around the loop-the-loop occurs when the normal force of the track on the marble tends to zero. Then the weight will provide the centripetal acceleration needed for the circular motion. For n → 0 N, mv 2 mg = ⇒ v12 = g( R − r ) ( R − r) Since rolling motion requires v12 = r 2ω 12 , we have g( R − r ) ω 12 r 2 = g( R − r ) ⇒ ω 12 = r2 The conservation of energy equation is 1 1 ( K f + U gf ) top of loop = ( K i + U gi ) initial ⇒ mv12 + Iω 12 + mgy1 = mgy0 = mgh 2 2 Using the above expressions and I = 25 mr 2 the energy equation simplifies to

g( R − r )  1 1 2 mg( R − r ) +   mr 2  + mg2( R − r ) = mgh ⇒ h = 2.7( R − r )    r2  2 2 5


Model: For the (bullet + block + rod) system, angular momentum is conserved. After the bullet is stuck in the block, the mechanical energy of the system is conserved. Assume the block is small. Visualize:

13.85.

The origin of the coordinate system was placed at the center-of-mass of the block as it freely hangs from the bottom of the rod. Solve: The initial angular momentum of the system about the pivot is due only to the bullet: Li = mb vb L = (0.010 kg)vb (1.0 m ) = (0.010 kg m )vb The angular momentum immediately after the bullet hits and sticks in the block is equal to Iω. The moment of inertia is

I = I rod + I block + I bullet =

1 1 mR L2 + mB L2 + mb L2 =  mR + mB + mb  L2   3 3

1 =  (1.0 kg) + 2.0 kg + 0.010 kg(1.0 m ) 2 = 2.343 kg m 2 3  Angular momentum is conserved in the collision: Lf = (2.343 kg m 2 )ω = Li = (0.010 kg m )vb We need to determine ω before we can find vb. To find ω we use the conservation of mechanical energy equation Kf + Ugf = Ki + Ugi as the pendulum swings out, which is 1 0 J + ( mb + mB )gyB + mR gyR = Iω 2 + 0 J 2 L 1 2 (0.010 kg + 2.0 kg)(9.8 m /s ) L(1 − cos 30°) + (1.0 kg)(9.8 m /s 2 ) (1 − cos 30°) = (2.343 kg m 2 )ω 2 2 2 The energy equation can be further simplified to

2.6390 kg m 2 /s 2 + 0.6565 kg m 2 /s 2 = (1.1715 kg m 2 )ω 2 ⇒ ω = 1.677 rad /s Finally, we can use the conservation of angular momentum equation to obtain the speed of the bullet:

vb =

(2.343 kg m 2 )(1.677 rad /s) = 393 m /s 0.010 kg m

Assess: A speed of 393.0 m/s for a bullet is reasonable.


13.86.

Model: Visualize:

For the (bullet + door) system, the angular momentum is conserved in the collision.

r r Solve: As the bullet hits the door, its velocity v is perpendicular to r. Thus the initial angular momentum about the rotation axis, with r = L, is Li = mBvBL = (0.010 kg)(400 m/s)(1.0 m) = 4.0 kg m2/s After the collision, with the bullet in the door, the moment of inertia about the hinges is I = Idoor + Ibullet =

1 1 mD L2 + mB L2 = (10.0 kg)(1.0 m ) 2 + (0.010 kg)(1.0 m ) 2 = 3.343 kg m2 3 3

Therefore, Lf = Iω = (3.343 kg m2)ω. Using the angular momentum conservation equation Lf = Li, (3.343 kg m2)ω = 4.0 kg m2/s and thus ω = 1.20 rad/s.


13.87.

Model: Because no external torque acts on the star during gravitational collapse, its angular momentum is conserved. Model the star as a solid rotating sphere. Solve: (a) The equation for the conservation of angular momentum is

2 2 Li = Lf ⇒ I iω i = I f ω f ⇒  mRi2  ω i =  mRf2  ω f 5  5  ⇒ Rf = Ri

ωi ωf

The angular velocity is inversely proportional to the period T. We can write

Rf = Ri

Tf = (3.5 × 10 8 m ) = Ti

0.10 s = 6.87 × 10 4 m = 68.7 km 2.592 × 10 6 s

(b) A point on the equation rotates with r = Rf. Its speed is

v=

2πRf 2π (68, 700 m ) = = 4.32 × 10 6 m /s 0.10 s T


Model: For the (turntable + bicycle wheel + professor) system the angular momentum is conserved because the turntable is frictionless and no external torques act on the system. Solve: (a) Nothing happens. The bicycle wheel already has an angular momentum and nothing changes for the wheel when it is handed to the professor. So, nothing happens to the professor. (b) The initial angular momentum is

13.88.

( Lwheel ) i = Iω = ( mR 2 )ω = ( 4.0 kg)(0.32 m ) 2  

180 × 2π rad  = 7.72 kg m 2 / s s  60

When the wheel is turned upside down, the angular momentum of the wheel becomes

( Lwheel ) f = −7.72 kg m 2 / s Since the initial and final total angular momentum should be equal, the professor must acquire some angular momentum. From the angular momentum conservation equation, Lprof + (Lwheel)f = (Lwheel)i ⇒ Lprof = (Lwheel)i – (Lwheel)f = 7.72 kg m2/s – (−7.72 kg m2/s) = 15.44 kg m2/s Let us now calculate the angular velocity of the turntable by using Lprof = Itotalω, where Itotal = Iturntable + Ibody + Iarms + Iwheel in hand. We assume that the axis of the professor and turntable also become the axis of the arms and wheel (i.e., no leaning).

1 1 mT RT2 = (5.0 kg)(0.25 m ) 2 = 0.156 kg m 2 2 2 1 = mbody R 2 = (70 kg)(0.125 m ) 2 = 0.547 kg m 2 2 1 2  = 2 marms R 2  = (2.5 kg)(0.45 m ) 2 = 0.338 kg m 2 3  3

I turntable = I body Iarms

I wheel in hand = mwheel r 2 = ( 4.0 kg)(0.45 m ) 2 = 0.810 kg m 2 Thus, Itotal = (0.156 + 0.547 + 0.338 + 0.810) kg m2 = 1.851 kg m2. Using this value for Itotal, we can now find ω to be

ω=

Lprof I total

=

15.44 kg m 2 / s = 8.34 rad /s = 79.7 rpm 1.851 kg m 2

Chapter 13  

Solve: The crankshaft at t = 0 s has an angular velocity of 250 rad/s. It gradually slows down to 50 rad/s in 2 s, maintains a constant angu...

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