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7.1. Solve: (a) From t = 0 s to t = 1 s the particle rotates clockwise from the angular position +4π rad to −2π rad. Therefore, ∆θ = −2π − ( +4π ) = −6π rad in one sec, or ω = −6π rad s . From t = 1 s to t = 2 s, ω = 0 rad/s. From t = 2 s to t = 4 s the particle rotates counterclockwise from the angular position −2π rad to 0 rad. Thus ∆θ = 0 − ( −2π ) = 2π rad and ω = +π rad s . (b)


7.2. Solve: Since ω = (dθ dt ) we have θf = θi + area under the ω-versus-t graph between ti and tf From t = 0 s to t = 2 s, the area is 1 2 (20 rad / s)(2 s) = 20 rad . From t = 2 s to t = 4 s, the area is (20 rad / s)(2 s) = 40 rad . Thus, the area under the ω-versus-t graph during the total time interval of 4 s is 60 rad or (60 rad) × (1 rev/2π rad) = 9.55 revolutions.


7.3. Model: Treat the record on a turntable as a particle rotating at 45 rpm. Solve: (a) The angular velocity is

ω = 45 rpm ×

1 min 2π rad × = 1.5π rad / s 60 s 1 rev

(b) The period is

T=

2π rad 2π rad = = 1.33 s 1.5π rad / s ω


7.4. Model: The airplane is to be treated as a particle. Visualize:

Solve:

(a) The angle you turn through is

θ 2 − θ1 =

s 5000 miles 180° = = 1.25 rad = 1.25 rad × = 71.6° π rad r 4000 miles

(b) The plane’s angular velocity is

ω=

θ 2 − θ1 1.25 rad rad 1 hr = = 0.139 rad / hr = 0.139 × = 3.86 × 10 −5 rad / s t2 − t1 9 hr hr 3600 s

Assess: An angular displacement of approximately one-fifth of a complete rotation is reasonable because the separation between Kampala and Singapore is approximately one-fifth of the earth’s circumference.


7.5. Model: The earth is a particle orbiting around the sun. Solve:

(a) The magnitude of the earth’s velocity is displacement divided by time:

v=

2π (1.5 × 1011 m ) 2πr = 3.0 × 104 m/s = 24 hr 3600 s T × 365 days × 1 day 1 hr

(b) Since v = rω, the angular acceleration is

ω=

v 3.0 × 10 4 m / s = = 2.0 × 10 −7 rad / s r 1.5 × 1011 m

(c) The centripetal acceleration is 4 v 2 (3.0 × 10 m / s) = = 6.0 × 10 −3 m / s 2 1.5 × 1011 m r 2

ar =

Assess: A tangential velocity of 3.0 × 10 4 m/s or 30 km/s is large, but needed for the earth to go through a displacement of 2π(1.5×1011 m) ≈ 9.4 × 108 km in 1 year.


7.6. Solve: Speed, radial velocity, radial acceleration, tangential acceleration, and the magnitude of the net force are constant. Furthermore, the radial velocity and tangential acceleration are zero.


7.7. Solve: The pebble’s angular velocity ω = (3.0 rev / s)(2π rad rev) = 18.85 rad / s. The speed of the pebble as it moves around a circle of radius r = 30 cm = 0.30 m is

v = ωr = (18.85 rad s)(0.30 m ) = 5.65 m / s The radial acceleration is

v 2 (5.65 m / s) = = 106 m / s 2 r 0.30 m 2

ar =


7.8. Solve: Let RE be the radius of the earth at the equator. This means RE + 300 m is the radius to the top of the tower. Letting T be the period of rotation, we have

vtop − vbottom =

2π ( RE + 300 m ) T

2πRE 2π (300 m ) 600π m = = = 2.18 × 10 −2 m / s 24 hrs 24(3600) s T


7.9. Model: The rider is assumed to be a particle. Solve:

Since ar = v 2 / r, we have

v 2 = ar r = (98 m / s 2 )(12 m ) ⇒ v = 34.3 m / s Assess: 34.3 m/s ≈ 70 mph is a large yet understandable speed.


7.10. Solve: The plane must fly as fast as the earth’s surface moves, but in the opposite direction. That is, the plane must fly from east to west. The speed is

2π rad  km km 1 mile v = ωr =  = 1680 × = 1040 mph 6.4 × 10 3 km ) = 1680  24 hr  ( hr hr 1.609 km


7.11. Model: Treat the block as a particle attached to a massless string that is swinging in a circle on a frictionless table. Visualize:

Solve:

(a) The angular velocity and speed are

ω = 75

rev 2π rad × = 471.2 rad / min min 1 rev

vt = rω = (0.5 m )( 471.2 rad min ) ×

(b) The radial component of Newton’s second law is

∑F

r

=T=

mv 2 r

Thus

T = (0.2 kg)

(3.93 m / s)2 0.5 m

= 6.18 N

1min = 3.93 m / s 60 s


7.12. Solve: Newton’s second law is F = mar = mrω 2 . Substituting into this equation yields: ω=

F = mr

9.2 × 10 −8 N (9.1 × 10 kg)(5.3 × 10 −11 m)

= 4.37 × 1016 rad / s = 4.37 × 1016

−31

rad 1 rev × = 6.95 × 1015 rev / s s 2π rad

Assess: This is a very high number of revolutions per second.


7.13. Model: The vehicle is to be treated as a particle in uniform circular motion. Visualize:

On a banked road, the normal force on a vehicle has a horizontal component that provides the necessary centripetal acceleration. The vertical component of the normal force balances the weight. Solve: From the physical representation of the forces in the r-z plane, Newton’s second law can be written mv 2 ∑ Fz = n cosθ − mg = 0 ⇒ n cosθ = mg ∑ Fr = n sinθ = r Dividing the two equations and making the conversion 90 km hr = 25 m / s yields:

v2 (25 m / s) = = 0.128 ⇒ θ = 7.27° rg (9.8 m / s 2 )500 m 2

tan θ =

Assess: Such a banking angle for a speed of approximately 55 mph is clearly reasonable and within our experience as well.


7.14. Model: We are using the particle model for the car in uniform circular motion on a flat circular track. There must be friction between the tires and the road for the car to move in a circle. Visualize:

Solve:

The centripetal acceleration is

v 2 (25 m / s) = = 6.25 m / s 2 r 100 m 2

ar =

The acceleration points to the center of the circle, so the net force is r r Fr = ma = (1500 kg) 6.25 m /s2 , toward center

(

= (9380 N, toward center ) This force is provided by static friction

fs = Fr = 9380 N

)


7.15. Model: The motion of the moon around the earth will be treated through the particle model. The circular motion is uniform. Visualize:

Solve:

The tension in the cable provides the centripetal acceleration. Newton’s second law is

∑ F = T = mrω r

2

 2π  = mr    Tmoon 

2

2

 2π 1 day 1 hr  20 = (7.36 × 10 22 kg)(3.84 × 10 8 m ) × ×  = 2.0 × 10 N  27.3 days 24 hrs 3600 s  Assess: This is a tremendous tension, but clearly understandable in view of the moon’s large mass and the large radius of circular motion around the earth.


7.16. Model: Model the ball as a particle in uniform circular motion. Rolling friction is ignored. Visualize:

Solve:

The track exerts both an upward normal force and an inward normal force. From Newton’s second law,

60 rev 2π rad 1min  = 0.237 N Fnet = n2 = mrω 2 = (0.030 kg)(0.20 m ) × × 1 rev 60 s   min 2


7.17. Model: The satellite is considered to be a particle in uniform circular motion around the moon. Visualize:

Solve: The radius of moon is 1.738 × 106 m and the satellite’s distance from the center of the moon is the same quantity. The angular velocity of the satellite is

ω=

2π 2π rad 1 min = × = 9.52 × 10 −4 rad / s T 110 min 60 s

and the centripetal acceleration is

ar = rω 2 = (1.738 × 10 6 m )(9.52 × 10 −4 rad / s) = 1.58 m / s 2 2

The acceleration of a body in orbit is the local “g” experienced by that body.


7.18. Model: The earth is considered to be a particle in uniform circular motion around the sun.

Solve: The earth orbits the sun in 365 days and is 1.5 × 1011 m from the sun. The angular velocity and centripetal acceleration are

ω=

2π rad 1 day 1hr × × = 2.0 × 10 −7 rad / s 365 days 24 hr 3600 s

ar = g = rω 2 = (1.5 × 1011 m )(2.0 × 10 −7 rad / s) = 6.0 × 10 −3 m / s 2 2

Assess: The smallness of this acceleration due to gravity is essentially due to the large earth-sun distance.


7.19. Model: Model the passenger in a roller coaster car as a particle in uniform circular motion. Visualize:

r Note that the normal force n of the seat pushing on the passenger is the passenger’s apparent weight. Solve: Since the passengers feel 50% heavier than their true weight, n = 1.50 w . Thus, from Newton’s second law: mv 2 mv 2 ∑ Fr = n − w = 1.50w − w = r ⇒ 0.50mg = r ⇒ v = 0.50 gr =

(0.50)(30 m)(9.8 m / s 2 ) = 12.1 m / s


7.20. Model: Model the roller coaster car as a particle at the top of a circular loop-the-loop undergoing uniform circular motion. Visualize:

Notice that the r-axis points downward, toward the center of the circle. r v Solve: The critical speed occurs when n goes to zero and w provides all the centripetal force pulling the car in the vertical circle. At the critical speed mg = mvc2 r , therefore vc = rg . Since the car’s speed is twice the critical speed, vt = 2vc and the centripetal force is

∑ Fr = n + w =

2 mv 2 m( 4vc ) m( 4rg) = = = 4 mg r r r

Thus the apparent weight is wapp = n = 3 mg. Consequently, wapp w = 3.


7.21. Model: Model the roller coaster car as a particle undergoing uniform circular motion along a loop. Visualize:

Notice that the r-axis points downward, toward the center of the circle. Solve: In this problem the apparent weight is equal to the weight: wapp = n = mg. We have

∑F

r

=n+w=

mv 2 = mg + mg ⇒ v = 2rg = 2(20 m )(9.8 m / s 2 ) = 19.8 m / s r


7.22. Model: Model the bucket of water as a particle in uniform circular motion. Visualize:

Solve: Let us say the distance from shoulder to cupped hand is 26 inches and the distance from the bucket handle to the top of the water in the bucket is 13 inches. This makes the shoulder to water distance ∼1 m. The minimum angular velocity for swinging a bucket of water in a vertical circle without spilling any water corresponds to the case when the speed of the bucket is critical. In this case, n = 0 N when the bucket is in the top position of the circular motion. We get

∑ F = n + w = 0 N + mg = r

⇒ ωc = g / r =

mvc 2 = mrω c2 r

9.8 m / s2 1 rev 60 s = 3.13 rad / s = 3.13 rad s × × = 30 rpm 1m 2π rad 1 min


7.23. Model: Model the car as a particle in nonuniform circular motion. Visualize:

Note that halfway around the curve, the tangent is 45° south of east. The perpendicular component of the acceleration is 45° north of east. Solve: The radial and tangential components of the acceleration are

ar = a cos 25° = (3.0 m s 2 ) cos 25° = 2.72 m s 2 at = a sin 25° = (3.0 m s 2 ) sin 25° = 1.27 m s 2


7.24. Model: Model the child on the merry-go-round as a particle in nonuniform circular motion. Visualize:

Solve: (a) The speed of the child is v0 = rω = (2.5 m)(1.57 rad/s) = 3.93 m/s. (b) The merry-go-round slows from 1.57 rad/s to 0 in 20 s. Thus

ω1 = 0 = ω 0 +

rω at (2.5 m)(1.57 rad /s) = –0.1965 m /s 2 t1 ⇒ at = − 0 = − 20 s r t1

During these 20 s, the wheel turns through angle

θ1 = θ 0 + ω 0 t1 +

at 2 0.1965 m /s 2 (20 s) 2 = 15.7 rad t1 = 0 + (1.57 rad /s) (20 s) − 2r 2(2.5 m)

In terms of revolutions, θ1 = (15.7 rad)(1 rev/2π rad) = 2.50 rev.


7.25. Model: Model the particle on the crankshaft as being in nonuniform circular motion. Visualize:

Solve: (a) The initial angular velocity is ω0 = 2500 rpm × (1 min/60 s) × (2π rad/rev) = 261.8 rad/s. The crankshaft slows from 261.8 rad/s to 0 in 1.5 s. Thus

ω1 = 0 = ω 0 +

rω at (0.015 m)(261.8 rad /s) = –2.618 m /s 2 t1 ⇒ at = − 0 = − 1.5 s r t1

(b) During these 1.5 s, the crankshaft turns through angle

θ1 = θ 0 + ω 0 t1 +

at 2 2.618 m /s 2 t1 = 0 + (261.8 rad /s) (1.5 s) − (1.5 s) 2 = 196.3 rad 2(0.015 m) 2r

In terms of revolutions, θ1 = (196.3 rad)(1 rev/2π rad) = 31.25 rev.


7.26. Model: Use the particle model for the dot in nonuniform circular motion. Visualize:

Solve:

After .5 seconds,

a a 400 2000 rev  rev  at  ω 1 = ω 0 +  t  (t1 − t 0 ) ⇒  =0 + rev / s 2 (0.5 s − 0 s) ⇒  t  =  60 s   r  r 6 s r Using this result,

a 400 rev  2 θ1 = θ 0 + ω 0 (t1 − t 0 ) + 12  t  (t1 − t 0 ) = 0 rad + 0 rad + 12  (0.5 s − 0 s)2 = 8.33 rev  6 s2   r For the next 0.5 s or at t2 = 1 s,

2000 rev  θ 2 = θ1 + ω 1 (t2 − t1 ) = 8.33 rev +  (1 s − 0.5 s) = 8.33 rev + 16.66 rev = 25 rev  60 s  So, the dot is at the same place it started, since the disk has rotated through an integral number of revolutions. The speed of the reference dot is

rev  2π rad 1min = 8.38 m/s × × v2 = rω 2 = (0.04 m ) 2000  min  1 rev 60 s


7.27. Model: Model the car as a particle in nonuniform circular motion. Visualize:

Note that the tangential acceleration stays the same at 1.0 m/s2 . As the tangential velocity increases, the radial acceleration increases as well. After a time t1, as the car goes through an angle θ1 − θ 0 , the total acceleration will increase to 2.0 m/s2. Our objective is to find this angle. Solve: Using v1 = v0 + at (t1 − t 0 ) , we get

v1 = 0 m / s + (1.0 m / s 2 )(t1 − 0 s) = (1.0 m / s 2 ) t1 2 2 rv 2 (1.0 m / s ) t1 t2 ⇒ ar = 21 = = 1 (m / s 4 ) 120 m 120 r 2

⇒ atotal = 2.0 m / s = a + a = 2 t

2

2 r

(1.0 m / s )

2 2

2

  t2 +  1 ( m / s 4 ) ⇒ t1 = 14.4 s 120  

We can now determine the angle θ1 using

a 2 θ1 = θ 0 + ω 0 (t1 − t 0 ) + 12  t  (t1 − t 0 )  r

= 0 rad + 0 rad +

2 1 (1.0 m / s ) (14.4 s)2 = 0.864 rad = 49.5° 2 (120 m )


7.28. Model: We will use the particle model for the test tube which is in nonuniform circular motion. Solve:

(a) The radial acceleration is

rev 1 min 2π rad  ar = rω 2 = (0.1 m ) 4000 × × = 1.75 × 10 4 m / s 2  min 60 s 1 rev  2

(b) An object falling 1 meter has a speed calculated as follows:

v12 = v02 + 2 a y ( y1 − y0 ) = 0 m + 2( −9.8 m / s 2 )( −1.0 m ) ⇒ v1 = 4.43 m / s When this object is stopped in 1 × 10 −3 s upon hitting the floor,

v2 = v1 + a y (t2 − t1 ) ⇒ 0 m s = −4.43 m s + a y (1 × 10 −3 s) ⇒ a y = 4.43 × 10 3 m / s 2

This result is one-fourth of the above radial acceleration. Assess: The radial acceleration of the centrifuge is large, but it is also true that falling objects are subjected to large accelerations when they are stopped by hard surfaces.


7.29. Model: We will use the particle model for the astronaut undergoing nonuniform circular motion. Solve:

(a) The initial conditions are ω 0 = 0 rad/s, θ 0 = 0 rad, t 0 = 0 s , and r = 6.0 m . After 30 s,

ω1 =

1 rev 1 rev 2π rad = × × = 4.83 rad / s 1.3 s 1.3 s rev

Using these values at t1 = 30 s,

ω 1 = ω 0 + ( at r )(t1 − t 0 ) = 0 + ( at r )t1 1  ⇒ at = (6.0 m )( 4.83 rad / s) = 0.967 m / s 2  30 s  (b) The radial acceleration is

ar = rω 12 = (6.0 m )( 4.83 rad / s)

2

g = 14.3g . 9 8 m / s2 ) (

Assess: The above acceleration is typical of what astronauts experience during liftoff.


7.30. Model: Assume the particle model for the satellite in circular motion. Visualize:

To be in a geosynchronous orbit means rotating at the same rate as the earth, which is 24 hours for one complete rotation. Because the altitude of the satellite is 3.58 × 107 m, r = 3.58 × 107 m, re = 3.58 × 107 m + 6.37 × 106 m = 4.22 × 107 m. Solve: (a) The period (T) of the satellite is 24 hours. (b) The acceleration due to gravity is

2π 2π 1 hr  = 0.223 m/s2 g = ar = rω 2 = r   = ( 4.22 × 10 7 m ) ×  T   24 hr 3600 s  2

2

(c) There is no normal force on a satellite, so the apparent weight is zero. It is in free fall.


7.31. Model: Treat the man as a particle. The man at the equator undergoes uniform circular motion as the earth rotates. Visualize:

Solve: The scale reads the man’s apparent weight wapp = n, the force of the scale pushing up against his feet. At the north pole, where the man is in static equilibrium, n P = (w app)P = mg = 735 N At the equator, there must be a net force toward the center of the earth to keep the man moving in a circle. The r-axis points toward the center, so

∑F

r

= w − nE = mω 2 r ⇒ nE = ( wapp ) E = mg − mω 2 r = ( wapp ) P − mω 2 r

The equator scale reads less than the north pole scale by the amount mω2r. The man’s angular velocity is that of the equator, or

ω=

2π 2π rad = = 7.27 × 10 −5 rad /s 24 hours × (3600 s / 1 hour) T

Thus the north pole scale reads more than the equator scale by

∆wapp = (75 kg)(6.37 × 10 6 m)(7.27 × 10 −5 rad /s) 2 = 2.5 N Assess: The man at the equator appears to have lost ∆m = ∆w app/g ≈ 0.25 kg, or the equivalent of ≈ 12 lb.


7.32. Model: We will use the particle model for the car which is in uniform circular motion. Visualize:

Solve:

The centripetal acceleration of the car is

v 2 (15 m / s) = = 4.5 m / s 2 r 50 m The acceleration is due to the force of static friction. The force of friction is fs = mar = (1500 kg) 4.5m s 2 = 6750 N . Assess: The model of static friction is fs max = nµs = mgµs ≈ mg ≈ 15, 000 N since µs ≈ 1 for a dry road surface. We see that fs < fs max , which is reasonable. 2

ar =

(

)


7.33. Model: Use the particle model for the (cart+child) system which is in uniform circular motion. Visualize:

Solve:

Newton’s second law along r and z directions can be written:

∑F

r

= T cos 20° − n sin 20° = mar

∑F

z

= T sin 20° − n cos 20° − mg = 0

The cart’s centripetal acceleration is

rev 1 min 2π rad  ar = rω 2 = (2.0cos20° m )13.5 × × = 3.756 m / s 2  min 60 s 1 rev  The above force equations can be rewritten as 2

0.94T − 0.342 n = (25 kg)(3.756 m / s 2 ) = 93.9 N 0.342T + 0.94n = (25 kg)(9.8 m / s 2 ) = 245 N

Solving these two equations yields T = 172 N for the tension in the rope. Assess: In view of the child+cart weight of 245 N, a tension of 172 N is reasonable.


7.34. Model: Model the ball as a particle which is in a vertical circular motion. Visualize:

Solve:

At the bottom of the circle,

∑F

r

=T−w=

(0.5 kg)v 2 ⇒ v = 5.5 m / s mv 2 ⇒ (15 N ) − (0.5 kg)(9.8 m / s 2 ) = r (1.5 m)


7.35. Model: We will use the particle model for the car, which is undergoing uniform circular motion on a banked highway, and the model of static friction. Visualize:

Note that we need to use the coefficient of static friction µs, which is 1.0 for rubber on concrete. Solve: Newton’s second law for the car is

∑F

r

= fs cosθ + n sin θ =

mv 2 r

∑F

z

= n cosθ − fs sin θ − w = 0 N

Maximum speed is when the static friction force reaches its maximum value fs max = µ s n. Then

mv 2 r Dividing these two equations and simplifying, we get n( µ s cos15° + sin 15°) =

n(cos15° − µ s sin 15°) = mg

µ s + tan 15° v 2 µ + tan 15° = ⇒ v = gr s 1 − µ s tan 15° gr 1 − µ s tan 15° =

.268) = 34.5 m / s (9.80 m / s )(70 m) (1(1.0−+00.268 ) 2

Assess: The above value of 34.5 m/s ≈ 70 mph is reasonable.


7.36. Model: Use the particle model for the rock which is undergoing uniform circular motion. Visualize:

Solve:

Newton’s second law is

mv 2 ∑ Fz = T sin10° − mg = 0 N r where the radius of the circular motion is r = (1.0 m ) cos10° = 0.985 m . Dividing these two equations, we get

∑F

r

tan 10° =

⇒ω =

= T cos10° =

gr ⇒v= v2

gr = tan 10°

(9.8 m / s )(0.985 m) = 7.40 m / s 2

tan 10°

v 7.40 m / s rad 1 rev 60 s = = 7.51 rad / s = 7.51 × × = 71.7 rpm r 0.985 m s 2π rad 1 min


7.37. Model: Use the particle model and static friction model for the coin which is undergoing circular motion. Visualize:

Solve: The force of static friction is fs = µ s n = µ s mg . This force is equivalent to the maximum centripetal force that can be applied without sliding. That is,

µ s mg = m

vt 2 2 = m(rω max ) ⇒ ω max = r

= 7.23

µs g = r

(0.80)(9.8 m / s 2 ) 0.15 m

= 7.23 rad / s

rad 1 rev 60 s × × = 69 rpm s 2π rad 1 min

So, the coin will stay still on the turntable. Assess: A rotational speed of approximately 1 rev per second for the coin to stay stationary seems reasonable.


7.38. Model: Use the particle model for the car which is in uniform circular motion. Visualize:

Solve:

Newton’s second law is

∑F

r

= T sin 20° = mar =

mv 2 r

∑F

z

= T cos20° − w = 0 N

These equations can be written as

T sin 20° =

mv 2 r

T cos20° = mg

Dividing these two equations gives

tan 20° = v 2 rg ⇒ v = rg tan 20° =

(4.55 m)(9.8 m / s 2 ) tan 20° = 4.03 m / s


7.39. Model: Use the particle model for the ball in circular motion. Visualize:

r Solve: (a) The mass moves in a horizontal circle of radius r = 20 cm. The acceleration a and the net force vector r point to the center of the circle, notr along the string. The only two forces are the string tension T , which does point along the string, and the weight w . These are shown in the free-body diagram. Newton’s second law for circular motion is ΣFz = T cosθ − w = T cosθ − mg = 0 N

ΣFr = T sin θ = mar =

From the z-equation,

T=

(0.5 kg)(9.8 m / s 2 ) mg = = 5.00 N cosθ cos11.54°

(b) We can find the rotation speed from the r-equation:

v=

rT sin θ = 0.633 m / s m

The rotation frequency is f = v 2πr = 0.503 rev / s . Converting to rpm,

f = 0.503

rev 60 sec × = 30.2 rpm sec 1 min

mv 2 r


7.40. Model: Use the particle model for the ball which is in uniform circular motion. Visualize:

Solve:

From Newton’s second law along r and z directions,

∑ Fr = n cosθ =

mv 2 r

∑F

z

= n sin θ − mg = 0 ⇒ n sin θ = mg

Dividing the two force equations gives

tan θ =

gr v2

From the geometry of the cone, tan θ = r y . Thus

r gr = ⇒ v = gy y v2


7.41. Model: Consider the passenger to be a particle and use the model of static friction. Visualize:

Solve: The passengers stick to the wall if the static friction force is sufficient to support their weight: fs = w. The minimum angular velocity occurs when static friction reaches its maximum possible value fs max = µsn. Although clothing has a range of coefficients of friction, it is the clothing with the smallest coefficient (µs = 0.6) that will slip first, so this is the case we need to examine. Assuming that the person is stuck to the wall, Newton’s second law is

∑F

r

= n = mω 2 r

∑ Fz =

fs − w = 0 ⇒ fs = mg

The minimum frequency occurs when 2 fs = fs max = µ s n = µ s mrω min

Using this expression for fs in the z-equation gives 2 fs = µ s mrω min = mg

⇒ ω min =

g = µs r

9.80 m /s 2 1 rev 60 s = 2.56 rad /s = 2.56 rad /s × × = 24.4 rpm 0.60(2.5 m) 2π rad 1 min

Assess: Note the velocity does not depend on the mass of the individual. Therefore, the minimum mass sign is not necessary.


7.42. Model: Use the particle model for the marble in uniform circular motion. Visualize:

Solve: The marble will roll in a horizontal circle if the static friction force is sufficient to support its weight: fs = w. If mg > fs max, then static friction is not sufficient and the marble will slip down the side as it rolls around the circumference. The r-equation of Newton’s second law is

∑F

r

2π rad 1 min  = n = mrω 2 = (0.010 kg)(0.060 m)150 rpm × × = 0.148 N  1 rev 60 s  2

Thus the maximum possible static friction is fs max = µsn = (0.80)(0.148 N) = 0.118 N. The friction force needed to support a 10 g marble is fs = mg = 0.098 N. We see that fs < fs max, therefore friction is sufficient and the marble spins in a horizontal circle. Assess: In reality, rolling friction will cause the marble to gradually slow down until fs max < mg. At that point, it will begin to slip down the inside wall.


7.43. Model: Use the particle model for the car and the model of kinetic friction. Visualize:

Solve: We will apply Newton’s second law to all three cars. Car A:

∑ F = n + ( f ) + w = 0 N − f + 0 N = ma ∑ F = n + ( f ) + y = n + 0 N − mg = 0 N x

y

x

k x

x

y

k y

k

x

y

The y-component equation means n = mg. Since fk = µ k n , we have fk = µ k mg. From the x-component equation,

− fk − µ k mg = = − µ k g = −9.8 m / s 2 m m Car B: Car B is in circular motion with the center of the circle above the car. ax =

mv 2 r + wt = 0 N − fk + 0 N = + mat

= n r + ( fk ) r + wr = n + 0 N − mg = mar =

∑F

r

= n t + ( fk )t

∑F

t

From the r-equation

n = mg +

 v2  mv 2 ⇒ f k = µ k n = µ k m g +  r r 

Substituting back into the t-equation,

at = −

 fk µ m v2  (25 m / s)2  2 = − k  g +  = − µ k  9.8 m / s 2 +  = −12.9 m / s 200 m m m  r  

Car C: Car C is in circular motion with the center of the circle below the car.

∑F

r

= n r + ( fk ) r + wr = − n + 0 N + mg = mar =

∑F

t

(

mv 2 r

= nt + ( fk )t + wt = 0 N − fk + 0 N = mat

)

From the r-equation n = m g − v 2 r . Substituting this into the t-equation yields

at =

− fk − µ k n = = − µ k ( g − v 2 r ) = −6.68 m / s 2 m m


7.44. Model: Use the particle model for the car which is undergoing circular motion. Visualize:

Solve: The car is in circular motion with the center of the circle below the car. Newton’s second law at the top of the hill is n mv 2 ∑ Fr = wr − nr = mg − n = mar = r ⇒ v 2 = r g − m  Maximum speed is reached when n = 0 and the car is beginning to lose contact with the road.

vmax = rg =

(50 m)(9.8 m / s 2 ) = 22.1 m / s

Assess: A speed of 22.1 m/s is equivalent to 49.5 mph, which seems like a reasonable value.


7.45. Model: Model the ball as a particle that is moving in a vertical circle. Visualize:

(

)

Solve: (a) The ball’s weight w = mg = (0.500 kg) 9.8 m / s 2 = 4.9 N . (b) Newton’s second law at the top is

∑F

r

= T1 + w = mar = m

v2 r

 ( 4.0 m / s) 2   v2  ⇒ T1 = m − g = (0.5 kg) − 9.8 m / s 2  = 2.9 N  r   1.02 m  (c) Newton’s second law at the bottom is

∑F

r

= T2 − w =

mv 2 r

  v2  (7.5 m / s)2  ⇒ T2 = m g +  = (0.5 kg)9.8 m / s 2 +  = 32.5 N 1.02 m  r  


7.46. Model: Use the particle model for yourself while in uniform circular motion. Visualize:

Solve:

(a) The speed and acceleration are

v=

2πr 2π (15 m ) = = 3.77 m / s T 25 s

v 2 (3.77 m / s) = = 0.95 m / s 2 r 15 m 2

ar =

(b) Newton’s second law at the top is

∑F

r

= w − n = mar =

mv 2 r

  v2  (3.77 m / s)2  2 ⇒ n = wapp = m g −  = m 9.8 m / s 2 −  = m(8.85 m / s ) r  15 m   ⇒

wapp w

=

8.85 m s 2 = 0.90 9.8 m s 2

(c) Newton’s second law at the bottom is

∑F

r

= n − w = mar =

mv 2 r

 (3.77 m / s2 )  = 10.75 m / s2 m  v2  ⇒ n = wapp = m g +  = m 9.8 m / s 2 + )  ( r 15 m    ⇒

wapp w

=

10.75 m s 2 = 1.10 9.8 m s 2


7.47. Model: Model a passenger as a particle rotating in a vertical circle. Visualize:

Solve:

(a) Netwon’s second law at the top is

∑F

r

= nT + w = mar =

The speed is

v=

mv 2 mv 2 ⇒ nT + mg = r r

2πr 2π (8 m ) = = 11.17 m / s T 4.5 s

 (11.17 m / s) 2   v2  ⇒ nT = m − g = (55 kg) − 9.8 m / s 2  = 319 N  r  8m   That is, the ring pushes on the passenger with a force of 319 N at the top of the ride. Newton’s second law at the bottom:  v2  mv 2 mv 2 ∑ Fr = nB − w = mar = r ⇒ nB = r + mg = m r + g

 (11.17 m / s) 2  + 9.8 m / s 2  = 1397 N = (55 kg) 8m   This is the force with which the ring pushes on the rider when she is at the bottom of the ring. (b) To just stay on at the top, n = 0 N in the r-equation at the top in part (a). Thus, 2

mg =

 2π  r 8m mv 2 = 2π = 5.68 s = mrω 2 = mr   ⇒ Tmax = 2π g 9.8 m / s 2 r  Tmax 


7.48. Model: Model the chair and the rider as a particle in uniform circular motion. Visualize:

Solve:

Newton’s second law along the r-axis is

∑F

r

= Tr + wr = mar ⇒ T sin θ + 0 N = mrω 2

Since r = L sin θ , this equation becomes

2π rad  T = mLω 2 = (150 kg)(9.0 m ) = 3330 N  4.0 s  2

Thus, the 3000 N chain is not strong enough for the ride.


7.49. Model: Model the ball as a particle in motion in a vertical circle. Visualize:

r Solve: If the ball moves in a complete circle, then there is a tension force T when the ball is at the top of the circle. The tension force adds to the weight to cause the centripetal acceleration. The forces are along the r-axis, and the center of the circle is below the ball. Newton’s second law at the top is mv 2 ( Fnet )r = T + w = T + mg = r rT m The tension T can’t become negative, so T = 0 N gives the minimum speed vmin at which the ball moves in a circle. If the speed is less than vmin , then the string will go slack and the ball will fall out of the circle before it reaches the top. Thus, ⇒ vtop = rg +

vmin = rg ⇒ ω min =

vmin = r

rg = r

g = r

(9.8 m / s ) = 3.13 rad / s = 29.9 rpm 2

(1.0 m)


7.50. Model: Model the person as a particle in uniform circular motion. Visualize:

Solve: The only force acting on the passengers is the normal force of the wall. Newton’s second law along the r-axis is:

∑F

r

= n = mrω 2

To create “normal” gravity, the normal force by the inside surface of the space station equals mg. Therefore,

mg = mrω 2 ⇒ ω =

2π = T

g r 500 m ⇒ T = 2π = 2π = 45.0 s r g 9.8 m / s 2

Assess: This is a fast rotation. The tangential speed is

v=

2πr 2π (500 m ) = = 70 m / s ≈ 140 mph T 45 s


7.51. Model: Model the ball as a particle swinging in a vertical circle. Visualize:

Solve: Initially, the ball is moving in a circle. Once the string is cut, it becomes a projectile. The final circularmotion velocity is the initial velocity for the projectile. The free body diagram for circular motion is shown at the bottom of the circle. Since T > w , there is a net force toward the center of the circle that causes the centripetal acceleration. The r-equation of Newton’s second law is mv 2 ( Fnet )r = T − w = T − mg = r

[

]

0.60 m 5.0 N − (0.10 kg)(9.8 m / s 2 ) = 4.91 m / s 0.100 kg r As a projectile the ball starts at y0 = 1.4 m with v0 = 4.91iˆ m / s . The equation for the y-motion is ⇒ vbottom =

r (T − mg) = m

y1 = 0 m = y0 + v0 y ∆t − 12 g( ∆t ) = y0 − 12 gt12 2

This is easily solved to find that the ball hits the ground at time

t1 =

2 y0 = 0.535 s g

During this time interval it travels a horizontal distance x1 = x 0 + v0 x t1 = ( 4.91 m / s)(0.535 s) = 2.63 m So the ball hits the floor 2.63 m to the right of the point where the string was cut.


7.52. Model: Model the ball as a particle undergoing circular motion in a vertical circle. Visualize:

Solve: Initially, the ball is moving in circular motion. Once the string breaks, it becomes a projectile. The final circular-motion velocity is the initial velocity for the projectile, which we can find by using the kinematic equation

v12 = v02 + 2 a y ( y1 − y0 ) ⇒ 0 m 2 s 2 = (v0 ) + 2( −9.8 m / s 2 )( 4.0 m − 0 m ) ⇒ v0 = 8.85 m / s 2

This is the speed of the ball as the string broke. The tension in the string at that instant can be found by using the r-component of the net force on the ball:

 v02y  (8.85 m / s)2 T = m ⇒ F = T = 0.100 kg = 13.1 N ( ) ∑ r  r  0.6 m  


7.53. Model: Model the car as a particle on a circular track. Visualize:

Solve:

(a) Newton’s second law along the t-axis is

∑F

t

= Ft = mat ⇒ 1000 N = (1500 kg)at ⇒ at = 2 3 m / s 2

With this tangential acceleration, the car’s tangential velocity after 10 s will be

v1t = v0 t + at (t1 − t 0 ) = 0 m s + (2 3 m / s 2 )(10 s − 0 s) = 20 3 m / s

The radial acceleration at this instant is

v12t (20 3 m / s) 16 = = m / s2 25 m 9 r 2

ar = The car’s acceleration at 10 s has magnitude

a1 = at2 + ar2 =

(2 3 m / s ) + (16 9 m / s ) 2 2

2 2

= 1.90 m / s 2

θ = tan −1

 23 at = tan −1   = 20.6° ar  16 9 

where the angle is measured from the r-axis. (b) The car will begin to slide out of the circle when the static friction reaches its maximum possible value fs max = µ s n . That is,

mv22t ⇒ v2 t = rg = (25 m )(9.8 m / s 2 ) = 15.7 m / s r In the above equation, n = mg follows from Newton’s second law along the r-axis. The time when the car begins to slide can now be obtained as follows:

∑F

r

= fs max = µ s n = µ s mg =

v2 t = v0 t + at (t2 − t 0 ) ⇒ 15.7 m / s = 0 m s + (2 3 m / s 2 )(t2 − 0) ⇒ t2 = 23.5 s 2


7.54. Model: Model the steel block as a particle and use the model of kinetic friction. Visualize:

Solve:

r (a) The components of thrust ( F ) along the r-, t-, and z-directions are Fr = F sin 20° = (3.5 N ) sin 20° = 1.20 N

Ft = F cos 20° = (3.5 N ) cos 20° = 3.29 N

Fz = 0 N

Newton’s second law is

( Fnet )r = T + Fr = mrω 2 ( Fnet )t = Ft − fk = mat ( Fnet ) z = n − mg = 0 N The z-component equation means n = mg. The force of friction is

fk = µ k n = µ k mg = (0.60)(0.5 kg)(9.8 m / s 2 ) = 2.94 N Substituting into the t-component of Newton’s second law

(3.29 N) − (2.94 N) = (0.5 kg)at ⇒ at = 0.70 m / s 2 Having found at, we can now find the tangential velocity after 10 revolutions = 20π rad as follows:

θ1 =

1  at  2 t1 ⇒ t1 = 2 r 

2rθ1 = 18.95 s at

a ω 1 = ω 0 +  t  t1 = 6.63 rad / s  r (b) Substituting ω1 into the r-component of Newton’s second law yields:

T1 + Fr = mrω 12 ⇒ T1 + (1.20 N ) = (0.50 kg)(2.0 m )(6.63 rad / s) ⇒ T1 = 42.8 N 2


7.55. Model: Model the block as a particle and use the model of kinetic friction. Visualize:

Solve: The only radial force is tension, so we can use Newton’s second law to find the angular velocity ωmax at which the tube breaks:

∑F

r

= T = mω 2 r ⇒ ω max =

Tmax = mr

50 N = 9.12 rad /s (0.50 kg)(1.2 m)

The compressed air and friction exert tangential forces, and the second law along the tangential direction is

∑F = F − f t

at =

t

k

= Ft − µ k n = Ft − µ k mg = mat

Ft 4.0 N − µk g = − (0.60)(9.80 m /s 2 ) = 2.12 m /s 2 0.50 kg m

The time needed to accelerate to 9.12 rad/s is given by a rω max (1.2 )(9.12 rad /s ) ω 1 = ω max = 0 +  t  t1 ⇒ t1 = = = 5.16 s  r 2.12 m /s 2 at During this interval, the block turns through angle

∆θ = θ1 − θ 0 = ω 0 t1 +

1 2

 2.12 m /s 2  1 rev  at  2 2 = 3.75 rev t1 = 0 + 12   (5.16 s) = 23.52 rad ×  r 2π rad  1.2 m 


7.56. Model: Assume the particle model for a ball in vertical circular motion. Visualize:

Solve:

(a) Newton’s second law in the r- and t-directions is

( Fnet )r = T + mg cosθ = mar =

mvt2 r

( Fnet )t = − mg sin θ = mat

Substituting into the r-component,

(20 N) + (2.0 kg)(9.8 m / s 2 ) cos30° = (2.0 kg)

vt2 ⇒ vt = 3.85 m / s (0.8 m)

(b) Substituting into the t-component,

−(9.8 m / s 2 ) sin30° = at ⇒ at = −4.9 m / s 2

The radial acceleration is

vt2 (3.85 m / s) = = 18.5 m / s 2 0.80 m r 2

ar = Thus, the magnitude of the acceleration is

a = ar2 + at2 =

(18.5 m / s ) + (−4.9 m / s ) 2 2

2 2

The angle of the acceleration vector from the r-axis is

φ = tan −1 The angle is below the r-axis.

at ar

= tan −1

4.9 = 14.8° 18.5

= 19.1 m / s 2


7.57. Solve: (a) You are spinning a lead fishing weight in a horizontal 1.0 m diameter circle on the ice of a pond when the string breaks. You know that the test weight (breaking force) of the line is 60 N and that the lead weight has a mass of 0.30 kg. What was the weight’s angular velocity in rad/s and in rpm? 60 N rev 60 s (b) ω2 = ⇒ ω = 20 rad / s × × = 191 rpm 2π rad min (0.3 kg)(0.5 m)


7.58. Solve: (a) At what speed does a 1500 kg car going over a hill with a radius of 200 m have an apparent weight of 11,760 N? 1500 kg v 2 â&#x2021;&#x2019; v = 19.8 m/s (b) 2940 N = 200 m


7.59. Solve: (a) A 1000 kg race car enters a 50 m radius curve and accelerates around the curve for 10.0 s. The forward force provided by the car’s wheels is 1500 N. After 10.0 s the car has moved 125 m around the track. Find the initial and final angular velocities. (b) From Newton’s second law, ∆s 125 m Ft = mat ⇒ 1500 N = (1000 kg)at ⇒ at = 1.5 m / s 2 ∆θ = = = 2.5 radians r 50 m a 1.5 m / s 2 θ f = θ i + ω i t + t t 2 ⇒ 2.5 rad = 0 rad + ω i (10 s) + (10 s)2 ⇒ ω i = 0.10 rad / s 2(50 m ) 2r ωf = ωi +

at 1.5 m / s 2 t = 0.1 rad / s + (10 s) = 0.40 rad / s 50 m r


7.60. Model: Assume the particle model for a sphere in circular motion at constant speed. Visualize:

Solve:

(a) Newton’s second law along the r and z axes is:

∑F

r

= T1 sin 30° + T2 sin 60° =

mvt2 r

∑F

z

= T1 cos 30° + T 2 cos 60° − w = 0 N

Since we want T1 = T2 = T, these two equations become

T (sin 30° + sin 60°) =

mvt2 r

T (cos 30° + cos 60°) = mg

Since sin 30° + sin 60° = cos 30° + cos 60° ,

mg =

mvt2 ⇒ vt = rg r

The triangle with sides L1, L2, and 1 m is isosceles, so L2 = 1 m and r = L2 cos30°. Thus

L2 cos 30°g = (b) The tension is

T=

(1 m) cos 30°g = (0.866 m)(9.8 m / s 2 ) = 2.91 m / s

(2.0 kg)(9.8 m / s 2 ) mg = = 14.35 N cos 30° + cos 60° 0.866 + 0.5


7.61. Model: Use the particle model for a sphere revolving in a horizontal circle. Visualize:

Solve:

Newton’s second law in the r- and z-directions is

mvt2 ∑ ( F ) z = T1 sin 30° − T2 sin 30° − w = 0 N r Using r = (1 m )cos30° = 0.886 m , these equations become

∑ ( F)

r

= T1 cos 30° + T2 cos 30° =

(0.300 kg)(7.5 m / s) = 22.50 N mvt2 = r cos 30° (0.866 m)(0.866) 2

T1 + T2 =

T1 − T2 =

(0.300 kg)(9.8 m / s 2 ) mg = 5.88 N = sin 30° (0.5)

Solving for T1 and T2 yields T1 = 14.19 N and T2 = 8.31 N.


7.62. Model: Use the particle model for a ball in motion in a vertical circle and then as a projectile. Visualize:

Solve:

For the circular motion, Newton’s second law along the r-direction is mv 2 ∑ Fr = T + w = r t Since the string goes slack as the particle makes it over the top, T = 0 N. That is,

mvt2 ⇒ vt = gr = (9.8 m / s 2 )(0.5 m ) = 2.21 m / s r The ball begins projectile motion as the string is released. The time it takes for the ball to hit the floor can be found as follows: w = mg =

y1 = y0 + v0 y (t1 − t 0 ) + 12 a y (t1 − t 0 ) ⇒ 0 m = 2.0 m + 0 m + 2

1 2

(−9.8 m / s )(t 2

1

− 0 s) ⇒ t1 = 0.639 s 2

The place where the ball hits the ground is

x1 = x 0 + v0 x (t1 − t 0 ) = 0 m + ( +2.21 m / s)(0.639 s − 0 s) = +1.41 m The ball hits the ground 1.41 m to the right of the point beneath the center of the circle.


7.63. Model: Use the particle model for the ball. Visualize:

Solve:

(a) Newton’s second law along the r- and z-directions is

∑F

r

= n cosθ = mrω 2

∑F

z

= n sin θ − w = 0 N

Using w = mg and dividing these equations yields:

tan θ =

g R− y = rω 2 r

where you can see from the figure that tan θ = ( R − y) r . Thus ω =

g . R− y

(b) ω will be minimum when (R – y) is maximum or when y = 0 m. Then ω min = g / R . (c) Substituting into the above expression,

ω=

g = R− y

9.8 m / s 2 rad 60 s 1 rev = 9.9 × × = 94.5 rpm 0.2 m − 0.1 m s 1 min 2π rad


7.64. Model: Use the particle model for the airplane. Visualize:

r Solve: In level flight, the lift force L balances the weight. When turning, the plane banks so that the radial component of the lift force can create a centripetal acceleration. Newton’s second law along the r- and z-directions is

∑F

r

= L sin θ =

mvt2 r

∑F

z

= L cosθ − mg = 0 N

These can be written:

sin θ =

mv 2 rL

cosθ =

mg L

Dividing the two equations gives: 2

400 miles × 1 hr × 1610 m   v v hour 3600 s 1 mile  = 18.5 km tan θ = ⇒r= =  gr g tan θ 9.8 m s 2 tan 10° 2

2

[

]

The diameter of the airplane’s path around the airport is 2 × 18.5 km = 37.0 km.


7.65. Model: Use the particle model for a small volume of water on the surface. Visualize:

Solve: Consider a particle of water of mass m at point C on the surface. Newton’s second law along the r- and z-directions is

( Fnet )r = n cosθ = mrω 2 ⇒ cosθ =

mrω 2 n

( Fnet ) z = n sin θ − mg = 0 N ⇒ sin θ =

mg n

Dividing both equations gives tan θ = g rω 2 . For a parabola z = ar2. This means

dz = 2ar slope of the curve at C = tan φ dr

⇒ 2 ar = tan φ = tan(90° − θ ) =

Equating the two equations for tanθ, we get

1 g ω2 = ⇒a= 2 2 ar rω 2g Thus the surface is described by the equation

z= which is the equation of a parabola.

ω2 2 r 2g

1 tan θ


Chapter 07