System dynamics and response 1st edition kelly solutions manual

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Chapter 2

2 Mechanical Systems 2.1 The mass moment of inertia of the annular cylinder is obtained by subtracting the moment of inertia of the hollow part from the moment of inertia of a solid cylinder. (a) I = Is − Ih The mass of a solid cylinder of radius r and length L is m = ρ πr 2 L (b) The mass of the solid cylinder is kg ⎞ ⎛ 2 ms = ⎜ 7600 3 ⎟π (0.4 m ) (1.2 m ) m ⎠ ⎝ = 4.58 x103 kg (c) The mass missing from the hollow part of the cylinder is kg ⎞ ⎛ 2 msh = ⎜ 7600 3 ⎟π (0.3 m ) (1.2 m ) m ⎠ ⎝ = 2.58 x10 3 kg (d) The total mass of the cylinder is m = ms − mh

( )

= 2.01x10 3 kg (e) (a) The mass moment of inertia of a solid cylinder about a centroidal axis parallel to the x axis is obtained using Table 2.1 as 1 I xx = m(3r 2 + L2 ) (f) 12 The moment of inertia of the solid cylinder about the cetnroidal axis is 1 I xxs = 4.58 x10 3 kg 3(0.4 m) 2 + (1.2 m) 2 12 = 733.5 kg ⋅ m 2 (g) The moment of inertia of the hollow part of the cylinder abut the centroidal x axis is 1 I xxh = 2.58 x10 3 kg 3(0.3 m) 2 + (1.2 m) 2 12 = 367.5 kg ⋅ m 2 (h) The moment of inertia of the cylinder about the centroidal x axis is I xx = I xxs − I xxh

)[

(

(

]

)[

]

= 366.0 kg ⋅ m 2 (i) The parallel axis theorem is used to determine the moment of inertia about the x axis as shown in Figure P2.1 is

27 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2 ⎛ L⎞ I xx = I xx + m⎜ ⎟ ⎝2⎠

2

(

)

= 366.0 kg ⋅ m 2 + 2.01x10 3 kg (0.6 m )

2

= 1.09 x10 3 kg ⋅ m 2 (j) (b) The mass moment of inertia of a cylinder about a longitudinal axis through its center is 1 I yy = mr 2 (k) 2 The mass moment of inertia of the solid cylinder about the y axis is 1 I yys = 4.58 x10 3 kg (0.4 m) 2 2 = 366.7 kg ⋅ m 2 (l) The moment of inertia of the hollow part of the cylinder about the y axis is 1 I yyh = 4.58 x10 3 kg (0.3 m) 2 2 = 116.0 kg ⋅ m 2 (l) The moment of inertia of the annular cylinder about the y axis is I yy = I yys − I yyh

(

)

(

)

= 250.7 kg ⋅ m 2

(m)

2.2 The total mass moment of inertia about an axis through A is the sum of the moments of inertia of the plate, the bar, and the cylinder, I A = I Ap + I Ab + I Ac (a) (a) The moment of inertia of the thin plate about an axis perpendicular to the page through its centroid is determined using Table 2.1 as 1 I zp = m( w 2 + h 2 ) 12 1 2 = (4.1 kg ) (0.2 m ) + (0.4 m) 2 12 = 6.83 x10 − 2 kg ⋅ m 2 (b) The parallel-axis theorem is used to determine its moment of inertia about an axis through A as I zAp = I zp + md 2

[

]

= (6.83 x10 − 2 kg ⋅ m 2 ) + (4.1 kg )(1.4 m )

2

= 8.10 kg ⋅ m 2 For the slender bar

28 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

(c)

Chapter 2 1 mL2 12 1 2 = (0.8 kg )(1.1 m ) 12 = 8.07 x10 − 2 kg ⋅ m 2

I zb =

I zAb = I zb + md

(d)

2

(

)

= 6.67 x10 − 2 kg ⋅ m 2 + (0.8 kg )(0.75 m) 2

= 0.531 kg ⋅ m 2

For the cylinder

(e)

(

)

1 m 3r 2 + L2 12 1 = (2.8 kg ) 3(0.1 m) 2 + (0.3 m) 2 12 = 2.80 x10 − 2 kg ⋅ m 2

I zc =

[

I zAc = I zc + md 2

(

]

)

= 2.80 x10 −2 kg ⋅ m 2 + (2.8 kg )(0.15 m )

(f)

2

= 0.091 kg ⋅ m 2 (g) The moment of inertia of the total assembly about an axis perpendicular to the page at A is obtained using Equations (a), (c), (e), and (g) as I zA = 8.10 kg ⋅ m 2 + (0.531 kg ⋅ m 2 ) + 0.091 kg ⋅ m 2

(

)

(

)

= 8.73 kg ⋅ m 2 (h) (a) A horizontal axis through A is a centroidal axis for each component. Thus application of the parallel axis theorem is not required. Using Table 2.1 for the plate 1 I xp = mh 2 12 1 2 = (4.1 kg )(0.4 m ) 12 = 0.0547 kg ⋅ m 2 (i) The moment of inertia of the slender bar about a horizontal axis is approximately zero.

For the cylinder 1 I xc = mr 2 2 1 2 = (2.8 kg )(0.1 m ) 2 = 0.014 kg ⋅ m 2 (j) Thus the moment of inertia of the assembly about a horizontal axis through A is

29 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2

(

) (

) (

I x = 5.4 x10 −2 kg ⋅ m 2 + 0 kg ⋅ m 2 + 1.4 x10 −2 kg ⋅ m 2 = 0.0687 kg ⋅ m

)

2

(k)

2.3 The total moment of inertia of the assembly of Figure P2.3 about an axis through O is the sum of the moment of the inertia of the bar about O and the moments of inertia of the particles about O. I O = I Ob + I Op1 + I Op 2 (a) The moment of inertia of the bar about its own centroidal axis is 1 I b = mL2 12 1 = (20 kg )(4 m) 2 12 = 26.7 kg ⋅ m 2 (b) The parallel axis theorem is used to calculate the moment of inertia of the bar about O as I Ob = I b + md 2 = 26.7 kg ⋅ m 2 + (20 kg )(1 m) 2

= 46.7 kg ⋅ m 2 (c) The moment of inertia of a particle about its own centroidal axis is zero. The parallel axis theorem is used to calculate the moment of inertia of a particle about another axis. For the 5-kg particle I Op1 = (5 kg)(1 m) 2 = 5 kg ⋅ m 2 The moment of inertia of the 10-kg particle about O is I Op 2 = (10 kg)(3 m) 2

(c)

= 90 kg ⋅ m 2 Combining Equations (a),(b),(c), and (d) leads to I O = 46.7 kg ⋅ m 2 + 5 kg ⋅ m 2 + 90 kg ⋅ m 2

(d)

= 141.7 kg ⋅ m 2

(e)

2.4 The velocity of the 2-kg particle is given as m (a) s (a) Noting that the particle is at the origin at t=0, the position vector of the particle at an arbitrary time t is obtained as v = [2 sin(3t )i + 4 cos(3t ) j + 2k ]

30 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2 t

r = ∫ vdt 0 t

= ∫ [2 sin(3t )i + 4 cos(3t ) j + 2k ]dt 0

t

4 ⎡ 2 ⎤ = ⎢− cos(3t )i + sin(3t ) j + 2tk ⎥ 3 ⎣ 3 ⎦ t =0 4 ⎫ ⎧2 = ⎨ [1 − cos(3t )]i + sin(3t ) j + 2tk ⎬ m 3 ⎭ ⎩3 (b) The acceleration of the particle at an arbitrary time is dv a= dt m = [6 cos(3t )i − 12 sin(3t ) j] 2 s (a) The velocity of the particle at t=1 is obtained from Equation (a) as m v (t = 1 s) = [2sin(3)i + 4 cos(3) j + 2k ] s m = [0.282i − 3.96 j + 2k ] s (b) The magnitude of the velocity at t=1 s is

[

(b)

(c)

(d)

]

1

v(t = 1 s) = (0.282) + (− 3.96) + (2) 2 2

= 4.45

2

2

m s

(e)

(c) The kinetic energy of the particle at t=1 s is 1 2 T = m v (t = 1 s) 2 2

1 (2 kg )⎛⎜ 4.45 m ⎞⎟ 2 s⎠ ⎝ = 19.8 N ⋅ m (d) The linear momentum of the particle at t=1 s is L = mv (t = 1 s) =

m⎤ ⎡ = (2 kg) ⎢(0.282i − 3.96 j + 2k ) ⎥ s⎦ ⎣ = (0.564i − 7.92 j + 4k ) N ⋅ s

(f)

(g)

2.5 (a) The velocity of the mass center of the bar at the instant shown is obtained using the relative velocity equation

v G = v O + ω k × rG / O

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(a)

Chapter 2 where O is located at the pin support. Since O is a fixed axis of rotation v o = a o = 0 .

Thus ⎛ r⎞ v G = 0 + ⎜10 ⎟k × (1.5 m )i ⎝ s⎠ ⎛ m⎞ (b) = ⎜15 ⎟ j ⎝ s⎠ (b) The acceleration of the mass center is obtained using the relative acceleration equation a G = a O + αk × rG / O + ωk × (ωk × rG / O ) r ⎞ ⎛ ⎛ r⎞ ⎛ m⎞ = 0 + ⎜ − 3 2 ⎟k × (1.5 m)i + ⎜10 ⎟k × ⎜15 ⎟ j s ⎠ ⎝ ⎝ s⎠ ⎝ s ⎠ m = (− 150 i − 4.5 j) 2 s (c) The kinetic energy of the bar at this instant is 1 1 2 T = m v G + Iω 2 2 2 2

r⎞ 1⎛ 1 ⎞ 1 ⎛ m⎞ 2⎛ = (6 kg )⎜15 ⎟ + ⎜ ⎟(6 kg )(5 m ) ⎜10 ⎟ 2 ⎝ 12 ⎠ 2 ⎝ s⎠ ⎝ s⎠ = 1300 N ⋅ m (d)The angular momentum of the bar about the centroidal axis is H G = Iω

(c)

2

1 (6 kg )(5 m) 2 ⎛⎜10 r ⎞⎟ 12 ⎝ s⎠ = 125 N ⋅ s ⋅ m

(d)

=

(e)

2.6 Since the center of the fan is a fixed axis of rotation the kinetic energy of each blade is 1 Tb = I O ω 2 (a) 2 The parallel axis theorem is used to determine the moment of inertia of each blade about O I O = I + md 2 = 0.4 kg ⋅ m 2 + (1.2 kg)(0.45 m) 2 = 0.592 kg ⋅ m 2 Using the result of Equation (b) in Equation (a) leads to

⎡⎛ 1 rev ⎞⎛ rad ⎞⎛ 1 min ⎞⎤ Tb = 0.592 kg ⋅ m 2 ⎢⎜ 250 ⎟⎜ 2π ⎟⎜ ⎟ 2 min ⎠⎝ rev ⎠⎝ 60 s ⎠⎥⎦ ⎣⎝

(

)

(b) 2

= 202.9 N ⋅ m 32 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

(c)

Chapter 2 The fan consists of four blades, thus the total kinetic energy of the fan is T = 4Tb = 811.5 N ⋅ m

(d)

2.7 (a) The total kinetic energy of the engine is the sum of the kinetic energies of the components T = Tcrsnk + Trod + T piston (a) The crank rotates about a fixed axis at O, thus its kinetic energy is 1 2 Tcrank = I O ω crank (b) 2 where I O = 0.4 kg ⋅ m 2 and ⎛ ⎝

ω crank = ⎜ 250

rev ⎞⎛ 2π rad ⎞⎛ 1 min ⎞ ⎟⎜ ⎟⎜ ⎟ min ⎠⎝ rev ⎠⎝ 60 sec ⎠

r (c) s Substituting these values into Equation (b) leads to 2 1 r⎞ ⎛ Tcrank = (0.5 kg ⋅ m 2 )⎜ 26.2 ⎟ 2 s⎠ ⎝ = 171.3 N ⋅ m (d) The connecting rod does not rotate about a fixed axis. Its kinetic energy is calculated by 1 1 Trod = mv 2 + I ω 2 (e) 2 2 The connecting rod is a slender rod of mass 0.4 kg and length 30 cm. Its centroidal mass moment of inertia is obtained using Table 2.1 as 1 I rod = mL2 12 1 2 = (0.4 kg )(0.3 m ) 12 = 3.0 x10 −3 kg ⋅ m 2 (f) Rigid-body kinematics is used to determine the angular velocity and velocity of the mass center of the connecting rod. The angle made by the connecting rod with the vertical is determined using the law of sines as illustrated below sin(b) sin(10°) = 0.3 m 0.1 m sin(b) = 3 sin(10°) = 0.5209 b = 31.4° (g) Let A be point at which the crank is pinned to the connecting rod. Application of the relative velocity equation leads to = 26.2

33 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2 v A = v O + ω crank k × rA / O r⎞ ⎛ = 0 + ⎜ − 26.2 ⎟k × (0.1 m )[− sin(31.4°)i + cos(31.4°) j] s⎠ ⎝ m = (2.23i + 1.36 j) (h) s The piston, P, is constrained to move in the y direction. Application of the relative velocity equation between A and P on the connecting rod gives v P j = v A + ω rod k × rP / A

m + ω rod k × (0.3 m )[sin(10°)i + cos(10°) j] s = (2.23 - 0.295ω rod )i + (1.36 + 0.0521ω rod ) j Equating the x components of the vectors on both sides of Equation (i) leads to 0 = 2.23 − 0.295ω rod = (2.23i + 1.36 j)

r s Equating the y components of the vectors on both sides of Equation (i) leads to v P = 1.36 + 0.0521ω rod

ω rod = 7.56

(i)

(j)

m (k) s The velocity of the mass center of the connecting rod is calculated using the relative velocity equation on the connecting rod between A and the mass center v rod = v A + ω rod k × rG / A = 1.76

m ⎛ r⎞ + ⎜ 7.56 ⎟k × (0.15 m )[sin(10°)i + cos(10°) j] s ⎝ s⎠ m = (1.18i + 1.56 j) s The magnitude of the velocity of the mass center of the connecting rod is = (2.23i + 1.36 j)

2

(l)

2

m⎞ ⎛ m⎞ ⎛ v rod = ⎜1.18 ⎟ + ⎜1.56 ⎟ s⎠ ⎝ s⎠ ⎝ m = 1.92 s The use of Equations (f), (j), and (m) in Equation (e) gives 2 2 1 m⎞ 1 r⎞ ⎛ −3 2 ⎛ Trod = (0.4 kg )⎜1.92 ⎟ + (3.0 x10 kg ⋅ m )⎜ 7.56 ⎟ 2 s⎠ 2 s⎠ ⎝ ⎝ = 0.823 N ⋅ m The kinetic energy of the piston is

34 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

(m)

(n)

Chapter 2 T piston = =

1 2 mv P 2 1 (0.3 kg )⎛⎜1.76 m ⎞⎟ 2 s⎠ ⎝

2

= 0.465 kg ⋅ m 2 (o) The total kinetic energy of the engine is calculated by substituting Equations (d), (n) and (o) in Equation (a) leading to T = 171.3 N ⋅ m + 0.823 N ⋅ m + 0.465 N ⋅ m = 172.6 N ⋅ m (p) (b) The acceleration of the piston is obtained using the relative acceleration equations.

The crank rotates with a constant angular velocity, thus its angular acceleration is zero.

Application of the relative acceleration equation between O and A on the crank leads to a A = a O + α crank k × rA / O −ω crank2 rA / O 2

r⎞ ⎛ = 0 + 0 − ⎜ 26.2 ⎟ (0.1 m )[− sin(31.4°)i + cos(31.4°) j] s⎠ ⎝ m (q) = (35.7 − 58.5 j) 2 s Application of the relative acceleration equation on the connecting rod between A and P leads to 2 a P j = a A + α rod k × rP / A − ω rod rP / A

m = (35.7i − 58.5 j) 2 + α rod k × (0.3 m )[sin(10°)i + cos(10°) j] s 2

r⎞ ⎛ − ⎜ 7.56 ⎟ (0.3 m )[sin(10°)i + cos(10°) j] s⎠ ⎝ = (32.7 − 0.295α rod )i + (− 75.4 + 0.0521α rod )j Equating x components of the vectors on both sides of Equation (r) leads to 32.7 − 0.295α rod = 0 r s Equating the y components of the vectors on both sides of Equation (r) leads to a P = −75.4 + (0.0521)(110.7 )

(r)

α rod = 110.7 2

(s)

m s2

(t)

= −69.6

2.8 The total kinetic energy of the system is the sum of the kinetic energy of the three system components (a) T = T1 + T2 + T3 where T1 is the kinetic energy of the 5-kg block, T2 is the kinetic energy of the disk and T3 is the kinetic energy of the 10-kg block.

35 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2 (a) At an arbitrary instant the kinetic energy of the 5-kg block is 1 T1 = m1 x& 2 2 1 = (5 kg) x& 2 2 = 2.5 x& 2 (b) The kinetic energy of the disk is 1 T2 = Iθ& 2 (c) 2 where θ& is the angular velocity of the disk. The velocity of the particle on the disk where the cable which connects the 5-kg block is attached must be the same as the velocity of the cable and thus the 5-kg block. Thus (0.1 m )θ& = x& x& θ& = = 10 x& (d) 0.1 Substitution of Equation (d) into Equation (c) leads to 1 2 T2 = (0.2)(10 x& ) 2 = 10 x& 2 (e) Letting y be the upward displacement of the 10-kg block, its kinetic energy is 1 T3 = m3 y& 2 (f) 2 Using the same kinematic arguments as above y& = (0.15 m)θ& = (0.15)(10 x& ) = 1.5 x& (g) Substitution of Equation (g) into Equation (f) leads to 1 2 T3 = (10 kg )(1.5 x& ) 2 = 11.25 x& 2 (h) Thus the total kinetic energy of the system is T = 2.5 x& 2 + 10 x& 2 + 11.25 x& 2

= 23.75x& 2 (b) Substitution of Equation (d) Into Equation (i) leads to 2 T = 23.75 0.1θ& = 0.2375θ& 2

(

(i)

)

2.9 The kinetic energy of a body in pure rotation is 1 T = I Oω 2 2 36 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

(j)

(a)

Chapter 2 The angular velocity of shaft AB of 250 rpm is converted to r/s as rev ⎞⎛ 2π r ⎞⎛ 1 min ⎞ ⎛ ω AB = ⎜ 250 ⎟⎜ ⎟⎜ ⎟ min ⎠⎝ rev ⎠⎝ 60 s ⎠ ⎝ = 26.2 The angular velocities of meshing gears are related by n1ω1 = n 2ω 2

(b)

n1 (c) n2 where n represents the number of teeth of a gear. The angular velocities of the shafts connected by gears to shaft AB are calculated using Equation (c). The angular velocity of the shaft with the 25 tooth gear is r ⎞⎛ 10 ⎞ ⎛ ω1 = ⎜ 26.2 ⎟⎜ ⎟ s ⎠⎝ 25 ⎠ ⎝ r = 10.47 (d) s The angular velocity of the remaining shaft is r ⎞⎛ 30 ⎞ ⎛ ω3 = ⎜ 26.2 ⎟⎜ ⎟ s ⎠⎝ 10 ⎠ ⎝ r = 78.5 (e) s The total kinetic energy of the system is 2 r⎞ 1 ⎛ T = 10 kg ⋅ m 2 + 2.1 kg ⋅ m 2 ⎜10.47 ⎟ s⎠ 2 ⎝

ω 2 = ω1

(

)

r⎞ 1 ⎛ + (1.4 kg ⋅ m 2 + 8 kg ⋅ m 2 + 3.5 kg ⋅ m 2 )⎜ 26.2 ⎟ s⎠ 2 ⎝ +

r⎞ 1 ⎛ ( 1 kg ⋅ m 2 + 6 kg ⋅ m 2 )⎜ 78.5 ⎟ s⎠ 2 ⎝

2

2

= 2.67 x10 4 J 2.10 The parallel combination of springs is replaced by a spring of stiffness N N⎞ N⎞ ⎛ ⎛ k1 = ⎜ 2 x10 5 ⎟ + ⎜1x10 5 ⎟ = 3 x10 5 m m⎠ m⎠ ⎝ ⎝ The two series combinations are replaced by springs of stiffnesses 1 N k2 = = 1.71x10 5 1 1 m + N N 3 x10 5 4 x10 5 m m

37 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

(f)

Chapter 2 k3 =

1 1

+

1

= 3.08 x10 5

N m

N N 8 x10 5 m m The remaining springs are in parallel leading to an equivalent stiffness of k eq = k 2 + k 3 5 x10 5

N⎞ ⎛ N⎞ ⎛ = ⎜1.71x10 5 ⎟ + ⎜ 3.08 x10 5 ⎟ m⎠ ⎝ m⎠ ⎝ N = 4.79 x10 5 m 2.11 Define θ as the clockwise angular displacement of the bar, measured from its equilibrium position. Let k t ,eq be the torsional stiffness of a torsional spring attached to a particle on the bar such that the potential energy of the torsional spring is equal to the potential energy of the system of springs at any instant assuming small θ . The potential energy of the equivalent torsional spring at any instant is 1 V = k t ,eqθ 2 (a) 2 The total potential energy of the existing springs is 1 1 1 2 2 V = (4000 )(3θ ) + (3000 )(2θ ) + (1800 )θ 2 2 2 2 1 = (4.98 x10 4 )θ 2 (b) 2 Equating the potential energies of Equations (a) and (b) leads to N⋅m k t ,eq = 4.98 x10 4 (c) r 2.12 Let θ be the clockwise angular displacement of bar AB and let φ be the counterclockwise angular displacement of bar CD. It is desired to replace the springs by a single spring of an equivalent stiffness located at A as illustrated.

38 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2

The potential energy of this spring, for an arbitrary θ is 1 2 V = k eq (0.1θ ) (a) 2 The potential energy of the springs as illustrated in Figure P2.12 is N⎞ 1⎛ N⎞ 1⎛ 2 2 V = ⎜ 5000 ⎟(0.3θ ) + ⎜ 8000 ⎟(1.0φ ) (b) m⎠ 2⎝ m⎠ 2⎝ The two bars are connected by a rigid massless link. The displacements of the particles at each end of the link are equal, 0.8θ = 0.9φ 8 φ= θ (c) 9 Substitution of Equation (c) into Equation (b) gives 2 1⎡ ⎛8⎞ ⎤ V = ⎢5000(0.3) 2 + 8000⎜ ⎟ ⎥θ 2 2 ⎢⎣ ⎝ 9 ⎠ ⎥⎦ 1 = 1.13x10 4 θ 2 (d) 2 Equating the left-hand sides of Equations (a) and (d) leads to 1.13 x10 4 = .01k eq

(

)

k eq = 1.13 x10 6

N m

(e)

2.13 Two springs in series are illustrated below.

Let x1 represent the change in length of the spring of stiffness k1 and let x 2 represent the change in length of the spring of stiffness k 2 . The total potential energy of the two springs is 1 1 V = k1 x12 + k 2 x 22 (a) 2 2 The displacement of the block, x, is the sum of the changes in length of the springs x = x1 + x2 (b)

39 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2 The force in each spring is the same,

k1 x1 = k 2 x2

(c)

Equation (c) is rearranged as

k1 x1 k2 Substitution of Equation (d) into Equation (b) leads to ⎛ k ⎞ x = ⎜⎜1 + 1 ⎟⎟ x1 ⎝ k2 ⎠ Equation (e) is rearranged resulting in k2 x1 = x k1 + k 2 Equation (f) is used in Equation (f) giving k1 x2 = x k1 + k 2 Substitution of Equations (f) and (g) in Equation (a) leads to x2 =

2

⎞ ⎞ 1 ⎛ k2 1 ⎛ k1 x ⎟⎟ x ⎟⎟ + k 2 ⎜⎜ V = k1 ⎜⎜ 2 ⎝ k1 + k 2 ⎠ 2 ⎝ k1 + k 2 ⎠ =

2 1 ⎡ k1 k 22 + k 2 k1 ⎤ 2 ⎥x ⎢ 2 ⎣ (k1 + k 2 )2 ⎦

=

1 ⎡ k1 k 2 (k1 + k 2 ) ⎤ 2 ⎢ ⎥x 2 ⎣ ( k1 + k 2 ) 2 ⎦

(d)

(e)

(f)

(g) 2

1 ⎛ k1 k 2 ⎞ 2 ⎟x ⎜ 2 ⎜⎝ k1 + k 2 ⎟⎠ The equivalent stiffness is obtained from Equation (h) as =

keq =

k1k2 k1 + k2

(h)

(i)

2.14 The spring of stiffness k1 is in series with the parallel combination of the spring of stiffness k 2 and the viscous damper. The force in each series component is the same as F, as illustrated below

40 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2

Let x1 represent the change in length of the spring of stiffness k1 and let x 2 represent the change in length of the parallel combination. Thus F = k1 x1 (a) and F = k 2 x2 + cx& 2 (b) The change in length of series components is the sum of the changes in lengths of the individual components x = x1 + x 2 (c) Equating the force in Equations (a) and (b) leads to k1 x1 = k 2 x 2 + cx& 2 (d) Solving Equation (c) for x 2 and substituting into Equation (d) leads to k1 x1 = k 2 ( x − x1 ) + c( x& − x&1 ) (k1 + k 2 ) x1 + cx&1 = k 2 x + cx& (e) F Noting from Equation (a) that x1 = , Equation (e) is rewritten as k1 & (k1 + k 2 ) F + c F = k 2 x + cx& k1 k1 ⎛ k2 ⎞ c ⎜⎜1 + ⎟⎟ F + F& = k 2 x + cx& k1 ⎠ k1 ⎝ 2.15 A free-body diagram of the disk is illustrated below.

Application of the rigid-body force equation to the disk gives 41 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

(f)

Chapter 2

∑ F = ma − F + P = ma Summing moments about the mass center of the disk leads to ∑ M G = Iα Fr =

1 2 mr α 2

(a)

(b)

If the disk rolls without slip then

a = rα (c) Substituting Equations (b) and (c) into Equation (a) leads to 1 − mrα + P = mrα 2 3 mrα = P 2 2P α= (d) 3mr Substitution of Equation (d) into Equation (b) gives 1 ⎛ 2P ⎞ F = mr ⎜ ⎟ 2 ⎝ 3mr ⎠ 2 (e) = P 3 (a) For the assumption of no slip to be valid, F < µmg . Thus the criterion for no slip is 2 P < µmg 3 3 P < µmg (f) 2 Substitution of given values into Equation (f) leads to the largest value of P such that the disk rolls without slip as 3 m⎞ ⎛ Pmax = (0.25)(1.8 kg)⎜ 9.81 2 ⎟ 2 s ⎠ ⎝ = 4.41 N (g) (b) The angular acceleration of the disk is calculated using Equation (d) as 2P α= 3mr 2(4.41 N) = 3(1.8 kg)(0.2 m)

r = 8.18 2 s

42 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

(h)

Chapter 2 2.16 Let θ be the counterclockwise angular rotation of the disk measured from the system’s equilibrium position. The work done by a viscous damper of damping coefficient ceq attached to the block of mass m A is W = − c rθ& d (rθ )

∫ ( ) = − ∫ (c r )θ&dθ eq

2

eq

(a)

It is shown in Example 2.4 that the two viscous dampers in series can be replaced by a single viscous damper of damping coefficient cc (b) c4 = 1 2 c1 + c 2 The work done by this equivalent viscous damper as the disk rotates through an angle θ is ⎛3 ⎞ ⎛3 ⎞ W = − ∫ c 4 ⎜ rθ& ⎟d ⎜ rθ ⎟ ⎝2 ⎠ ⎝2 ⎠ ⎛9 ⎞ (c) = − ∫ ⎜ c 4 r 2 ⎟θ&dθ ⎝4 ⎠ The work done by the viscous damper of damping coefficient c3 as the disk rotates through an angle θ is ⎛3 ⎞ ⎛3 ⎞ W = − ∫ c3 ⎜ rθ& ⎟d ⎜ rθ ⎟ ⎝2 ⎠ ⎝2 ⎠ ⎛9 ⎞ = − ∫ ⎜ c3 r 2 ⎟θ&dθ (d) ⎝4 ⎠ The total work done by the attached viscous dampers is obtained by adding Equations (c) and (d) ⎡9 ⎛ cc ⎞ ⎤ W = − ∫ ⎢ ⎜⎜ c3 + 1 2 ⎟⎟r 2 ⎥θ&dθ (e) c1 + c 2 ⎠ ⎦ ⎣4 ⎝ The equivalent viscous damping coefficient is obtained by requiring that the work done by the equivalent viscous damper, Equation (a) is equal to the work done by that attached viscous dampers, Equation (e), for all θ and θ& . Thus cc ⎞ 9⎛ (f) ceq = ⎜⎜ c3 + 1 2 ⎟⎟ 4⎝ c1 + c 2 ⎠ 2.17 A free-body diagram of the block is illustrated below.

43 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2 (a) In order to initiate motion the applied force P must be greater than the friction force developed using the static coefficient of friction, P > µ s mg m⎞ ⎛ = (0.30)(45 kg)⎜ 9.81 2 ⎟ s ⎠ ⎝ = 132.4 N When motion occurs the friction force is F = µ k mg Application of Newton’s second law to the free-body diagram leads to P = F = ma The acceleration of the particle is obtained from Equation (c) as 1 a = (P − µ k mg ) m 1 ⎡ m ⎞⎤ ⎛ = 132.4 N − (0.27)(45 kg)⎜ 9.81 2 ⎟⎥ ⎢ 45 kg ⎣ s ⎠⎦ ⎝ = 0.294

m s2

(a) (b) (c)

(d)

2.18 The solution follows that of Example 2.6, but with n1 =3 (a) n2 (a) The power delivered to the system is the same as calculated in Example 2.6 P = 209.4 kW (b) (b) Use of Equation (c) of Example 2.6 with ω = 1000 rpm gives ω CD = 3000 rpm . Then from Equation (g) of Example 2.6 TCD =

ω AB T AB ω CD

⎛ 1000 rpm ⎞ ⎟⎟(2000 N ⋅ m ) = ⎜⎜ ⎝ 3000 rpm ⎠ = 666.7 N ⋅ m 2.19 The force of Figure P2.19 has the following superposition

44 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

(c)

Chapter 2

The mathematical representation of the force is F (t ) = 20tu (t ) − 20tu (t − 1) + ( 40 − 20t )u (t − 1) − ( 40 − 20t )u (t − 2) = 20tu (t ) + ( 40 − 40t )u (t − 1) − ( 40 − 20t )u (t − 2) (a) Assuming that F (t ) is the only force applied to the particle, application of Newton’s second law to the particle leads to Noting that a = t

dv the velocity of the particle is obtained by integrating Equation (b) dt t

dv ∫0 dt dt = ∫0 [20tu (t ) + (40 − 40t )u(t − 1) − (40 − 20t )u(t − 2)]dt Noting that t

(b)

t

∫ g (t )u(t − a)dt = u(t − a)∫ g (t )dt 0

(c)

a

the integrals in Equation (b) are evaluated leading to

[

]

[

t

]

t

v(t ) − v(0) = 10t 2 u (t ) + u (t − 1) 40t − 20t 2 t =1 − u (t − 2) 40t − 10t 2 t = 2 v(t ) = 2.5 + 10t 2 u (t ) + ( 40t − 20t 2 − 20)u (t − 1) − ( 40t − 10t 2 − 40)u (t − 2) (a) Evaluation of Equation (a) for t=0.5 s gives v(0.5) = 2.5 + 10(0.5) 2

= 5.0

m s

(e)

(b) Evaluation of Equation (d) at t=2.5 s gives v(2.5) = 2.5 + 10(2.5) 2 + 40(2.5) − 20(2.5) 2 − 20 − 40(2.5) − 10(2.5) 2 − 40

[

= 22.5

(d)

m s

2.20 The superposition of the displacement over one period is illustrated below

45 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

] (f)

Chapter 2

(a) The mathematical representation of the displacement over one period is x(t ) = 0.02tu (t ) − 0.02tu (t − 0.1) + 0.002u (t − 0.1) − 0.002u (t − 0.3) + (0.008 − 0.02t )u (t − 0.3) − (0.008 − 0.2t )u (t − 0.4) = 0.02tu (t ) + (0.002 − 0.02t )u (t − 0.1) + (0.006 − 0.02t )u (t − 0.3) − (0.008 − 0.02t )u (t − 0.4) (b) The velocity over one period is obtained as

(a)

dx dt = 0.02 + 0.02tδ (t ) − 0.02u (t − 0.1) + (0.002 − 0.02t )δ (t − 0.1)

v=

− 0.02u (t − 0.3) + (0.006 − 0.02t )δ (t − 0.3) + 0.02u (t − 0.4) (b) − (0.008 − 0.02t)δ (t − 0.4) Note that the terms multiplying the unit impulse functions are zero at the times where the impulses are applied and thus these terms are zero. Thus the velocity reduces to v (t ) = 0.02 − 0.02u (t − 0.1) − 0.02u (t − 0.3) + 0.02u (t − 0.4) (c) 2.21 A free-body diagram of the system at an arbitrary instant is shown below. It is assumed the block slides on a frictionless surface.

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Chapter 2

Application of Newton’s second law to the free-body diagram leads to ∑ F = ma − 1x10 5 x − 4 x10 5 x − 2000 x& = 250 &x& Equation (a) is rearranged leading to the second-order differential equation 250 &x& + 2000 &x& + 5 x10 5 x = 0

(a)

(b)

2.22 Let x be the distance traveled by the vehicle along the incline after the driver spots the stalled car. Thus from the definition of x and the information given appropriate initial conditions are (a1) x(0) = 0 m x& (0) = 60 (a2) s A free-body diagram of the vehicle, drawn at an arbitrary instant after the braking force is applied, is illustrated below.

Application of Newton’s second law to the free-body diagram of the vehicle in the direction of motion leads to ∑ Fx = ma x

− FB + mg sin θ = m&x& Rearranging Equation (b) and substituting given values leads to F &x& = − B + g sin θ m F m⎞ ⎛ = − B + ⎜ 9.81 2 ⎟ sin(8°) 2000 ⎝ s ⎠ −4 = −5 x10 FB + 1.365

47 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

(b)

(c)

Chapter 2 Since the braking force is constant, Equation (c) shows that the acceleration is a constant.

Integration of Equation (c) leads to the time-dependent velocity as x& = (1.365 − 5 x10 −4 FB )t + C1 (d) The constant of integration is evaluated through application of Equation (a2) leading to m C1 = 60 . Integration of Equation (d) then gives s 1 x = (1.365 − 5 x10 − 4 FB )t 2 + 60t + C 2 (e) 2 Application of Equation (a1) leads to C 2 = 0 . Thus 1 x(t ) = 1.365 − 5 x10 − 4 FB t 2 + 60t (f) 2 x& (t ) = 1.365 − 5 x10 − 4 FB t + 60 (g) The vehicle stops when its velocity is equal to zero. Thus the minimum braking force required is the value of FB such that x& = 0 when x = 300 m . Define a = 1.365 − 5 x10 −4 FB . The time required for the vehicle to stop is obtained from Equation (b) as 60 ts = − (h) a Requiring x(t s ) = 300 , using Equation (h) in Equation (f) gives

(

)

(

)

2

1 ⎛ 60 ⎞ ⎛ 60 ⎞ a⎜ − ⎟ + 60⎜ − ⎟ 2 ⎝ a ⎠ ⎝ a ⎠ 1 3600 =− 2 a

300 =

Equation (i) is solved giving a = −6

(i)

m . Thus the minimum braking force is determined s2

from

− 6 = 1.365 − 5 x10 −4 FB FB = 1.47 x10 4 N 2.23 The free-body diagram of the bolt is illustrated below

48 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

(j)

Chapter 2 As it turns the bolt rotates about a fixed axis through its center. Application of the moment equation to the free-body diagram of the bolt, assuming positive moments counterclockwise, leads to ∑ M O = I Oα

500 N ⋅ m − 480 N ⋅ m = (0.035 kg ⋅ m 2 )α r α = 571.4 2 (a) s The angular acceleration is constant. The angular velocity of the bolt at an arbitrary time, noting that the angular velocity is zero at t=0, is t t dω dt = ∫0 dt ∫0 αdt

ω = 571.4t

(b)

The angular displacement of the bolt is dθ ∫0 dt dt = ∫0 ωdt t

t

θ = 285.7t 2

(c)

It takes 50 revolutions to remove the bolt, thus

(50 rev)⎛⎜ 2π r ⎞⎟ = 285.7t 2

⎝ rev ⎠ 314.2 r = 285.7t 2 t = 1.05 s

2.24 The projectile motion equations obtained in Example 2.11 are x = (v0 cosθ )t

(d)

(a)

1 y = − gt 2 + (v0 sin θ )t (b) 2 (a) The projectile lands when y = −100 m . Substituting given values into Equation (b) leads to 1 − 100 = − (9.81)t 2 + 800 sin(30°)t 2 2 4.90t − 400t − 100 = 0 (c) The quadratic formula is applied to solve Equation (c) for the time 400 ± (−400) 2 − 4(4.90)(−100) 2(4.90) = −0.249, 81.9 (d) The negative root is discarded implying that the particle impacts the ground at t=81.9 s. (a) The horizontal range of the projectile is the value of x at the time the particle impacts the ground. From Equation (a) t=

49 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2 m⎞ ⎛ x(81.9) = ⎜ 800 ⎟ cos(30°)(81.9 s) s⎠ ⎝ = 5.67 x10 4 m (e) (b) The projectile achieves its maximum altitude when the y component of velocity is zero, 0 = − gt + v0 sin θ t=

v0 sin θ g

m⎞ ⎛ ⎜ 800 ⎟ sin(30°) s⎠ =⎝ m 9.81 2 s t = 40.77 s The maximum altitude is the value of y at 40.77 s. From Equation (b) m⎞ 1⎛ m⎞ ⎛ 2 y (40.77) = − ⎜ 9.81 2 ⎟(40.77 s ) + ⎜ 800 ⎟ sin(30°)(40.77 s) s⎠ 2⎝ s ⎠ ⎝ 3 = 8.15 x10 m

(f)

(g)

2.25 A free-body diagram of the projectile at an arbitrary instant shows the gravity force and the drag force.

(a) Summing forces in the x direction leads to ∑ Fx = ma x

− 2 N = (0.2 kg )&x&

m (a) s2 Assuming the projectile is fired at the origin of the coordinate system the initial conditions for determining x(t) are x(0) = 0 (b)

&x& = −10

m⎞ ⎛ x& (0) = ⎜ 600 ⎟ cos(40°) s⎠ ⎝ m = 459.6 s Integrating Equation (a) with respect to time leads to 50 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

(c)

Chapter 2

x& (t ) = −10t + C1 (d) Application of Equation (c) to Equation (d) leads to C1 = 459.6 and x& = −10t + 459.6 (e) Integration of Equation (e) with respect to time and application of Equation (b) leads to x(t ) = −5t 2 + 459.6t (f) Summing forces in the y direction leads to ∑ Fy = ma y − mg = m&y& &y& = 0 The initial conditions which are applied to determine y(t) are y (0) = 0

(g) (h)

m⎞ ⎛ y& (0) = ⎜ 600 ⎟ sin(40°) s⎠ ⎝ m = 385.7 (i) s Integration of Equation (g) with application of Equation (i) leads to y& (t ) = −9.81t + 385.7 (j) Integration of Equation (j) with application of Equation (h) leads to y (t ) = −4.9t 2 + 385.7t (k) (b) The maximum altitude occurs when y& = 0 which from Equation (j) occurs at t = 39.3 s . The maximum altitude is obtained from Equation (k) as y (39.3) = −4.9(31.3) 2 + 385.7(31.3)

= 7.59 x103 m (l) The time at which the projectile lands is when y=0, which is obtained using Equation (k) as t = 78.6 s . The range of the projectile is obtained from Equation (f) as x(78.6) = −5(78.6) 2 + 459.6 = 5.24 x103 m 2.26 A free-body diagram of the disk, drawn at an arbitrary instant, is shown below

51 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2

Application of Newton’s second law to the free-body diagram gives − F − kx − cx& = m&x&

(a)

1 2 mr and that 2 since the disk rolls without slip its angular acceleration is related to the acceleration of the mass center of the disk by &x& = rα , summation of moments about the mass center of the disk leads to ∑ M G = Iα

Noting that the centroidal mass moment of inertia of the thin disk is I =

⎛1 ⎞⎛ &x& ⎞ M (t ) + Fr = ⎜ mr 2 ⎟⎜ ⎟ ⎝2 ⎠⎝ r ⎠ Elimination of F between Equations (a) and (b) leads to 3 mr&x& + crx& + krx = M (t ) 2

(b)

(c)

2.27 The free-body diagram of the system at an arbitrary instant is shown below.

Application of Newton’s second law in the vertical direction to the free-body diagram leads to ∑ F = ma

− kx + k ( y − x) + c( y& − x& ) = m&x& Rearrangement of Equation (a) leads to m&x& + cx& + 2kx = cy& + ky

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(a) (b)

Chapter 2 2.28 Let y(t) represent the displacement of the center of the disk, measured downward from the system’s equilibrium position. Let θ represent the clockwise angular rotation of the disk, measured from the system’s equilibrium position

When the center of the disk has moved distance of y downward, the cable has decreased in length by 2y. When the block has moved downward a distance x, the cable has increased in length by x. Since the cable is inextensible, x − 2y = 0 x y= (a) 2 It is further assumed that there is no slip between the disk and the cable, thus x − y = rθ

x = rθ 2 x θ= 2r x−

(b)

Free-body diagrams of the disk and the block, drawn at an arbitrary instant are shown below.

53 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2

Application of Newton’s second law to the free-body diagram of the block leads to ∑ F = ma − T2 − kx = m&x& T2 = −m&x& − kx Application of Newton’s second law to the free-body diagram of the disk leads to − ky + T1 + T2 = m&y& Application of the moment equation to the free-body diagram of the disk gives ∑ M G = Iα T r − T r = Iθ&& 2

1

Substitution of Equations (b) and (c) into Equation (e) leads to && (− m&x& − kx )r − T1r = I ⎛⎜ x ⎞⎟ ⎝ 2r ⎠ I ⎞ ⎛ T1 = −⎜ m + 2 ⎟ &x& − kx 2r ⎠ ⎝ Substitution of Equations (a),(c), and (f) into Equation (d) gives I ⎞ ⎛ x⎞ ⎛ ⎛ &x& ⎞ − k ⎜ ⎟ − ⎜ m + 2 ⎟ &x& − kx − m&x& − kx = m⎜ ⎟ 2r ⎠ ⎝2⎠ ⎝ ⎝2⎠ I ⎞ 5 ⎛5 ⎜ m + 2 ⎟ &x& + kx = 0 2r ⎠ 2 ⎝2

(c )

(d )

( e)

(f)

(g)

2.29 A free-body diagram of the system at an arbitrary instant is shown below. Assuming small θ the system is linear and the static spring forces cancel with the gravity forces and neither are included on the free-body diagrams.

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Chapter 2

The bar rotates about a fixed axis through O, thus the moment equation may be applied as (a) ∑ M O = I Oα with clockwise moments taken as positive. Application of Equation (a) leads to − 2k (3lθ )(3l) − klθ (l) = I Oθ&& I θ&& + 19kl 2θ = 0

(b) The mass moment of inertia of abut O is the sum of the moments of inertia of each of the particles. The moment of inertia of a particle about an axis through the particle is zero.

The moment of inertia of each particle about O is then obtained using the parallel axis theorem. The moment of inertia of the particle of mass 2m about O is (c) I P1 = 2m(3l) 2 = 18ml 2 The moment of inertia of the particle of mass m about an axis through O is I p 2 = ml 2 (d) O

Thus

I O = I p1 + I pw

= 19ml 2 Substitution of Equation (e) into Equation (b) leads to 19ml 2θ&& + 19kl 2θ = 0 mθ&& + kθ = 0

(e)

(f)

2.30 A free-body diagram of the system at an arbitrary instant is shown below. Assuming small θ the system is linear and the static spring forces cancel with the gravity forces and neither are included on the free-body diagrams.

The bar rotates about a fixed axis through O, thus the moment equation may be applied as (a) ∑ M O = I Oα 55 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2 with clockwise moments taken as positive. Application of Equation (a) leads to − 2k (3lθ )(3l) − klθ (l) = I Oθ&& I θ&& + 19kl 2θ = 0

(b) The moment of inertia of the assembly about O is the sum of the moments of inertia of the two particles and the moment of inertia of the bar, I O = I Ob + I Op1 + I Op 2 (c) The moment of inertia of the bar about its cenroidal axis is obtained from Table 2.1 as 1 I b = m ( 4l ) 2 12 4 = ml 2 (d) 3 The parallel axis theorem is used to determine the moment of inertia of the bar about O, I Ob = I b + md 2 O

4 2 ml + ml 2 3 7 = ml 2 3 =

(e)

The moment of inertia of a particle about an axis through the particle is zero. The moment of inertia of each particle about O is then obtained using the parallel axis theorem. The moment of inertia of the particle of mass 2m about O is I OP1 = 2m(3l) 2 = 18ml 2 (f) The moment of inertia of the particle of mass m about an axis through O is I Op 2 = ml 2 (g) Substitution of Equations (e), (f), and (g) into Equation (c) gives 7 I O = ml 2 + 18ml 2 + ml 2 3 64 2 = ml 3 Use of Equation (h) in Equation (b) leads to 64 2 && ml θ + 19kl 2θ = 0 3

(h)

(i)

2.31 For the system shown in Figure P2.31, let θ be the counterclockwise angular rotation of the disk, measured from the system’s equilibrium position, and let y be the displacement of the particle on the outer radius of the disk. Kinematics is used to give x x = rθ ⇒ θ = (a) r ⎛ x⎞ y = 3rθ = 3r ⎜ ⎟ = 3x (b) ⎝r⎠

56 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2 Free-body diagrams of the disk and the block, drawn at an arbitrary instant are shown below

Application of Newton’s second law to the free-body diagram of the block leads to ∑ F = ma

− T − 2kx = m&x& (c) Application of the moment equation, assuming moments are positive counterclockwise, to the free-body diagram of the disk gives ∑ M G = Iα ⎛ &x& ⎞ Tr − 3kx(3r ) = I ⎜ ⎟ ⎝r⎠ Elimination of T between Equations (c) and (d) leads to I⎞ ⎛ ⎜ mr + ⎟ &x& + 11krx = 0 r⎠ ⎝

(d)

(e)

2.32 Define θ as the clockwise angular rotation of the upper bar measured from the system’s equilibrium position and define φ as the counterclockwise rotation of the lower bar, measured from its equilibrium position. The link connecting the two bars is rigid.

Thus the displacements of the particles on each bar where the link is attached must be the same. Assuming small displacements this leads to L 2L θ= φ 2 3 3 φ= θ (a) 4 Free-body diagrams of each bar, drawn at an arbitrary instant, are shown below.

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Chapter 2

1 mL2 , 12 application of the moment equation, assuming positive moments clockwise, to the upper bar gives ∑ M A = Iα

Noting that the centroidal mass moment of inertia of a slender bar is I =

⎛ L :⎞ 1 ⎛ L ⎞⎛ L ⎞ (b) − ⎜ k θ ⎟⎜ ⎟ + F ⎜ ⎟ = mL2θ&& ⎝ 2 ⎠ 12 ⎝ 2 ⎠⎝ 2 ⎠ The parallel-axis theorem is used to determine the moment of inertia of the lower bar 2 1 1 ⎛L⎞ 2 about its pin support at B as I B = mL + m⎜ ⎟ = mL2 . Application of the moment 12 3 ⎝2⎠ equation to the lower bar, assuming positive moments are clockwise, gives ∑ M B = I Bα ⎛3 ⎞ ⎛ 2L ⎞ 1 ⎛3 ⎞ − ⎜ cLθ& ⎟(L ) − F ⎜ ⎟ = mL2 ⎜ θ&&⎟ ⎝4 ⎠ ⎝ 3 ⎠ 3 ⎝4 ⎠ Equation (a) is used to solve for the force in the rigid link as 1 1 F = mL2θ&& + kLθ 6 2 Substitution of Equation (c) into Equation (d) leads to 1 3 ⎛1 ⎞⎛ 2 L ⎞ 1 − ⎜ mL2θ&& + kLθ ⎟⎜ ⎟ = mL2θ&& + cL2θ& 2 4 ⎝6 ⎠⎝ 3 ⎠ 4 Equation (e) is rearranged to give 13 2 && 3 2 & 1 2 mL θ + cL θ + kL θ = 0 36 4 3

58 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

(c)

(d)

(e)

(f)

Chapter 2 2.33 The free-body diagram of the bar at an arbitrary instant follows

Since the pin support is a fixed axis of rotation the appropriate form of the moment equation is (a) ∑ M O = I Oα where moments are taken positive counterclockwise to be consistent with the chosen positive direction for θ . Application of Equation (a) to the free-body diagram, assuming small θ , leads to L − k tθ − mg θ = I Oθ&& (b) 2 The moment of inertia of a slender bar about its centroidal axis is obtained from Table 2.1 as 1 I = mL2 (c) 12 Application of the parallel axis theorem leads to 2

1 ⎛L⎞ mL2 + m⎜ ⎟ 12 ⎝2⎠ 1 = mL2 3 Use of Equation (d) in Equation (b) and subsequent rearrangement leads to 1 2 && ⎛ L⎞ mL θ + ⎜ k t + mg ⎟θ = 0 3 2⎠ ⎝ IO =

(d)

(e)

2.34 Free-body diagrams of the blocks, drawn at an arbitrary instant, are illustrated below

59 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2 Application of Newton’s second law to the left block gives ∑ F = ma

− 3000 x1 − 200 x&1 + 1000( x 2 − x1 ) = 30 &x&1 Application of Newton’s second law to the right block gives − 1000( x2 − x1 ) − 4000x2 = 20&x&2 Equations (a) and (b) are rearranged and written in a matrix form as ⎡30 0 ⎤ ⎡ &x&1 ⎤ ⎡ 4000 − 1000⎤ ⎡ x1 ⎤ ⎡0⎤ ⎢ 0 20⎥ ⎢ &x& ⎥ + ⎢− 1000 5000 ⎥ ⎢ x ⎥ = ⎢0⎥ ⎣ ⎦⎣ 2 ⎦ ⎣ ⎦⎣ 2 ⎦ ⎣ ⎦

(a)

(b) (c)

2.35 Free-body diagrams of the bar and the block, drawn at an arbitrary instant are shown below assuming small θ . Static spring forces and gravity cancel are not included as they will cancel with each other in the governing equations.

Summing forces on the free-body diagram of the block assuming positive forces act downward leads to ∑ F = ma − k ( x − Lθ ) = m&x& m&x& + kx − kLθ = 0 (a) Since the bar rotates about a fixed axis through the pin support the appropriate moment equation to be applied to the free-body diagram of the bar is (b) ∑ M O = I Oα The moment of inertia of the slender bar about its own centroidal axis is obtained from 1 Table 2.1 as I = mL2 . Application of the parallel axis theorem gives 12 2 1 ⎛L⎞ 2 I O = mL + m⎜ ⎟ 12 ⎝2⎠ 1 = mL2 (c) 3 60 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2 Application of Equation (b) to the free-body diagram of the bar, taking clockwise moments as positive, leads to 1 − kLθ ( L) + k ( x − Lθ )( L) = mL2θ&& 3 1 2 && mL θ − kLx + 2kL2θ = 0 (d) 3 Equations (a) and (d) provide the mathematical model for the system.

2.36 Free-body diagrams of each of the blocks at an arbitrary instant are illustrated.

Application of Newton’s second law to each of the blocks, assuming forces are positive to the right, leads to − kx1 + k ( x 2 − x1 ) = m&x&1 (a) − k ( x 2 − x1 ) + 2k ( x3 − x 2 ) + c( x& 3 − x& 2 ) = 2m&x&2 (b) − 2k ( x3 − x 2 ) − c( x& 3 − x& 2 ) + F (t ) = m&x&3 Equations (a), (b), and (c) are rearranged as m&x&1 + 2kx1 − kx2 = 0 2m&x&2 + cx& 2 − cx& 3 − kx1 + 2kx2 − kx3 = 0 m&x&3 − cx& 2 + cx& 3 − kx2 + kx3 = F (t ) Equations (d), (e), and (f) are summarized in matrix form as 0 ⎤ ⎡ x&1 ⎤ ⎡ 2k − k 0 ⎤ ⎡ x1 ⎤ ⎡ 0 ⎤ ⎡m 0 0 ⎤ ⎡ &x&1 ⎤ ⎡0 0 ⎢ 0 2m 0 ⎥ ⎢ &x& ⎥ + ⎢0 c − c ⎥ ⎢ x& ⎥ + ⎢− k 2k − k ⎥ ⎢ x ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ 2 ⎥ ⎢ ⎥⎢ 2 ⎥ ⎢ ⎥⎢ 2 ⎥ ⎢ ⎥ ⎢⎣ 0 0 m ⎥⎦ ⎢⎣ &x&3 ⎥⎦ ⎢⎣0 − c c ⎥⎦ ⎢⎣ x& 3 ⎥⎦ ⎢⎣ 0 − k k ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣ F (t )⎥⎦

(c)

(d) (e) (f)

(g)

2.37 Free-body diagrams of the link and the block are illustrated assuming small x.

Since the link is massless, application of the moment equation to its free-body diagram leads to

61 Copyright © 2008 Nelson Thomson Learning. All rights reserved.

Chapter 2

∑M = 0 O

F (t )l + G (t )(3l ) = 0

1 G (t ) = − F (t ) 3 Application of Newton’s second law to the free-body diagram of the block gives G (t ) − kx − cx& = m&x& 1 m&x& + cx& + kx = − F (t ) 3

(a)

(b)

2.38 Assuming small displacements and using similar triangles, as illustrated below, the displacement of the end of the spring attached to the bar of length 3l is 2 y (t ) such that the spring is stretched.

A free-body diagram of the bar at an arbitrary instant, assuming small θ , is illustrated.

Note that the since θ is assumed positive clockwise rotation of the bar in the positive direction of θ leads to an increase in length of the spring. The bar rotates about a fixed axis at O, thus the appropriate form of the moment equation is (a) ∑ M O = I Oα The moment of inertia of the slender bar about its own centroidal axis is obtained from 1 Table 2.1 as I = ml 2 . Application of the parallel axis theorem gives 12 2

1 ⎛l⎞ ml 2 + m⎜ ⎟ 12 ⎝2⎠ 1 = ml 2 (b) 3 Application of Equation (a) to the free-body diagram of the bar, taking clockwise moments as positive, leads to 1 l − k (2 y + lθ )l − clθ&(l) + mg θ = ml 2θ&& 2 3 1 2 && l⎞ ⎛ ml θ + cl 2θ& + ⎜ kl 2θ − mg ⎟θ = −2kl 2 y (t ) (c) 3 2⎠ ⎝ IO =

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Chapter 2 2.39 Define θ (t ) as the clockwise angular displacement of the crank from the aligned position, as illustrated below.

Define x p (t ) as the distance between the piston and the center of the crank at any instant.

From the above diagram

x p = 0.3 sin θ + 0.1 cos θ

The velocity of the piston is calculated as x& p = 0.3θ& cosθ − 0.1θ& sin θ Noting that θ& = ω , which is a constant, differentiation of Equation (b) leads to &x& = −0.3θ& 2 sin θ − 0.1θ& 2 cosθ p

(a) Since the angular velocity is constant and the system is aligned at t=0, θ = ωt Use of Equation (d) in Equations (a), (b), and (c) leads to x p = 0.3 sin(ωt ) + 0.1 cos(ωt )

v p = 0.3ω cos(ωt ) − 0.1ω sin(ωt )

(a)

(b) (c ) (d )

( e) (f)

a p = −0.3ω 2 sin(ωt ) − 0.1ω 2 cos(ωt ) (g) (b)Assume the inertia effects of the crank and connecting rod are negligible. A free-body diagram of the engine, drawn at an arbitrary instant, is illustrated below

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Chapter 2

Application of Newton’s second law, in the form of D’Alembert’s principle to the freebody diagram leads to ∑ Fext = ∑ Feff − 9000 x − 1000 x& = 10 &x& + 0.3( &x& + &x&p ) 10.3&x& + 1000 x& + 9000 x = 0.3&x&p Use of Equation (g) in Equation (h) leads to 10.3&x& + 1000 x& + 9000 x = 0.3(− 0.3ω 2 sin(ωt ) − 0.1ω 2 cos(ωt ) ) Noting that ω = 250 rpm = 26.2 r / s , Equation (i) becomes 10.3 &x& + 1000 x& + 9000 x = −61.8 sin(ωt ) − 20.6 cos(ωt )

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( h) (i ) ( j)