Chapter 7 Homework solutions

Chapter 7 Homework solutions

28. Strategy Use the component form of the definition of center of mass. Solution Find the location of the center of mass.

Find x_CM and y_CM. CM CM A A B B A B A A B B A B (5.0 g)(0) (1.0 g)(25 cm) 4.2 cm 5.0 g 1.0 g (5.0 g)(0) (1.0 g)(0) 0 5.0 g 1.0 g m x m x x m m m y m y y m m + + = = = + + + + = = = + +

The location of the center of mass is (x_CM,y_CM)= (4.2 cm, 0) .

y (cm) x (cm) 0 0 25 25 Α Β CM?

36. Strategy The total momentum of the system is equal to the total mass of the system times the velocity of the center of mass. Let east be in the positive direction.

Solution Find the total momentum.

CM A A B B M m m = = + pr vr vr vr since pr=pr_A+pr_B. Thus, CM A A B B A B . m m m m + = + v v v r r r

Find the velocity of the center of mass.

B A 10 m/s 10 m/s N 15 kg 5.0 kg CM CM (5.0 kg)(10 m s) (15 kg)( 10 m s) 5 m s , so 5 m s west . 5.0 kg 15 kg v = + − = − = + v r

43. Strategy Use conservation of momentum. The collision is perfectly inelastic, so v_1f =v_2f =v_f. Also, the block is initially at rest, so v_2i =0.

Solution Find the speed of the block of wood and the bullet just after the collision. 1 1f 2 2f 1 2 f 1 1i 2 2i 1 1i 2 1 f 1i 1 2 ( ) (0), so 0.050 kg (100.0 m s) 5.0 m s . 0.050 kg 0.95 kg m v m v m m v m v m v m v m m v v m m + = + = + = + = = = + +

48. My Strategy Use the equations for one-dimensional elastic collisions derived in class July 3. Strategy Use conservation of linear momentum. Since the collision is perfectly elastic, kinetic

energy is conserved.

Solution The 100-g ball is (1) and the 300-g ball is (2). Note that m₂=3m₁. 2 1 1i 2 2i 1 1i 2 1 1i 1 1f 2 2f 1i 1f 2f 1f 2f 1 (0) , so m 3 . m v m v m v m m v m v m v v v v v v m + = + = = + = + = + 2 2 2 2 2 2 2 2 2 2 1 1i 2 1 1i 1 1f 2 2f 1 1f 1 2f 1i 1f 2f 1 1 1 1 1 1 1 (0) (3 ) , so 3 . 2m v +2m =2m v = 2m v +2m v =2m v +2 m v v =v + v Substitute for v_1i.

(

)

1f 3 2f 1f 6 1f 2f 9 2f 1f 3 2f , so 61f 2f 6 2f , or 1f 2f.

v + v =v + v v + v =v + v v v = − v v = −v

Find the final velocities of each ball.

1i 1f 2f 2f 2f 2f 2f 1i

1 1

3 3 2 , so (5.00 m s) 2.50 m s.

2 2

v =v + v = −v + v = v v = v = =

Since v_1f = −v_2f, v_1f = −2.50 m s. So, the 300-g ball moves at 2.50 m s in the + -directionx

and the 100-g ball moves at 2.50 m s in the −x-direction .

57. Strategy Use conservation of momentum. Let each of the first two pieces be 45° from the positive x-axis (one CW, one CCW).

Solution Find the speed of the third piece. Find v_3x. 1 2 3 1 2 3 0, so 3 1 2 cos 45 cos( 45 ) 2. 2 2 x x x x x x x x x v v p +p +p =mv +mv +mv = v = −v −v = −v ° −v − ° = − − = −v Similarly,

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91. Strategy Use conservation of momentum and energy. The collision is elastic, so kinetic energy is conserved.

Solution Find the speed of bob B immediately after the collision. Momentum conservation:

Af Bf Ai Bi Ai 0, so Bf Ai Af.

mv +mv =mv +mv =mv + v =v −v Perfectly elastic collision (K_i =K_f):

2 2 2 2 2 2 2 2 Af Bf Ai Bi Ai Af Bf Ai 1 1 1 1 1 0, so . 2mv +2mv =2mv +2mv =2mv + v +v =v Energy conservation: 2 Ai Ai 1 , so 2 . 2mv =mgh v = gh Find v_Af in terms of v_Ai. 2 2 2 2 2 2 2 2 Ai Af Bf Af ( Ai Af) Af Ai 2 Ai Af Af , so Af( Af Ai) 0. v =v +v =v + v −v =v +v − v v +v v v −v = Thus, v_Af =0 or v_Ai. The only way v_Af could equal v_Ai is if bob B didn’t exist, so v_Af =0. Calculate v_Bf.

2

Bf Ai Af 2 0 2(9.80 m s )(5.1 m) 10 m s

Chapter 8 Homework Solutions

2. Strategy The rotational inertia of a solid disk is 1 2

2 .

I= MR Solution Find the rotational inertial of the solid iron disk.

2 2 2

1 1

(49 kg)(0.200 m) 0.98 kg m

2 2

I = MR = = ⋅

5. Strategy Find the rotational inertia in each case by using Eq. (8-2). Solution

(a) I=m r( 2+02+02+r2)=2mr2=2(3.0 kg)(0.50 m)2= 1.5 kg m⋅ 2

(b) I=m(02+ +r2 02+r2)=2mr2=2(3.0 kg)(0.50 m 2)2= 0.75 kg m⋅ 2

(c) I=m r( 2+r2+r2+r2)=4mr2=4(3.0 kg)(0.50 m 2)2= 1.5 kg m⋅ 2

13. Strategy Use Eq. (8-3).

Solution Find the magnitude of the torque.

(40.0 kg)(9.80 N kg)(2.0 m) 780 N m

F r mgr

τ = _⊥ = = = ⋅

19. Strategy Use Eq. (8-3) to find the torque.

Solution Let the axis of rotation be a the hinge of the trap door. Since the door is in equilibrium, the magnitude of the torque exerted on the door by the rope is the same as that exerted by gravity. Compute the torque due to the rope.

2 cos 65.0 2 1.65 m (16.8 kg)(9.80 m s ) cos 65.0 2 57.4 N m rF L mg τ = _⊥ = ° = ° = ⋅ L 65.0° 65.0° mg

25. (a) Strategy Use the work-kinetic energy theorem. Solution Find the work done spinning up the wheel.

2 2 2 f f 2 2 1 1 ( ) 2 2 1

(182 kg)(0.62 m) [(120 rev min)(1 60 min s)(2 rad rev)] 5.5 kJ 2 W K Iω MR ω π = ∆ = = = = 0.62 m ω =f 120 rpm

(b) Strategy Use the equations for rotational motion with constant acceleration and the relationship between work, torque, and angular displacement.

Solution Find the torque.

3 av

av

5.5 10 J

( ), so 29 N m .

(120 rev min)(1 60 min s)(2 rad rev)(30.0 s) 2 W W t t τ θ τ ω τ ω π × = ∆ = ∆ = = = ⋅ ∆

28. Strategy Choose the axis of rotation at the fulcrum. Use Eqs. (8-8). Solution Find F. 0 F(3.0 m) (1200 N)(0.50 m), τ Σ = = − + so (1200 N)(0.50 m) 200 N . 3.0 m F= =

33. Strategy Use Eqs. (8-8).

Solution Choose the axis of rotation at the point of contact between the driveway and the ladder.

w w 0 , so . x F f N f N Σ = = − = w l p 3.0 m 0 (4.7 m) (2.5 m) cos (5.0 m) cos , 4.7 m N W W τ θ   θ Σ = = − −     so w l p cos 15 m (2.5 m) . 4.7 m 4.7 N = θ _W +W _ __     Find θ. 14.7 4.7 m (5.0 m) sin , so sin . 5.0 θ θ − = = Calculate f. 1 4.7 5.0 w cos sin _{15 m} (120 N)(2.5 m) (680 N) 180 N 4.7 m 4.7 f N − _{  } = =  +  =    

So, the force of friction is 180 N toward the wall .

37. Strategy Use Eqs. (8-8). Choose the axis of rotation at the point where the beam meets the store. Solution The tension in the cable cannot exceed 417 N. Sum the

torques.

0 Tsin (1.50 m) (50.0 N)(0.75 m) (200.0 N)(1.00 m)

τ θ

Σ = = − −

Solve for θ and substitute 417 N (the breaking strength) for T.

1(50.0 N)(0.75 m) (200.0 N)(1.00 m)

sin 22.3

(417 N)(1.50 m)

θ ₌ − + _{= °}

The minimum angle is 22.3° .

200.0 N 0.75 m 1.50 m 50.0 N T 1.00 m θ

48. Strategy Use the rotational form of Newton’s second law. Solution Find the frictional torque.

2 0 20.0 rad s (400.0 kg m ) 26.7 N m 300.0 s I I t ω τ α ∆ − Σ = = = ⋅ = − ⋅ ∆

49. Strategy Use the rotational form of Newton’s second law and Eq. (5-21). Solution Find the torque that the motor must deliver.

2 1 2

I = MR for a uniform disk, so

(

)

(

)

2 2 4 4(2.0 rev)(2 rad rev)

MR I MR ω ω ω τ α θ θ π  ₋  Σ = = _ _= = = ⋅ ∆ ∆  