Chapter 7 Homework solutions

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1 Chapter 7 Homework solutions 8 Strategy Use the component form of the definition of center of mass Solution Find the location of the center of mass Find x and y ma xa + mbxb (50 g)(0) + (10 g)(5 cm) x = = = 4 cm ma + mb 50 g + 10 g ma ya + mb yb (50 g)(0) + (10 g)(0) y = = = 0 m + m 50 g + 10 g A B The location of the center of mass is ( x, y ) = (4 cm, 0) y (cm) Α? Β 5 x (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 p r = Mv r = m Av r A + mbv r B since p r = p r A + p r B 10 m/s Thus, r r r ma va + mbvb v = 50 kg A ma + mb Find the velocity of the center of mass (50 kg)(10 m s) + (15 kg)( 10 m s) v = = 5 m s, so v r = 5 m s west 50 kg + 15 kg 15 kg B 10 m/s N 43 Strategy Use conservation of momentum The collision is perfectly inelastic, so v 1f = v f = v f Also, the block is initially at rest, so v i = 0 Solution Find the speed of the block of wood and the bullet just after the collision m1v 1f + mvf = ( m1 + m ) vf = m1v 1i + mvi = m1v 1i + m (0), so m kg vf = v1i = (1000 m s) = 50 m s m1 + m 0050 kg kg 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 () Note that m = 3 m1 m m1v1i + mvi = m1v1i + m (0) = m1v1i = m1v1f + mvf, so v1i = v1f + vf = v1f + 3 vf m1 1 m1v 1i + 1 m (0) = 1 m1v 1i = 1 m v1f + mvf = m1v 1f + (3 m1 ) vf, so v1i = v1f + 3 vf Substitute for v 1i

2 ( ) v + 3v = v + 6v v + 9v = v + 3 v, so 6v v = 6 v, or v = v 1f f 1f 1f f f 1f f 1f f f 1f f Find the final velocities of each ball 1 1 v1i = v1f + 3vf = vf + 3vf = vf, so vf = v1i = (500 m s) = 50 m s = = So, the 300-g ball moves at 50 m s in the + x -direction Since v1f vf, v1f 50 m s and the 100-g ball moves at 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 3 x p1x + px + p3x = mv1x + mvx + mv3x = 0, so v3x = v1 x vx = v cos 45 v cos( 45 ) = v v = v Similarly, v3 y = v1 y v y = v sin 45 v sin( 45 ) = v v + = 0, so v3 = v3x = v = ( ) = 170 m s 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: mvaf + mvbf = mvai + mvbi = mvai + 0, so vbf = vai vaf Perfectly elastic collision ( Ki = Kf ): 1 mvaf + 1 mvbf = 1 mv 1 1 Ai + mvbi = mvai + 0, so vaf + vbf = vai Energy conservation: 1 mvai mgh, so vai gh = = Find v Af in terms of v Ai v Ai = vaf + vbf = vaf + ( vai vaf ) = vaf + vai vaivaf + vaf, so vaf ( vaf vai ) = 0 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 v Bf = vai vaf = gh 0 = (980 m s )(51 m) = 10 m s

3 Chapter 8 Homework Solutions Strategy The rotational inertia of a solid disk is 1 I = MR Solution Find the rotational inertial of the solid iron disk 1 1 (49 kg)(000 m) 098 kg m I = MR = = 5 Strategy Find the rotational inertia in each case by using Eq (8-) Solution (a) I = m( r r ) = mr = (30 kg)(050 m) = 15 kg m (b) I = m(0 + r r ) = mr = (30 kg)(050 m ) = 075 kg m (c) I = m( r + r + r + r ) = 4mr = 4(30 kg)(050 m ) = 15 kg m 13 Strategy Use Eq (8-3) Solution Find the magnitude of the torque τ = F r = mgr = (400 kg)(980 N kg)(0 m) = 780 N m 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 τ = rf L = mg cos m = (168 kg)(980 m s )cos 650 = 574 N m L mg 5 (a) Strategy Use the work-kinetic energy theorem Solution Find the work done spinning up the wheel 1 1 W = K = Iωf = ( MR ) ωf 1 (18 kg)(06 m) [(10 rev min)(1 60 min s)( rad rev)] = π = 55 kj ω f = 10 rpm 06 m

4 (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 W J W = τ θ = τ ( ωav t), so τ = = = 9 N m ωav t (10 rev min)(1 60 min s)( π rad rev)(300 s) 8 Strategy Choose the axis of rotation at the fulcrum Use Eqs (8-8) Solution Find F Σ τ = 0 = F(30 m) + (100 N)(050 m), so 33 Strategy Use Eqs (8-8) (100 N)(050 m) F = = 30 m 00 N Solution Choose the axis of rotation at the point of contact between the driveway and the ladder Σ Fx = 0 = f Nw, so f = Nw 30 m Σ τ = 0 = Nw(47 m) Wl(5 m) cos θ Wp (50 m) cos θ, so 47 m cos 15 m N θ w = Wl (5 m) + Wp 47 m 47 Find θ m = (50 m)sin θ, so θ = sin 50 Calculate f 1 cossin m f = Nw = (10 N)(5 m) + (680 N) = 180 N 47 m 47 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 Σ τ = 0 = T sin θ (150 m) (500 N)(075 m) (000 N)(100 m) Solve for θ and substitute 417 N (the breaking strength) for T 1 (500 N)(075 m) + (000 N)(100 m) θ = sin = 3 (417 N)(150 m) The minimum angle is m 100 m T 150 m θ 500 N 000 N 48 Strategy Use the rotational form of Newton s second law Solution Find the frictional torque ω 0 00 rad s Σ τ = Iα = I = (4000 kg m ) = 67 N m t 3000 s The torque is 67 N m opposite the flywheel s rotation

5 49 Strategy Use the rotational form of Newton s second law and Eq (5-1) Solution Find the torque that the motor must deliver 1 I = MR for a uniform disk, so 0305 m ( ) ( ) 1 (0 kg) 349 rad s f i MR f I MR ω ω Σ τ = α = = ω = = 0001 N m θ 4 θ 4(0 rev)( π rad rev)

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