HW6 Solutions

# Athe speed of the box is related to the angular speed

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(a)The speed of the box is related to the angular speed of the wheel by v = R , so that K m v v K m box box box box m/ s 1 2 2 141 2 . implies that the angular speed is = 1.41/0.20 = 7.07 rad/s. Thus, the kinetic energy of rotation is 1 2 2 100 I . J. (b) Since it was released from rest at what we will consider to be the reference position for gravitational potential, then (with SI units understood) energy conservation requires

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0 0 box 0 0 6.0 10.0 . K U K U m g h (b) Since it was released from rest at what we will consider to be the reference position for gravitational potential, then (with SI units understood) energy conservation requires 0 0 box 0 0 6.0 10.0 . K U K U m g h Therefore, h = 16.0/58.8 = 0.27 m. 17. (a) We apply Eq. 10-34: 2 2 2 2 2 2 2 1 1 1 1 2 2 3 6 1 (0.42 kg)(0.75 m) (4.0 rad/s) 0.63 J. 6 K I mL mL (b) Simple conservation of mechanical energy leads to K = mgh . Consequently, the center of mass rises by 2 2 2 2 2 2 2 (0.75 m) (4.0 rad/s) 0.153 m 0.15 m. 6 6 6(9.8 m/s ) K mL L h mg mg g

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