A 2.5 kg solid cylinder (radius = 0.10 m , length = 0.70 m ) is released from re

Question

A 2.5 kg solid cylinder (radius = 0.10 m , length = 0.70 m ) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.90 m high and 5.0 m long.

Part A When the cylinder reaches the bottom of the ramp, what is its total kinetic energy? (Express your answer using two significant figures.)

(answer in J)

Part B When the cylinder reaches the bottom of the ramp, what is its rotational kinetic energy?(Express your answer using two significant figures.)

(answer in J)

Part C When the cylinder reaches the bottom of the ramp, what is its translational kinetic energy?(Express your answer using two significant figures.)

(answer in J)

Explanation / Answer

m = 2.5 Kg

r = 0.10 m

part A) for the total kinetic energy

total kinetic energy = decrease in potential energy

total kinetic energy = 2.5 * 9.8 * 0.90

total kinetic energy = 22.1 J

part B) Now , for the rotational kinetic energy

rotational kinetic energy = total kinetic energy * (I/r^2)/(m + I/r^2)

rotational kinetic energy = 22.1 * (0.50 * m )/(m + 0.50 m)

rotational kinetic energy = 7.4 J

part C)

translational kinetic energy = 22.1 - 7.35

translational kinetic energy = 15 J

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