Question

A 2.4 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.

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

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

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

Answer 1

note : in previous attempt i mistakenly took the object as solid sphere instead of solid cylinder, my bad plz review

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a)

using conservation of energy

total kE = mgh = 2.4* 9.8* 0.9 = 21 J

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b) translational kE = 0.5 m v^2

rotational kE = 0.5 I w^2 = 0.5* (0.5) m r^2 (v^2/r^2)

rotational kE = (0.5) 0.5 m v^2

kEr = 0.5* kEt

now

total kE = kEr + kEt

21.168 = kE r + kEr / 0.5

kE r = 7 J

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c)

kEt = 6.048 / 0.5 = 14 J

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