When the cylinder reaches the bottom of the ramp, what is its total kinetic energy?

Question

A 2.7 kg solid cylinder (radius = 0.10 m , length = 0.50 m ) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.80 m high and 5.0 m  long. 

b.) When the cylinder reaches the bottom of the ramp, what is its rotational kinetic energy?

c.)When the cylinder reaches the bottom of the ramp, what is its translational kinetic energy? 

Can some one work this problem out step by step.  Thanks!

Physicsquest · Accepted Answer

First find the velocity.

sq-root((gh)/0.75) = velocity m/s

sq-rt((9.8 x 0.8)/0.75) = 3.233 m/s

b.) When the cylinder reaches the bottom of the ramp, what is its rotational kinetic energy?

ω = v/r = 3.233/0.1 = 32.33 rad/s

Inertia = 1/2mr^2 = 0.5 x 2.7 x 0.1^2 = 0.0135 kg-m^2

Rotational KE = 1/2Iω^2 = 0.5 x 0.0135 x 32.33^2 = 7.0553 J

c.)When the cylinder reaches the bottom of the ramp, what is its translational kinetic energy? 

1/2mv^2 

0.5 x 2.7 x 3.233^2 = 14.1105 J

When the cylinder reaches the bottom of the ramp, what is its total kinetic energy?

0.75mv^2 = mgh

mgh = 2.7 x 9.8 x 0.8 = 21.168 J

Translational KE = 14.1105 J

Rotational = 7.0553 J

Total KE = 21.168 J

When the cylinder reaches the bottom of the ramp, what is its total kinetic energy?

1 Antwort