A ring 1 (5 kg, r=1 m) and ring 2 (5 kg, r=5 m) are placed beside each other at the top of an inclined ramp. They are both released at the same time and roll without slipping down the ramp. Which object reaches the bottom first?

Question

anonymous · Answer

guys i really need help with this

anonymous · Answer

They will reach the bottom simultaneously.

Kinetic energy at bottom, distance h from top, comes from the change in potential energy, mgh, and is divided into translational and rotational KE:

mgh = (m/2) v^2 + (I/2) omega^2.

Moment of inertia I for hoop is mR^2, where m is mass and R is radius.

Omega, angular velocity, is v/R.


Thus, mgh = (m/2)v^2 +(m/2)R^2 (v/R)^2

R cancels out and mgh= m v^2,

or v= sqrt(gh) 

rather than sqrt(2gh) that holds if there were only sliding and no rotation.

anonymous · Answer

thank you so much, great explanation :)

alekos · Answer

i agree. They should reach the bottom at the same time because the speed is not dependent on the radius or mass. Only depends on the moment of inertia and the height of the ramp which are the same

anonymous · Answer

@rahaf You are quite welcome.