A block of mass m slides down an inclined plane into a loopthe loop of radius r.

Question

A block of mass m slides down an inclined plane into a loopthe loop of radius r. At what vertical height on the inclinedplane (in terms of the radius of the loop) must the block bereleased if it is to have the required minimum speed at the top ofthe loop? I know that the minimum velocity that the mass has at the topof the loop is equal to V = gr and that the potential energy at the beginning of theinclined plane must be equal to the total energy that the mass hasat the top of the loop. However, I cannot seem to get beyondthis point. Help!! A block of mass m slides down an inclined plane into a loopthe loop of radius r. At what vertical height on the inclinedplane (in terms of the radius of the loop) must the block bereleased if it is to have the required minimum speed at the top ofthe loop? I know that the minimum velocity that the mass has at the topof the loop is equal to V = gr and that the potential energy at the beginning of theinclined plane must be equal to the total energy that the mass hasat the top of the loop. However, I cannot seem to get beyondthis point. Help!!

Explanation / Answer

At the top the total energy is PE+KE=mg*2r+1/2*m*v2 KE at bottom=2mgr+1/2*m*gr KE at the bottom=3/2*mgr Let it be released from height h then, mgh=3/2*mgr h=3/2*r Hence we get by it. "Hope this helps!Best of luck for the rest of yourcoursework." KE at bottom=2mgr+1/2*m*gr KE at the bottom=3/2*mgr Let it be released from height h then, mgh=3/2*mgr h=3/2*r Hence we get by it. "Hope this helps!Best of luck for the rest of yourcoursework."

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