Asked by kassy
a block slides down a frictionless ramp. at the bottom of the ramp is a vertical circular loop with a radius of 75 cm. from what minimum height does the block need to be released in order to make it around the loop?
Answers
Answered by
bobpursley
let h be the height of the release.
Then h-2r is the altitude of the top of the loop.
Initial PE=finalPE+final KE
mgh=mg( 2r)+1/2 m v^2
but v^2/r= g to keep it at the top of the loop.
mgh= 2mgr+1/2 mg*r
h= 2r+1/2 r=2.5 r
check my math.
Then h-2r is the altitude of the top of the loop.
Initial PE=finalPE+final KE
mgh=mg( 2r)+1/2 m v^2
but v^2/r= g to keep it at the top of the loop.
mgh= 2mgr+1/2 mg*r
h= 2r+1/2 r=2.5 r
check my math.
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