A cart slides down a frictionless inclined track to a circular loop of radius R

Question

A cart slides down a frictionless inclined track to a circular loop of radius R = 9.0 m. In order for the cart to negotiate the loop safely, the normal force acting on the cart at the top of the loop, due to the track, must be at least equal to the cart's weight. (Note: This is different from the conditions needed to just negotiate the loop.

Vmin = 13.28 m/s
3) When the car is descending vertically in the loop (point (c) in the picture), what is its speed |v|?
|v| = ??
4) At the bottom of the loop, on the flat part of the track, the cart must be stopped in a distance of d = 10.0 m. What retarding acceleration |a| is required?
|a| = ??

Explanation / Answer

mv^2/R=mg+mg=2mg v=13.28 the picture is not given u can find velocity at any point by change in kinetic energy = work done=mgh at bottom of loop work done =0 =change in kinetic energy so, v=13.28 v^2= u^2-2as a=9.522

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