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Rotational Work and Kinetic Energy - Speed and Height

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A steel ball rolls down an incline into a loop-the-loop of radius R. (a) What minimum speed must the ball have at the top of the loop in order to stay on the track? (b) At what vertical height (h) on the incline, in terms of the radius of the loop, must the ball be released in order for it to have the required minimum speed at the top of the loop? (Neglect frictional losses)

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Solution Summary

This solution is provided in 173 words. It uses equation for normal reaction of a loop surface to find speed and height.

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a) Let us assume the radius of the ball = r

Normal reaction of the loop surface = N (downward)

Hence,
N + m*g = m*v^2/(R - r)

For minimum v, N = 0
Therefore,
m*g = m*v^2/(R - r)
=> v^2 = ...

Solution provided by:
Amrit Lal Ahuja, PhD (IP)
Education
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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