Asked by S.U.
A body is allowed to slide down a frictionless track freely under gravity.The track ends in a semicircular shaped part of diameter d.What should be the height(minimum) from which the body must fall so that it completes the circle?
Answers
Answered by
MathMate
Assuming the "semicircular shape" is like the loop of a roller-coaster, and we want the body to complete the circle.
At the top of the loop, of diameter d (radius d/2), the velocity must be such that the centripetal force exerted by the rails ≥0, or
mv²/(d/2)-mg ≥0
At this position, the kinetic energy is (1/2)mv^2=mgh, where h is the height of release of the body above the top of the loop.
Substitute in above, we find:
2mgh/(d/2)-mg ≥0
or simply
h > d/4.
So the body must be released at a point h> d/4+d, above the lowest part of the loop, or
h > 5d/4
At the top of the loop, of diameter d (radius d/2), the velocity must be such that the centripetal force exerted by the rails ≥0, or
mv²/(d/2)-mg ≥0
At this position, the kinetic energy is (1/2)mv^2=mgh, where h is the height of release of the body above the top of the loop.
Substitute in above, we find:
2mgh/(d/2)-mg ≥0
or simply
h > d/4.
So the body must be released at a point h> d/4+d, above the lowest part of the loop, or
h > 5d/4
Answered by
Biren
Anything falling from a height of h will rise to a height of h so h = d Period
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