Question

A body initially at rest and sliding along a frictionless track from a heigjt h just completes a vertical circle of diameter ab equal to

d.the height is equal to

Answer 1

Answer:5/4D

Explanation:

1/2mvsquare=mgh

h=1vsquare/2g

h=5r/2

r=D/2

h=5/4D

Answer 2

ANSWER:

The height from which the frictionless track is sliding along is $h = \frac{5 D}{4}$

EXPLANATION:

If only conservative forces act on a system, total mechanical energy remains constant.

K + U = E (constant)

Delta K + Delta U = 0

Delta K = - Delta U

For particles to complete circle  

V = sqrt{5gR}

Hence, $h = \frac{5 D}{4}$.

The height from which the frictionless track is sliding along can be manipulated using the above equation.