Question

A block of mass m slides down a frictionless track, then around the inside of a circular loop-the-loop of radius R. From what minimum height h must the block start to make it around the loop without falling off? Give your answer as a multiple of R.

Answer 1

Answer:

$H = \frac{5}{2}R$

Explanation:

As we know that if the block will complete the circular motion of the path then the speed at the bottom most part of the path must be equal to

$v = \sqrt{5Rg}$

now we know that

velocity at the bottom of the path is due to conversion of potential energy to kinetic energy

so we can say it is given as

$U = KE$

$mgH = \frac{1}{2}mv^2$

now we have

$mgH = \frac{1}{2}m(5Rg)$

$H = \frac{5}{2}R$