Question

A ball is released from the top of vertical circular
pipe. Find angle theta with vertical where the ball will
lose contact with inner side wall of pipe and start
moving with outer side wall. (Thickness of pipe is
small as compared to radius of circle)​

Answer 1

Answer:

$1) cos^{-1}(\frac{2}{3})$

Explanation:

1) When ball starts from top and reach at angle $\theta$ with the vertical .

Using Energy Conservation ,

$mgR(1-cos(\theta))=\frac{1}{2}mv^2\\ \\=> 2mg(1-cos(\theta))=\frac{mv^2}{R}$

2) And Since, Ball loses contact with the inner wall .

Its normal reaction from wall will be 0.

Hence, Force equation will be

$\frac{mv^2}{R}=mgcos(\theta)\\ \\=>2mg(1-cos(\theta))=mgcos(\theta)\\ \\=>2(1-cos(\theta))=cos(\theta)\\ \\=>cos(\theta)=\frac{2}{3}\\ \\=>\theta= cos^{-1}(\frac{2}{3})$

Answer 2

Explanation:

hope this will help you