# Numerical of Circular Motion

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A stone of mass $500$g is attached to a string of length $50$cm which will break if the tension in it exceeds $20$N. The stone is whirled in a vertical circle, the axis of rotation being at a height of $100$cm above the ground. The angular speed is very slowly increased until the string breaks. In what position is this break most likely to occur, and at what angular speed? Where will the stone hit the ground?

My attempt :
Here, mass of stone (m)=$0.5$kg
Length of string (l)=$0.5$m
Height above the ground (h')=$1$m
Let, $\omega$ be the angular speed when the stone breaks.
Then maximum tension in string =$m\omega^2 r+mg$
$$20=0.5\times \omega^2 \times 0.5+10$$
$$\omega =7.75 \textrm {rads^-1}$$

So far you are correct.
For the 2nd part imagine the string just broken and the stone now only has its velocity which is horizontal and gravity acting on it. This can be thought as a projectile motion ,

$y=1/2gt^2$
where $y$ is vertical height, $g$is gravitational acceleration, $t$ is time of flight.

This gives you the time of flight on solving which on multiplication with horizontal velocity shall give horizontal distance.

$t=\sqrt{2y/g}$
$v*t=s$

$s= r\omega \sqrt{2*0.5/9.8}$
So distance comes out to be 1.237

answered Oct 28, 2017 by (2,320 points)
selected Dec 23, 2017