# How to find the angular velocity and tension in the string?

A light string with a particle of mass $m$ at one end, wraps itself around a fixed vertical cylinder of radius $a$. The entire motion is in the horizontal plane(neglect gravity). The angular speed of the string (and the particle) is $\omega_0$ when the distance of particle from point of contact between the string and the cylinder is $b$. If the angular speed is $\omega$ and the tension in the string is $T$ after the string has turned through the additional angle $\theta$.

we have to find $\omega$ and $T$

what I have tried doing is to conserve the angular momentum. But I am facing difficulty in writing angular momentum when the string had turned by an angle $\theta$

$r\omega = b\omega_0$
where $r$ is the length of the string at any time.
The particle is always in circular motion about the instantaneous point of contact, so the tension in the string provides centripetal force $mr\omega^2$.
When the rope has turned through angle $\theta$ it has shortened by $a\theta$. The length is then $r=r_0-a\theta$.
• will the mass $m$ traverse a spiral path? – biswa prakash mishra May 8 '17 at 4:17
• The string does not slip as the mass rotates around the pole, so the string gets shorter. ... Note that $r$ is not the radius $R=\sqrt{a^2+r^2}$ from the centre of the pole. – sammy gerbil May 8 '17 at 11:29