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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 $T$ enter image description here

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 $\theta$

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The force (tension in the string) is not always directed towards the same point, so angular momentum is not conserved. However, the force is always perpendicular to the velocity of the particle, so kinetic energy - and therefore velocity - is conserved. That is
$r\omega = b\omega_0$
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 $r=r_0-a\theta$.

See similar question A rope wrapping around a cylinder.

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  • $\begingroup$ will the mass $m$ traverse a spiral path? $\endgroup$ – biswa prakash mishra May 8 '17 at 4:17
  • $\begingroup$ 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. $\endgroup$ – sammy gerbil May 8 '17 at 11:29

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