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## Homework Statement

A particle of mass m at the end of a light string wraps itself about a fixed vertical cylinder of radius a. All the motion is in the horizontal plane (disregard gravity). The angular velocity of the cord is w0 when the distance from the particle to the point of contact of the string and cylinder is b. Find the angular velocity and tension in the string after the cord has turned through an additional angle theta.

Hints from my professor:

-There is a torque acting on m so angular momentum is not conserved. But something even more simple is.

-The force on m due to the string is always perpendicular to the velocity vector.

## Homework Equations

T=mv^2/r

w= d(theta)/dt

w=vsin(theta)/r

???

The answer given is:

w=w0/(1-(a/b)theta), T=m*b*w0*w

## The Attempt at a Solution

I saw that v*sin(theta) = w0 so I got as far as w=w0/r but that's it. And I'm confused in the tension equation as to how one would obtain a as b*w0*w.