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WubbaLubba Dubdub

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## Homework Statement

A uniform circular disc of mass 2m and radius R placed freely on a horizontal smooth surface as shown in the figure. A particle of mass m is connected to the circumference of the disc with a mass less string. Now an impulse J is applied on the particle in the directions shown in the figure. The acceleration of center of mass of the disc just after application of impulse is

## Homework Equations

$$F = ma$$

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## The Attempt at a Solution

assuming tension T in the string

Force acting on particle = T

Force acting on disc, ##T = 2ma## , a is the acceleration of center of mass of disc

Torque on disc ##TR = Iq## , q is the angular acceleration

hence ##TR = \frac {2mR^2} {2}q##

thus ##T = mRq##

also since impulse is change in momentum, velocity of particle, ##v= \frac{J}{m}##

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I can't seem to find a relation between velocity and tension. Does the particle follow a circular path? If so, do i take tension to be equal to the centripetal force?