Question

A student is standing on the edge of a stationary turntable, made from a uniform disc of mass M and radius R on a frictionless bearing at its center. She starts to run with speed v around the perimeter of the table. (This is her speed relative to the ground not the turntable). If her mass is m, what is the magnitude of the angular velocity of the turntable (relative to the ground)?​

Answer 1

The angular velocity of the turntable is  $\boxed{\frac{mvr}{M+m}}$

Explanation:

Given:

A turntable having mass M and radius R and a student of mass m standing on its edge

To find out:

The angular velocity of the turntable when the students starts moving with speed v

Solution:

From the conservation of momentum, if the speed of the turntable is v' then

Initial momentum = Final momentum (as there is no external force)

Initially all are in rest therefore initial momentum is zero

$0=mv-(m+M)v'$

$\implies v'=\frac{m}{M+m}v$

This will be the linear speed of the turntable

Therefore, the angular velocity of the turntable

$\omega=\frac{v'}{r}$

$\implies \omega=\frac{mvr}{M+m}$

Hope this answer is helpful.

Know More:

Q: A particle is kept fixed on a turntable rotating uniformly. As seen from the ground the particle goes in a circle, its speed is 20 cm/s and acceleration is 20 cm/s2. The particle is now shifted to a new position to make the radius half of the original value. The new value of the speed and acceleration will be:

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