Question

A man turns on rotating table with an angular speed $\omega$. He is holding two equal masses at arms length. Without moving his arms, he just drops the two masses. How will his angular speed change?

• A. less than $\omega$
• B. more than $\omega$
• C. it will be equal to $\omega$
• D. it will be more than $\omega$ if the dropped mass is more than $9.8Kg$ and it will be less than $\omega$ if the mass dropped is less than $9.8Kg$

Answer 1

The correct option is (B) more than $\omega$

Step 1,

When the man is rotating with the table with angular velocity ω its angular momentum can be given as

Angular momentum = Iω

When the person drops two masses its angular momentum can be given as

Angular momentum = $I_{new}{Angularvelocity}_{new}$

Step 2,

Since no external torque is acting on the system so the angular momentum of the system will be conserved.

So,

$I\times Angularvelocity=I_{new}\times {Angularvelocity}_{new}$

After dropping the mass moment of inertia decreases and so angular momentum must have to increase to keep the momentum conserved.

Hence, the new angular velocity will be more than the previous angular velocity.