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# Engineering

A thin circular ring of mass m and radius R is rotating   about its axis with a constant angular velocity  \omega  .Two objects each of mass    M  are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity \omega {}'  =

Option 1)

\frac{\omega m}{m+2M}

 Option 2)

\frac{\omega \left ( m+2M \right )}{m}
Option 3)

\frac{\omega \left ( m-2M \right )}{\left ( m+2M \right )}

 Option 4)

\frac{\omega m}{\left ( m+M \right )}

Answers (1)
A Aadil

As we learnt in

Law of conservation of angular moment -

\vec{\tau }= \frac{\vec{dL}}

- wherein

If net torque is zero

i.e. \frac{\vec{dL}}= 0

\vec{L}= constant

angular momentum is conserved only when external torque is zero .

 

 From conservation of Angular Momentum 

L_{1}=L_{2}

I_{1}\omega_{1}=I_{2}\omega_{2}    \left \lfloor \because\ \;\omega_{1}=\omega, \;\omega_{2}=\omega^{'} \right \rfloor

mR^{2}\omega=\left ( m+2M \right )R^{2}\omega{'} \; \Rightarrow\ \; \omega{'}=\frac{m\omega}{m+2M}


Option 1)

\frac{\omega m}{m+2M}

This is the correct option.

Option 2)

\frac{\omega \left ( m+2M \right )}{m}

This is an incorrect option.

Option 3)

\frac{\omega \left ( m-2M \right )}{\left ( m+2M \right )}

This is an incorrect option.

Option 4)

\frac{\omega m}{\left ( m+M \right )}

This is an incorrect option.

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