Initial angular velocity of a circular disc of mass  M  is  \omega _{1}  . Then two small spheres of mass  m are attached gently to two diametrically opposite points on the edge of the disc What is the final angular velocity of the disc ?

Option 1)

\left ( \frac{M+m}{M} \right )\omega _{1}

Option 2)

\left ( \frac{M+m}{m} \right )\omega _{1}

Option 3)

\left ( \frac{M}{M+4m} \right )\omega _{1}

Option 4)

\left ( \frac{M}{M+2m} \right )\omega _{1}

Answers (1)

As we learn 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 .

 

 L_{1}=L_{2}\; \Rightarrow I_{1}\omega {}_{1}=I_{2}\omega {}_{2}

I_{1}=\frac{1}{3}mR^{2}

I_{2}=\frac{1}{2}mR^{2}+2mR^{2}

\frac{1}{2}mR^{2}\omega_{1}=\frac{MR^{2}+4mR^{2}}{2} \;\omega {}_{2}

\omega{}_{2}=\frac{M\omega{}_{1}}{M+4m}

 


Option 1)

\left ( \frac{M+m}{M} \right )\omega _{1}

This is an incorrect option.

Option 2)

\left ( \frac{M+m}{m} \right )\omega _{1}

This is an incorrect option.

Option 3)

\left ( \frac{M}{M+4m} \right )\omega _{1}

This is the correct option.

Option 4)

\left ( \frac{M}{M+2m} \right )\omega _{1}

This is an incorrect option.

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