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Physics Question on System of Particles & Rotational Motion

A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity ω.\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 ω.\omega.' = :

A

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

B

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

C

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

D

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

Answer

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

Explanation

Solution

As no external torque is acting about the axis, angular momentum of system remains conserved. I1ω=I2ω′I_{1} \omega=I_{2} \omega' ⇒ mR2ω=(mR2+2MR2)ω′\Rightarrow\,\quad mR^{2}\omega=\left(mR^{2}+2MR^{2}\right)\omega' ⇒ ω′=(mm+2M)ω\Rightarrow\,\quad\omega'=\left(\frac{m}{m+2M}\right)\omega