@Pranjal Jain
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$\text{angular velocity}=\dfrac{\text{Relative velocity perpendicular to line joining them}}{\text{Distance between them}}$

By simple geometry, $\angle AOP=30^\circ$

Case I:$A$ is fixed to ground, so, the velocity of $P$ will be $\omega R$ at angle $120^\circ$ to the line joining $AP$, thus, $\omega_1=\dfrac{\omega R\cos 30^\circ}{\sqrt{3}R}$

Case II:$A$ is fixed to disc, in this case, velocity of $A$ will add up in relative velocity. In this case also, velocity of $A$ will be $\omega R$ at angle $60^\circ$ to the line joining $AP$. Thus, $\omega_2=\dfrac{\omega R\cos 30^\circ+\omega R\cos 30^\circ}{\sqrt{3}R}=2\omega_1\Rightarrow \dfrac{\omega_1}{\omega_2}=\dfrac{1}{2}$

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TopNewest@Raghav Vaidyanathan the problem is in the middle.

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I have rotated the image. $\huge\ddot\smile$

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U can tell me the solution . ???

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$\text{angular velocity}=\dfrac{\text{Relative velocity perpendicular to line joining them}}{\text{Distance between them}}$

By simple geometry, $\angle AOP=30^\circ$

Case I:$A$ is fixed to ground, so, the velocity of $P$ will be $\omega R$ at angle $120^\circ$ to the line joining $AP$, thus, $\omega_1=\dfrac{\omega R\cos 30^\circ}{\sqrt{3}R}$Case II:$A$ is fixed to disc, in this case, velocity of $A$ will add up in relative velocity. In this case also, velocity of $A$ will be $\omega R$ at angle $60^\circ$ to the line joining $AP$. Thus, $\omega_2=\dfrac{\omega R\cos 30^\circ+\omega R\cos 30^\circ}{\sqrt{3}R}=2\omega_1\Rightarrow \dfrac{\omega_1}{\omega_2}=\dfrac{1}{2}$Log in to reply

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$\omega$

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