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6.System of Particles and Rotational Motion
hard
A hemisphere of radius $R$ and of mass $4m$ is free to slide with its base on a smooth horizontal table. A particle of mass $m$ is placed on the top of the hemisphere. The angular velocity of the particle relative to hemisphere at an angular displacement $\theta $ when velocity of hemisphere $v$ is
A$\frac{{5v}}{{R\,\cos \,\theta }}$
B$\frac{{2v}}{{R\,\cos \,\theta }}$
C$\frac{{3v}}{{R\,\sin \,\theta }}$
D$\frac{{5v}}{{R\,\sin \,\theta }}$
Solution
Let $v_{r}$ be the velocity of particle relative to hemisphere and $v$ be the linear velocity of hemisphere at this moment. Then from conservation of linear momentum, we have
${4 m v=m\left(v_{r} \cos \theta-v\right)}$
Or ${5 v=v_{r} \cos \theta}$
$\therefore \quad v_{r}=\frac{5 v}{\cos \theta}$
$\therefore \quad \omega=\frac{v_{r}}{R}=\frac{5 v}{R \cos \theta}$
${4 m v=m\left(v_{r} \cos \theta-v\right)}$
Or ${5 v=v_{r} \cos \theta}$
$\therefore \quad v_{r}=\frac{5 v}{\cos \theta}$
$\therefore \quad \omega=\frac{v_{r}}{R}=\frac{5 v}{R \cos \theta}$
Standard 11
Physics
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