$A$ uniform rod of length $L$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $\theta$,its angular velocity $\omega$ is given as

  • A
    $\sqrt{\frac{6g}{L}} \sin \theta$
  • B
    $\sqrt{\frac{6g}{L}} \sin \frac{\theta}{2}$
  • C
    $\sqrt{\frac{6g}{L}} \cos \frac{\theta}{2}$
  • D
    $\sqrt{\frac{6g}{L}} \cos \theta$

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$A$ thin uniform rod $AB$ of mass $m$ and length $l$ is hinged at one end $A$ to the ground level. Initially, the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its $B$ end strikes the ground is $(g = \text{acceleration due to gravity})$

Match the linear motion formulas in Column-$I$ with their corresponding rotational motion formulas in Column-$II$.
Column-$I$ Column-$II$
$(1)$ $W = F \Delta x$ $(a)$ $P = \tau \omega$
$(2)$ $P = Fv$ $(b)$ $W = \tau \Delta \theta$
$(c)$ $L = I \omega$

$A$ disc of mass $1\,kg$ and radius $R$ is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of the disc. $A$ body of the same mass as that of the disc is fixed at the highest point of the disc. Now the system is released. When the body comes to the lowest position,its angular speed will be $4 \sqrt{\frac{x}{3 R}} \text{ rad s}^{-1}$ where $x=$ (Given $g = 10 \text{ m s}^{-2}$)

$A$ flywheel of moment of inertia $I$ is rotating at $n$ revolutions per second. The work needed to double the frequency would be

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Three objects,$A$ (a solid sphere),$B$ (a thin circular disk),and $C$ (a circular ring),each have the same mass $M$ and radius $R$. They all spin with the same angular speed $\omega$ about their own symmetry axes. The amounts of work $(W)$ required to bring them to rest would satisfy the relation:

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