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Q.
A body is rolling down an inclined plane. Its translational and rotational kinetic energies are equal. The body is a
System of Particles and Rotational Motion
Solution:
When a body rolls down an inclined plane, it is accompanied by rotational and translational kinetic energies.
Rotational kinetic energy, $K_{R}=\frac{1}{2} I \omega^{2}$
where, $I$ is the moment of inertia and $\omega$ is the angular velocity.
Translational kinetic energy for pure rolling,
$=K_{T}=\frac{1}{2} m v_{ CM }^{2}=\frac{1}{2} m(r \omega)^{2}$
$\left(\because v_{ CM }=r \omega\right)$
where, $m$ is mass of the body, $v_{ CM }$ is the velocity and $\omega$ is the angular velocity.
Given, translational kinetic energy = rotational kinetic
$\therefore \frac{1}{2} m\left(r^{2} \omega^{2}\right)=\frac{1}{2} I \omega^{2}$
$\Rightarrow I=m r^{2}$
We know that, $m r^{2}$ is the moment of inertia of hollow cylinder about its axis, where $m$ is the mass of hollow cylindrical body and $r$ is the radius of the cylinder.