Question

A body of moment of inertia $5kg{{m}^{2}}$ rotating with angular velocity $6{{s}^{-1}}$ has the kinetic as a mass of 20 kg moving with velocity of
A. $5m{{s}^{-1}}$
B. $4m{{s}^{-1}}$
C. $3m{{s}^{-1}}$
D. $2m{{s}^{-1}}$

Answer 1

If the moment of inertia of the body along this axis is I and its angular velocity is $\omega$ , about the same axis, then the kinetic energy of the body is equal to $K=\dfrac{1}{2}I{{\omega }^{2}}$.
The kinetic energy of a body of mass m, moving velocity v is $K=\dfrac{1}{2}m{{v}^{2}}$

Formula used:
${{K}_{R}}=\dfrac{1}{2}I{{\omega }^{2}}$
${{K}_{T}}=\dfrac{1}{2}m{{v}^{2}}$

Complete step by step answer:
We know that if a body is in a translation motion, it possesses some energy, which is called kinetic energy. Similarly, when a body is in a rotational motion, it possesses some kinetic energy. Although, the body is not moving as a whole, but every point (consisting an element of the body) is in motion. When we sum up all the kinetic energies of these smallest elements, we get the rotational kinetic energy of the body.

Suppose a body in pure rotational motion and is rotating about a fixed axis. If the moment of inertia of the body along this axis is $I$ and its angular velocity is $\omega$ , about the same axis, then the kinetic energy of the body is equal to $K=\dfrac{1}{2}I{{\omega }^{2}}$.It is given that the moment of inertia of the body is $5kg{{m}^{2}}$ and its angular velocity is $6{{s}^{-1}}$.
$K=\dfrac{1}{2}I{{\omega }^{2}}\\\ \Rightarrow K =\dfrac{1}{2}(5){{(6)}^{2}}\\\ \Rightarrow K =90J$.
Consider another body of mass m moving with velocity v. It is under translation motion. Then the kinetic energy of this body is equal to $K=\dfrac{1}{2}m{{v}^{2}}$ …. (ii).
It is said that the kinetic energy of this body is the same as the kinetic energy of the body in rotational motion and the mass of this body is 20 kg. Substitute $K=90J$ and $m=20kg$ in (ii).
$90=\dfrac{1}{2}(20){{v}^{2}}$
$\Rightarrow {{v}^{2}}=9$
$\therefore v=3m{{s}^{-2}}$

Therefore, the correct option is C.

Note: When a body is translational motion as well as rotational motion, its total kinetic energy is equal to the sum of its translational kinetic energy and rotational kinetic energy. i.e. $K={{K}_{T}}+{{K}_{R}}$. Here, ${{K}_{R}}=\dfrac{1}{2}I{{\omega }^{2}}$ and ${{K}_{T}}=\dfrac{1}{2}mv_{com}^{2}$.Note that ${{v}_{com}}$ is the velocity of the centre of mass of the body.