Question

When a sphere of moment of inertia $ I $ about an axis through centre of gravity and mass m rolls from rest down an inclined plane without slipping, its kinetic energy is :

• A. $ \frac{1}{2}I{{\omega }^{2}} $
• B. $ \frac{1}{2}m{{v}^{2}} $
• C. $ I\omega +mv $
• D. $ \frac{1}{2}I{{\omega }^{2}}+\frac{1}{2}m{{v}^{2}} $

Answer 1

Answer: (D)

When a sphere is in motion, it has kinetic energy due to rotation (the energy due to rotational motion) and translational (the energy due to motion from one location to another).
The translational kinetin energy is.
$(K E)_{T}=\frac{1}{2} m v^{2}$
where $m$ is mass of object and $v$ is speed.
The rolling kinetic energy is
$(K E)_{R}=\frac{1}{2} I \omega^{2}$
where $I$ is moment of inertia and co is angular speed.
Hence, total kinetic energy is
$K E=(K E)_{T}+(K E)_{R}$
$K E=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$