Question

A circular disc of mass 0.41 kg and radius 10 m rolls without slipping with a velocity of 2 m/s. The total kinetic energy of disc is:

• A. 0.41 J
• B. 1.23 J
• C. 0.82 J
• D. 2.4 J

Answer 1

The correct option is B 1.23 J

A rolling disc has kinetic energy of rotation as well as of translation.
When a disc rolls on a plane surface, it rotates about an axis through its centre of mass and perpendicular to its plane, while the centre of mass of the disc moves forward along the surface. Therefore, it has kinetic energy of rotation as well as, of translation. The kinetic energy of rotation is
$K{E}_{rot}=\frac{1}{2}I{\omega }^2$
where I is the moment of inertia and $\omega$ the angular velocity of disc about the axis of rotation.
The moment of inertia I of a disc of mass M and radius R about an axis through its centre of mass is $\frac{1}{2}M{R}^2$. Therefore,
$K{E}_{rot}=\frac{1}{4}M{R}^2{\omega }^2$ ...(i)
The kinetic energy of translation is
$K{E}_{trans}=\frac{1}{2}M{\nu }_{CM}^2$ ...(ii)
As, ${\nu }_{CM}=R\omega$
$\therefore K{E}_{trans}=\frac{1}{2}M{R}^2{\omega }^2$ ...(iii)
$\therefore K{E}_{total}=K_{rot}+K_{trans}$
$=\frac{1}{4}M{R}^2{\omega }^2+\frac{1}{2}M{R}^2{\omega }^2$
$=\frac{3}{4}M{R}^2{\omega }^2$
$=\frac{3}{4}M{\nu }^2$
$=\frac{3}{4}\times 0.41\times (2{)}^2$
$=1.23J$