Question

A uniform sphere of mass $m$ and radius $r$ rolls without sliding over a horizontal plane, rotating about a horizontal axle $OA$ (figure shown above). In the process, the centre of the sphere moves with velocity $v$ along a circle of radius $R$. Find the kinetic energy of the sphere.

• A. $T=\frac{7}{10}m{v}^2(1+\frac{2r^2}{7R^2})$
• B. $T=\frac{7}{10}m{v}^2(1-\frac{2r^2}{7R^2})$
• C. $T=\frac{7}{20}m{v}^2(1-\frac{2r^2}{7R^2})$
• D. None of these

Answer 1

The correct option is A $T=\frac{7}{10}m{v}^2(1+\frac{2r^2}{7R^2})$

The sphere has two types of motion, one is the rotation about its own axis and the other is motion in a circle of radius $R$. Hence the sought kinetic energy
$T=\frac{1}{2}{I}_1{\omega }_1^2+\frac{1}{2}{I}_2{\omega }_2^2$ (1)
where, $I_1$ is the moment of inertia about its own axis, and $I_2$ is the moment of inertia about the vertical axis, passing through $O$,
But, $I_1=\frac{2}{5}m{r}^2$ and $I_2=\frac{2}{5}m{r}^2+m{R}^2$ (using parallel axis theorem,) (2)
In addition to ${\omega }_1=\frac{v}{r}$ and ${\omega }_2=\frac{v}{R}$ (3)
Using (2) and (3) in (1), we get $T^{\prime }=\frac{7}{10}m{v}^2(1+\frac{2r^2}{7R^2})$