Question

A solid sphere having mass m and radius r rolls down an inclined plane. Then its kinetic energy is

• A. $\frac{5}{7}$ rotational and $\frac{2}{7}$translational
• B. $\frac{2}{7}$ rotational and $\frac{5}{7}$translational
• C. $\frac{2}{5}$ rotational and $\frac{3}{5}$translational
• D. $\frac{1}{2}$ rotational and $\frac{1}{2}$translational

Answer 1

The correct option is A $\frac{5}{7}$ rotational and $\frac{2}{7}$translational

When a solid sphere having

$mass=m$

Radius$=r$

Now the Kinetic energy wllbe

The total Kinetic energy (K.E) of an object is given by

$K.E=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

Where $I$ is the moment of inertia of the object and $\omega$ is the angular momentum. For a solid sphere, the moment of inertia is given by

$I=\frac{2}{5}m{r}^2$

Angular velocity would be derived from its radius and linear velocity

$\omega =\frac{v}{r}$

So the total equation would

$K.E=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}m{r}^2\left(\frac{v^2}{r}\right)=\frac{1}{2}m{v}^2+\frac{2}{10}m{v}^2$

To get the percentage attributed to rotational energy, we would divide the rotational part of the energy by the total energy

$\frac{\frac{2}{10}m{v}^2}{\frac{1}{2}m{v}^2+\frac{2}{10}m{v}^2}=\frac{\frac{2}{10}m{v}^2}{\frac{7}{10}m{v}^2}=\frac{2}{7}$

Hence $\frac{5}{7}$ is rotational and $\frac{2}{7}$ translational.