Question

A uniform sphere of mass $500gm$ rolls without slipping on a plane horizontal surface with its centre moving at a speed of $5cm/s$. Its kinetic energy is

• A. $8.75\times {10}^{-4}$
• B. $6.25\times {10}^{-4}$
• C. $8.75\times {10}^{-3}$
• D. $1.13\times {10}^{-3}$

Answer 1

The correct option is A

$8.75\times {10}^{-4}$

Step 1. Given data:

Mass of sphere, $M=500gm=0.5kg$

Speed of sphere about the center of mass$v=5cm/s=0.05m/s$

Step 2. Finding the kinetic energy:

The total kinetic energy is the sum of translation energy and rotational energy, i.e.

$K.E_{total}=K.E_{translational}+K.E_{rotational}$

$K.E_{total}=\frac{m{v}^2}{2}+\frac{I{\omega }^2}{2}$

Moment of inertia of sphere is $I=\frac{2m{r}^2}{5}$ and angular velocity is $\omega =\frac{v}{r}$

$K.E_{total}=\frac{m{v}^2}{2}+\frac{2m{v}^2}{5\times 2}$

$\Rightarrow K.E_{total}=\frac{m{v}^2}{2}\left(1+\frac{2}{5}\right)$

$\Rightarrow K.E_{total}=\frac{0.5\times {\left(0.05\right)}^2\times 7}{2\times 5}$

$\Rightarrow K.E_{total}=8.75\times {10}^{-4}J$

Hence, option A is correct.