JEE Physics 2023 Question with Solution

Given: mass of disc $m = 0.5 \, \text{kg}$, initial speed $u = 18 \, \text{m/s}$, coefficient of friction $\mu = 0.3$, $g = 10 \, \text{m/s}^2$, and the disc starts with pure sliding.

Find: total kinetic energy after $2 \, \text{s}$ when the disc acquires pure rolling motion.

Due to kinetic friction, the translational acceleration is

$$a = -\mu g = -3$$

So the speed after $2 \, \text{s}$ is

$$V = 18 - 3 \times 2$$ $$V = 12 \, \text{m/s}$$

For a uniform disc in pure rolling motion,

$$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

Using $I = \frac{1}{2}mr^2$ and $\omega = \frac{v}{r}$,

$$KE = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v^2}{r^2}\right)$$ $$KE = \frac{3}{4}mv^2$$

Substituting $m = 0.5$ and $v = 12 \, \text{m/s}$,

$$KE = \frac{3}{4} \times 0.5 \times 12^2$$ $$KE = 54 \, \text{J}$$

Therefore, the total kinetic energy after $2 \, \text{s}$ is $54 \, \text{J}$.

Given: after $2 \, \text{s}$, the disc is in pure rolling motion.

Find: how the total kinetic energy is written for a uniform disc.

The total kinetic energy is the sum of translational and rotational parts:

$$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

For a uniform disc,

$$I = \frac{1}{2}mr^2$$

and for pure rolling,

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

Hence,

$$KE = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2$$ $$KE = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$$

Now with $v = 12 \, \text{m/s}$ and $m = 0.5 \, \text{kg}$,

$$KE = \frac{3}{4} \times 0.5 \times 144 = 54 \, \text{J}$$

Thus, the required value is $54$.