QJEEM24Ph6636: JEE Main 2024 Physics - Class 11 JEEM24Ph6636 Solution: A disc of radius and mass is rolling horizontally without…

1. Initial Energy (At the bottom, h=0)

The disc is rolling without slipping on the horizontal surface. Its total initial energy ( $E_i$ ) is the sum of its translational kinetic energy ( $K_{\text{trans}}$ ) and its rotational kinetic energy ( $K_{\text{rot}}$ ). We set the initial potential energy $U_i$ to zero.

$E_i = K_{\text{trans}} + K_{\text{rot}} + U_i$

The translational kinetic energy is:

$K_{\text{trans}} = \frac{1}{2} M v^2$

The rotational kinetic energy is:

$K_{\text{rot}} = \frac{1}{2} I \omega^2$

For a solid disc, the moment of inertia is $I = \frac{1}{2} M R^2$ . Since it's rolling without slipping, $\omega = \frac{v}{R}$ .

$K_{\text{rot}} = \frac{1}{2} \left( \frac{1}{2} M R^2 \right) \left( \frac{v}{R} \right)^2 = \frac{1}{4} M v^2$

So, the total initial energy is:

$E_i = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2$

2. Final Energy (At maximum height $h$ )

The disc moves up a smooth incline. "Smooth" means there is no friction.

Because there is no friction, there is no external torque to change the disc's rotation. Therefore, its angular velocity $\omega$ remains constant as it moves up the incline.

At the maximum height $h$ , the disc's translational motion stops, so its final translational speed $v_f$ is 0. However, it is still spinning with its original angular velocity $\omega$ .

The final energy ( $E_f$ ) is the sum of its final potential energy ( $U_f$ ), final translational kinetic energy ( $K_{\text{trans, f}}$ ), and final rotational kinetic energy ( $K_{\text{rot, f}}$ ).

$E_f = K_{\text{trans, f}} + K_{\text{rot, f}} + U_f$

$U_f = M g h$

$K_{\text{trans, f}} = \frac{1}{2} M (v_f)^2 = 0$

$K_{\text{rot, f}} = \frac{1}{2} I \omega^2 = \frac{1}{4} M v^2 \quad \text{(This value is unchanged from the start)}$

So, the total final energy is:

$E_f = 0 + \frac{1}{4} M v^2 + M g h$

3. Conservation of Energy

We set the total initial energy equal to the total final energy:

$E_i = E_f$

$\frac{3}{4} M v^2 = \frac{1}{4} M v^2 + M g h$

Subtract $\frac{1}{4} M v^2$ from both sides:

$\frac{3}{4} M v^2 - \frac{1}{4} M v^2 = M g h$

$\frac{2}{4} M v^2 = M g h$

$\frac{1}{2} M v^2 = M g h$

Cancel $M$ from both sides:

$\frac{1}{2} v^2 = g h$

Solving for the maximum height $h$ :

$h = \frac{v^2}{2g}$

Final Answer:

The maximum height that the disc can go up the incline is:

$\boxed{h = \frac{v^2}{2g}}$