A disc of radius R and mass M is rolling horizontally without slipping with speed v . It then moves up an inclined smooth surface as shown in figure. The maximum height that the disc can go up the incline is :

Rotational Motion PYQ — JEE Main 2024 · 1 Feb · Shift 2 · Q80 with Solution

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Correct answer: D: $\FRAC{1}{2}\FRAC{V^2}{G}$

Initial motion on horizontal surface

The disc rolls without slipping with speed $v$ .

So initially it has:

translational kinetic energy
rotational kinetic energy

For a solid disc, $I_{\text{cm}}=\frac{1}{2}MR^2$

Since it is rolling without slipping, $\omega=\frac{v}{R}$

Hence, $K_{\text{trans}}=\frac{1}{2}Mv^2$

and

=\frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v}{R}\right)^2 =\frac{1}{4}Mv^2$$ Therefore total initial kinetic energy is $$K_i=\frac{1}{2}Mv^2+\frac{1}{4}Mv^2=\frac{3}{4}Mv^2$$ --- 2. **Motion on the smooth incline** The incline is **smooth**, so there is **no friction** on the incline. Therefore: - no external torque acts about the center due to contact force, - angular speed remains constant, - rotational kinetic energy remains unchanged. So only the **translational kinetic energy** can be converted into gravitational potential energy. Initially translational kinetic energy is $$\frac{1}{2}Mv^2$$ At the maximum height, the center of mass momentarily comes to rest in translation, so translational kinetic energy becomes zero. Thus, $$Mg h = \frac{1}{2}Mv^2$$ So, $$h=\frac{v^2}{2g}$$ --- 3. **Check options** Thus the maximum height is $$\boxed{\frac{1}{2}\frac{v^2}{g}}$$ This corresponds to **Option D**. --- 4. **Comparison with stored answer** Stored correct answer is **A**: $\frac{3}{4}\frac{v^2}{g}$. But this would be true only if the disc continued to roll without slipping on the incline, so that both translational and rotational kinetic energies were converted into potential energy. Since the incline is explicitly **smooth**, there is no friction to enforce rolling, and rotational kinetic energy remains constant. Hence the stored answer appears incorrect.

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View written solutionFree

Correct answer: D: $\FRAC{1}{2}\FRAC{V^2}{G}$

Initial motion on horizontal surface

The disc rolls without slipping with speed $v$ .

So initially it has:

translational kinetic energy
rotational kinetic energy

For a solid disc, $I_{\text{cm}}=\frac{1}{2}MR^2$

Since it is rolling without slipping, $\omega=\frac{v}{R}$

Hence, $K_{\text{trans}}=\frac{1}{2}Mv^2$

and

=\frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v}{R}\right)^2 =\frac{1}{4}Mv^2$$ Therefore total initial kinetic energy is $$K_i=\frac{1}{2}Mv^2+\frac{1}{4}Mv^2=\frac{3}{4}Mv^2$$ --- 2. **Motion on the smooth incline** The incline is **smooth**, so there is **no friction** on the incline. Therefore: - no external torque acts about the center due to contact force, - angular speed remains constant, - rotational kinetic energy remains unchanged. So only the **translational kinetic energy** can be converted into gravitational potential energy. Initially translational kinetic energy is $$\frac{1}{2}Mv^2$$ At the maximum height, the center of mass momentarily comes to rest in translation, so translational kinetic energy becomes zero. Thus, $$Mg h = \frac{1}{2}Mv^2$$ So, $$h=\frac{v^2}{2g}$$ --- 3. **Check options** Thus the maximum height is $$\boxed{\frac{1}{2}\frac{v^2}{g}}$$ This corresponds to **Option D**. --- 4. **Comparison with stored answer** Stored correct answer is **A**: $\frac{3}{4}\frac{v^2}{g}$. But this would be true only if the disc continued to roll without slipping on the incline, so that both translational and rotational kinetic energies were converted into potential energy. Since the incline is explicitly **smooth**, there is no friction to enforce rolling, and rotational kinetic energy remains constant. Hence the stored answer appears incorrect.

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