Q67 / 90 · JEE Main 2024 (01 Feb Shift 2) · Physics

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:

A
$\frac{v^{2}}{g}$
B
$\frac{3}{4} \frac{v^{2}}{g}$
Correct answer
C
$\frac{1}{2} \frac{v^{2}}{g}$
D
$\frac{2}{3} \frac{v^{2}}{g}$

Solution

Only the translational kinetic energy of disc changes into gravitational potential energy and rotational $KE$ remains unchanged as there is no friction.

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

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

Hence, the correct answer is option: 'C'.

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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:

Solution

Relevant Questions

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: