Question 138: Centre of Mass and Rotational Motion

Question

A disc of radius $\mathrm{R}$ and mass $\mathrm{M}$ 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 :

Accepted Answer

Initial Conditions:
Consider a solid disc of mass $ M $ and radius $ R $ rolling without slipping on a horizontal surface. The center of mass of the disc has an initial speed $ v $.
Kinetic Energy of the Rolling Disc:
For an object that rolls without slipping, its total kinetic energy $ K $ is the sum of its translational kinetic energy and its rotational kinetic energy.
Translational Kinetic Energy:
$      K_{	ext{trans}} = \frac{1}{2} M v^2.      $
Rotational Kinetic Energy:
The moment of inertia $ I $ of a solid disc about its center is:
$      I = \frac{1}{2} M R^2.      $
Since the disc rolls without slipping, the angular velocity $\omega$ is related to the linear speed by:
$      v = \omega R \implies \omega = \frac{v}{R}.      $
Therefore, the rotational kinetic energy is:
$      K_{	ext{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{2} M R^2ight) \left(\frac{v^2}{R^2}ight) = \frac{1}{4} M v^2.      $
Total Initial Kinetic Energy:
Summing these gives:
$    K_{	ext{total}} = K_{	ext{trans}} + K_{	ext{rot}} = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2.    $
Energy Conservation as the Disc Climbs the Incline:
As the disc moves up the incline, there are no dissipative forces (assuming a smooth incline and no slipping), so energy is conserved. The total kinetic energy at the bottom will be converted entirely into gravitational potential energy at the maximum height $ h $:
$    K_{	ext{total}} = M g h.    $
Substitute $ K_{	ext{total}} = \frac{3}{4} M v^2 $:
$    \frac{3}{4} M v^2 = M g h.    $
Canceling $ M $ from both sides:
$    \frac{3}{4} v^2 = g h.    $
Thus,
$    h = \frac{3}{4} \frac{v^2}{g}.    $
Final Answer:
$ \boxed{\frac{3}{4} \frac{v^2}{g}} $
This corresponds to
Option A
.

Answer

$\frac{3}{4} \frac{v^2}{\mathrm{~g}}$

Answer

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

Answer

$\frac{2}{3} \frac{v^2}{\mathrm{~g}}$

Answer

$\frac{1}{2} \frac{v^2}{\mathrm{~g}}$

Centre of Mass and Rotational Motion

A disc of radius \mathrm{R} and mass \mathrm{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 :

Choose the correct answer: