Question

A disc of mass M and radius R rolls on a horizontal
surface and then rolls up an inclined plane as shown
in the figure. If the velocity of the disc is v, the height
to which the disc will rise will be :-
1)3v^2/2g 2) 3v^2/4g​

Answer 1

The height to which the disc will rise be $h = \frac{3v^2}{4g}$

Explanation:

Given statement

v= initial velocity

Radius of gyration of disc, $k = \frac{R}{\sqrt {2}}$

Potential Energy = Translational Kinetic Energy + Rotational Kinetic Energy

where Translational kinetic energy = $\frac{1}{2}mv^2$, m is the mass and v is the velocity and

Rotational Kinetic Energy = $\frac{1}{2} Iw^2$,   ω is the angular velocity and I is the moment of inertia around the axis of rotation.

$mgh = \frac{1}{2}mv^2 + \frac{1}{2} Iw^2$

$mgh = \frac{1}{2}mv^2(1 + \frac {k^2}{R^2})$

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

Thus, the height to which the disc will rise be $h = \frac{3v^2}{4g}$