Question

A disc of radius R and mass M is rolling horizontally without slipping with speed . 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. $\frac{3v^2}{4g}$
• C. $\frac{v^2}{2g}$
• D. $\frac{2v^2}{3g}$

Answer 1

Answer: (C)

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

Answer 2

Given: - Mass of the disc: $M$ - Radius of the disc: $R$ - Speed of the disc: $v$ - Gravitational acceleration: $g$

Step 1: Total Kinetic Energy of the Rolling Disc

For a disc rolling without slipping, the total kinetic energy ($K$) is the sum of translational kinetic energy and rotational kinetic energy:

$K = \text{Translational K.E.} + \text{Rotational K.E.}$ $K = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2$

The moment of inertia ($I$) of the disc about its center is:

$I = \frac{1}{2} M R^2$

The angular velocity ($\omega$) is related to the linear speed by:

$\omega = \frac{v}{R}$

Substituting these values:

$K = \frac{1}{2} M v^2 + \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v}{R}\right)^2$ $K = \frac{1}{2} M v^2 + \frac{1}{4} M v^2$ $K = \frac{3}{4} M v^2$

Step 2: Conservation of Energy

As the disc moves up the incline, its kinetic energy is converted into potential energy ($U$) at the maximum height $h$:

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

Solving for $h$:

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

Conclusion:

The maximum height that the disc can go up the incline is $\frac{3}{4} \frac{v^2}{g}$.