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

The Correct Option is C

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 \( \frac{3}{4} \frac{v^2}{g} \).