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:

The object is rolling with velocity \( V_0 \) and angular velocity \( \omega \). The reference line is taken such that the potential energy \( U = 0 \) at the surface of point B/C, which is smooth. Also, the relationship between velocity and angular velocity for rolling motion is given by: \[ V_0 = \omega R \]

Step 1: Conservation of Energy:

The principle of conservation of energy is applied: \[ K_i + U_i = K_f + U_f \] where: - \( K_i \) and \( U_i \) are the initial kinetic and potential energies, - \( K_f \) and \( U_f \) are the final kinetic and potential energies.

Step 2: Applying the energy equation:

The initial and final energies are expressed as: \[ \frac{1}{2}mv_0^2 + \frac{1}{2} \left( \frac{MR^2}{2} \right) \left( \frac{v_0}{R} \right)^2 + 0 = mgh + \frac{1}{2} \left( \frac{MR^2}{2} \right) \left( \frac{v_0^2}{R^2} \right) \]

Step 3: Simplifying the equation:

Simplifying the above equation: \[ \frac{1}{2} mv_0^2 = mgh_{\text{max}} \]

Step 4: Finding the maximum height:

Solving for \( h_{\text{max}} \): \[ h_{\text{max}} = \frac{v_0^2}{2g} \]

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