A wheel of mass M and radius a and M.I. `I_G` (about centre of mass) is set rolling with angular velocity `omega` up a rough inclined plane of inclination ``theta`. The distance traveled by it up the plane is

Options

A. `(I_Gomega^2)/(2mgsintheta)`
B. `(omega^2(Ma^2+I_G))/(2Mgsin theta)`
C. `(I_Gomega^2)/(2omegaMgsin theta)`
D. `(I_Gomega)/(2Mgsin theta)`

Explanation

To solve the problem of a wheel rolling up a rough inclined plane, we will use the principle of conservation of energy. The wheel has both translational and rotational kinetic energy initially, and as it rolls up the incline, it will convert this energy into gravitational potential energy until it comes to a stop.

### Step-by-Step Solution

1. **Identify the Initial Conditions:**
- Mass of the wheel: \( M \)
- Radius of the wheel: \( a \)
- Moment of inertia about the center of mass: \( I_G \)
- Initial angular velocity: \( \omega \)
- Inclination of the plane: \( \theta \)

2. **Determine the Initial Kinetic Energy:**
The wheel has both translational and rotational kinetic energy.
- Translational kinetic energy (\( K_t \)):
\[ K_t = \frac{1}{2} M v^2 \]
- Rotational kinetic energy (\( K_r \)):
\[ K_r = \frac{1}{2} I_G \omega^2 \]
- The relationship between translational velocity \( v \) and angular velocity \( \omega \) for pure rolling is:
\[ v = a \omega \]
- Substituting \( v \) into the translational kinetic energy:
\[ K_t = \frac{1}{2} M (a \omega)^2 = \frac{1}{2} M a^2 \omega^2 \]
- Therefore, the total initial kinetic energy \( K_i \) is:
\[ K_i = K_t + K_r = \frac{1}{2} M a^2 \omega^2 + \frac{1}{2} I_G \omega^2 \]

3. **Determine the Final Conditions:**
When the wheel comes to a stop at distance \( L \) up the incline, all kinetic energy is converted into potential energy:
- Potential energy (\( U_f \)) at height \( h \):
\[ U_f = Mgh = MgL \sin \theta \]
- Since the wheel comes to rest, the final kinetic energy \( K_f = 0 \).

4. **Apply Conservation of Energy:**
The initial kinetic energy equals the potential energy at the maximum height:
\[ K_i = U_f \]
\[ \frac{1}{2} M a^2 \omega^2 + \frac{1}{2} I_G \omega^2 = MgL \sin \theta \]

5. **Rearranging the Equation:**
Factor out \( \frac{1}{2} \omega^2 \):
\[ \frac{1}{2} \omega^2 \left( M a^2 + I_G \right) = MgL \sin \theta \]
Rearranging for \( L \):
\[ L = \frac{\frac{1}{2} \omega^2 (M a^2 + I_G)}{Mg \sin \theta} \]

6. **Final Expression for Distance \( L \):**
Thus, the distance traveled by the wheel up the inclined plane is:
\[ L = \frac{(M a^2 + I_G) \omega^2}{2Mg \sin \theta} \]