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 follow these steps:

### Step 1: Understand the Problem
We have a wheel of mass \( M \), radius \( a \), and moment of inertia \( I_G \) about its center of mass. The wheel is rolling up an inclined plane with an angle \( \theta \) and an initial angular velocity \( \omega \). We need to find the distance \( S_0 \) that the wheel travels up the incline before coming to a stop.

### Step 2: Set Up the Coordinate System
Draw the inclined plane and identify the angle \( \theta \). The height \( h \) that the wheel rises can be expressed in terms of the distance \( S_0 \) traveled along the incline:
\[ h = S_0 \sin \theta \]

### Step 3: Analyze the Forces Acting on the Wheel
The forces acting on the wheel are:
- The gravitational force \( Mg \) acting downwards.
- The normal force \( N \) perpendicular to the incline.
- The frictional force \( f \) acting up the incline.

Since the wheel is rolling without slipping, the point of contact with the incline is momentarily at rest.

### Step 4: Apply Work-Energy Principle
The work done by gravity as the wheel moves up the incline is equal to the change in kinetic energy (both translational and rotational). The work done by gravity when the wheel rises a height \( h \) is:
\[ W = -Mgh = -Mg(S_0 \sin \theta) \]

The initial kinetic energy of the wheel consists of both translational and rotational components:
\[ KE_{\text{initial}} = \frac{1}{2} M v^2 + \frac{1}{2} I_G \omega^2 \]
Since the wheel is rolling without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ v = \omega a \]
Substituting this into the kinetic energy expression gives:
\[ KE_{\text{initial}} = \frac{1}{2} M (\omega a)^2 + \frac{1}{2} I_G \omega^2 = \frac{1}{2} \omega^2 (M a^2 + I_G) \]

### Step 5: Set Up the Energy Equation
Setting the work done by gravity equal to the change in kinetic energy, we have:
\[ -Mg(S_0 \sin \theta) = \frac{1}{2} \omega^2 (M a^2 + I_G) \]

### Step 6: Solve for \( S_0 \)
Rearranging the equation to solve for \( S_0 \):
\[ S_0 \sin \theta = -\frac{\frac{1}{2} \omega^2 (M a^2 + I_G)}{Mg} \]
\[ S_0 = -\frac{\frac{1}{2} \omega^2 (M a^2 + I_G)}{Mg \sin \theta} \]

Since the distance traveled cannot be negative, we take the absolute value:
\[ S_0 = \frac{\frac{1}{2} \omega^2 (M a^2 + I_G)}{Mg \sin \theta} \]

### Final Answer
The distance traveled by the wheel up the incline is:
\[ S_0 = \frac{\omega^2 (M a^2 + I_G)}{2Mg \sin \theta} \]