The acceleration of the centre of mass of uniform solid disc rolling down an inclined plane of angle `alpha` is `xg sin alpha`. Find x

Explanation

To solve the problem of finding the value of \( x \) in the acceleration of the center of mass of a uniform solid disc rolling down an inclined plane, we can follow these steps:

### Step 1: Identify the forces acting on the disc
When the disc is rolling down the inclined plane, the forces acting on it include:
- The gravitational force \( mg \) acting downwards.
- The normal force \( N \) acting perpendicular to the inclined plane.
- The frictional force \( f \) acting up the incline.

### Step 2: Resolve the gravitational force
We can resolve the weight of the disc into two components:
- The component parallel to the incline: \( mg \sin \alpha \)
- The component perpendicular to the incline: \( mg \cos \alpha \)

### Step 3: Apply Newton's second law along the incline
According to Newton's second law, the net force acting on the disc along the incline can be expressed as:
\[ F_{\text{net}} = mg \sin \alpha - f \]
This net force is also equal to the mass of the disc multiplied by its acceleration \( a \):
\[ ma = mg \sin \alpha - f \]

### Step 4: Calculate the torque due to friction
The frictional force \( f \) creates a torque about the center of the disc. The torque \( \tau \) can be expressed as:
\[ \tau = f \cdot R \]
where \( R \) is the radius of the disc. The torque is also related to the angular acceleration \( \alpha_n \) of the disc by the equation:
\[ \tau = I \alpha_n \]
For a solid disc, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} m R^2 \]
Thus, we can write:
\[ f \cdot R = \frac{1}{2} m R^2 \alpha_n \]

### Step 5: Relate linear acceleration to angular acceleration
For pure rolling motion, the relationship between linear acceleration \( a \) and angular acceleration \( \alpha_n \) is given by:
\[ a = R \alpha_n \]
From this, we can express \( \alpha_n \) as:
\[ \alpha_n = \frac{a}{R} \]

### Step 6: Substitute \( \alpha_n \) into the torque equation
Substituting \( \alpha_n \) into the torque equation gives:
\[ f \cdot R = \frac{1}{2} m R^2 \left(\frac{a}{R}\right) \]
This simplifies to:
\[ f = \frac{1}{2} m a \]

### Step 7: Substitute \( f \) back into the force equation
Now, substituting \( f \) back into the force equation:
\[ ma = mg \sin \alpha - \frac{1}{2} ma \]
Rearranging gives:
\[ ma + \frac{1}{2} ma = mg \sin \alpha \]
\[ \frac{3}{2} ma = mg \sin \alpha \]

### Step 8: Solve for acceleration \( a \)
Now, we can solve for \( a \):
\[ a = \frac{2}{3} g \sin \alpha \]

### Step 9: Compare with the given form
The problem states that the acceleration is given as \( x g \sin \alpha \). By comparing:
\[ \frac{2}{3} g \sin \alpha = x g \sin \alpha \]
We find:
\[ x = \frac{2}{3} \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{\frac{2}{3}} \]