A sphere starts rolling down can incline of inclination theta. Find the speed of its centre when it has covered a distance l.

Explanation

To find the speed of the center of a sphere rolling down an incline after covering a distance \( l \), we can follow these steps:

### Step 1: Understand the Geometry of the Problem
The sphere rolls down an incline with an angle \( \theta \). When it has traveled a distance \( l \) along the incline, the vertical height \( h \) from which it has descended can be calculated using the sine function:
\[ h = l \sin \theta \]

### Step 2: Apply the Work-Energy Theorem
The work-energy theorem states that the work done by the gravitational force is equal to the change in kinetic energy of the sphere. The gravitational force does work as the sphere descends, converting potential energy into kinetic energy.

The work done by gravity as the sphere descends a height \( h \) is:
\[ W = mgh = mg(l \sin \theta) \]

### Step 3: Write the Kinetic Energy Expression
The kinetic energy of the sphere consists of two parts: translational kinetic energy and rotational kinetic energy. The total kinetic energy \( K \) when the sphere is rolling is given by:
\[ K = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
For a solid sphere, the moment of inertia \( I \) is:
\[ I = \frac{2}{5} m r^2 \]
Also, since the sphere rolls without slipping, we have the relationship:
\[ v = \omega r \quad \text{or} \quad \omega = \frac{v}{r} \]

### Step 4: Substitute and Simplify
Substituting \( I \) and \( \omega \) into the kinetic energy expression:
\[ K = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ K = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2 \]

### Step 5: Set Work Equal to Kinetic Energy
Now, according to the work-energy theorem:
\[ mg(l \sin \theta) = \frac{7}{10} mv^2 \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g(l \sin \theta) = \frac{7}{10} v^2 \]

### Step 6: Solve for \( v \)
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{10}{7} g(l \sin \theta) \]
Taking the square root gives us the speed of the center of the sphere:
\[ v = \sqrt{\frac{10}{7} g l \sin \theta} \]

### Final Answer
Thus, the speed of the center of the sphere when it has covered a distance \( l \) down the incline is:
\[ v = \sqrt{\frac{10}{7} g l \sin \theta} \]

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