The velocity of a sphere rolling down an inclined plane of height h at an inclination `theta` with the horizontal, will be :

Options

A. `(gh)/(1+k^(2)//r^(2))`
B. `(2gh)/(1+k^(2)//r^(2))`
C. `sqrt((gh)/(1+k^(2)//r^(2)))`
D. `sqrt((2gh)/(1+k^(2)//r^(2)))`

Answer

To find the velocity of a sphere rolling down an inclined plane of height \( h \) at an inclination \( \theta \) with the horizontal, we can use the work-energy theorem. Here’s a step-by-step solution:

### Step 1: Understand the System
When a sphere rolls down an inclined plane, it converts potential energy into kinetic energy. The potential energy at height \( h \) is given by \( mgh \), where \( m \) is the mass of the sphere and \( g \) is the acceleration due to gravity.

**Hint**: Identify the types of energy involved in the motion (potential and kinetic).

### Step 2: Apply the Work-Energy Theorem
According to the work-energy theorem, the work done by gravity equals the change in kinetic energy (both translational and rotational). The equation can be expressed as:
\[ mgh = \Delta KE_{translational} + \Delta KE_{rotational} \]
Since the sphere starts from rest, the initial kinetic energy is zero.

**Hint**: Remember that the total kinetic energy includes both translational and rotational components.

### Step 3: Write the Kinetic Energy Expressions
The translational kinetic energy (\( KE_{translational} \)) of the sphere is given by:
\[ KE_{translational} = \frac{1}{2} mv^2 \]
The rotational kinetic energy (\( KE_{rotational} \)) is given by:
\[ KE_{rotational} = \frac{1}{2} I \omega^2 \]
For a solid sphere, the moment of inertia \( I \) is \( \frac{2}{5} mr^2 \) and the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ v = r\omega \quad \Rightarrow \quad \omega = \frac{v}{r} \]

**Hint**: Recall the formulas for kinetic energy and the moment of inertia for a solid sphere.

### Step 4: Substitute the Expressions
Substituting \( \omega \) into the rotational kinetic energy expression:
\[ KE_{rotational} = \frac{1}{2} \left(\frac{2}{5} mr^2\right) \left(\frac{v}{r}\right)^2 = \frac{1}{5} mv^2 \]
Now, substituting both kinetic energy expressions back into the work-energy equation:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 \]

**Hint**: Combine the kinetic energy terms carefully.

### Step 5: Combine and Simplify
Combining the kinetic energy terms:
\[ mgh = \left(\frac{1}{2} + \frac{1}{5}\right) mv^2 = \left(\frac{5}{10} + \frac{2}{10}\right) mv^2 = \frac{7}{10} mv^2 \]
Now, cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{7}{10} v^2 \]

**Hint**: Make sure to isolate \( v^2 \) correctly.

### Step 6: Solve for Velocity
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{10gh}{7} \]
Taking the square root gives:
\[ v = \sqrt{\frac{10gh}{7}} \]

**Hint**: Remember to take the square root carefully to find the final expression for velocity.

### Final Result
The velocity of the sphere rolling down the inclined plane is:
\[ v = \sqrt{\frac{10gh}{7}} \]