A solid sphere rolls down a smooth inclined plane of height h. If it stats from rest then the speed of the sphere when it reaches the bottom is given by

Options

A. `sqrt(gh)`
B. `sqrt((10)/(7)gh)`
C. `sqrt((4)/(7)gh)`
D. `sqrt((5)/(4)gh)`

Explanation

To solve the problem of a solid sphere rolling down a smooth inclined plane of height \( h \), we can use the principles of energy conservation. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify Initial Energy
The initial energy of the solid sphere at the height \( h \) is purely gravitational potential energy, which can be expressed as:
\[ E_{\text{initial}} = mgh \]
where \( m \) is the mass of the sphere, \( g \) is the acceleration due to gravity, and \( h \) is the height of the inclined plane.

**Hint:** Remember that potential energy is calculated based on height and mass.

### Step 2: Identify Final Energy
When the sphere reaches the bottom of the inclined plane, all of the potential energy will have converted into kinetic energy. The kinetic energy of a rolling object consists of two parts: translational kinetic energy and rotational kinetic energy. The total kinetic energy \( E_{\text{final}} \) can be expressed as:
\[ E_{\text{final}} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

**Hint:** Kinetic energy has two components for rolling objects: translational and rotational.

### Step 3: Moment of Inertia and Angular Velocity
For a solid sphere, the moment of inertia \( I \) about its center is given by:
\[ I = \frac{2}{5} m r^2 \]
The relationship between linear velocity \( v \) and angular velocity \( \omega \) for rolling without slipping is:
\[ \omega = \frac{v}{r} \]

**Hint:** Use the relationship between linear and angular velocity for rolling objects.

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

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

### Step 5: Apply Conservation of Energy
According to the conservation of energy, the initial energy equals the final energy:
\[ mgh = \frac{7}{10} mv^2 \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{7}{10} v^2 \]

**Hint:** Use conservation of energy to equate initial and final energies.

### Step 6: Solve for \( v^2 \)
Rearranging the equation gives:
\[ v^2 = \frac{10}{7} gh \]

**Hint:** Isolate \( v^2 \) to find the expression for the speed.

### Step 7: Find \( v \)
Taking the square root of both sides, we find the speed \( v \) of the sphere at the bottom of the incline:
\[ v = \sqrt{\frac{10}{7} gh} \]

**Hint:** Remember to take the square root to find the final speed.

### Final Answer
Thus, the speed of the sphere when it reaches the bottom of the inclined plane is:
\[ v = \sqrt{\frac{10}{7} gh} \]