A sphere rolls down an inclined plane through a height h. Its velocity at the bottom would be

Options

A. `sqrt(2 gh)`
B. `sqrt(7/10 gh)`
C. `sqrt(10/7 gh)`
D. `(sqrt(10/7))gh`

Answer

To find the velocity of a sphere rolling down an inclined plane through a height \( h \), we can use the principles of energy conservation. Here’s a step-by-step solution:

### Step 1: Understand the Energy Conservation Principle
When the sphere rolls down the inclined plane, its potential energy at the height \( h \) will convert into kinetic energy at the bottom of the incline. The total mechanical energy is conserved.

### Step 2: Write the Expression for Potential Energy
The potential energy (PE) at height \( h \) is given by:
\[ PE = mgh \]
where \( m \) is the mass of the sphere, \( g \) is the acceleration due to gravity, and \( h \) is the height.

### Step 3: Write the Expression for Kinetic Energy
The kinetic energy (KE) of the sphere at the bottom consists of two parts: translational kinetic energy and rotational kinetic energy.
- Translational Kinetic Energy (TKE):
\[ TKE = \frac{1}{2} mv^2 \]
- Rotational Kinetic Energy (RKE):
\[ RKE = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the sphere and \( \omega \) is the angular velocity.

### Step 4: Find the Moment of Inertia and Relate Linear and Angular Velocity
For a solid sphere, the moment of inertia \( I \) is:
\[ I = \frac{2}{5} m r^2 \]
The relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ \omega = \frac{v}{r} \]
Substituting \( \omega \) into the RKE expression gives:
\[ RKE = \frac{1}{2} \left( \frac{2}{5} m r^2 \right) \left( \frac{v}{r} \right)^2 = \frac{1}{5} mv^2 \]

### Step 5: Combine Kinetic Energies
Now, we can combine the translational and rotational kinetic energies:
\[ KE = TKE + RKE = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 \]
Finding a common denominator:
\[ KE = \left( \frac{5}{10} mv^2 + \frac{2}{10} mv^2 \right) = \frac{7}{10} mv^2 \]

### Step 6: Set Potential Energy Equal to Kinetic Energy
Setting the potential energy equal to the total kinetic energy gives:
\[ mgh = \frac{7}{10} mv^2 \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{7}{10} v^2 \]

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

### Final Answer
Thus, the velocity of the sphere at the bottom of the incline is:
\[ v = \sqrt{\frac{10}{7} gh} \]