The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height `h` from rest without slipping will be.

Options

A. `sqrt(10/7 gh)`
B. `sqrt(gh)`
C. `sqrt(6/5 gh)`
D. `sqrt(4/3 gh)`

Answer

To find the speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height \( h \) from rest without slipping, we can use the principles of energy conservation and the moment of inertia of the sphere. Here’s the step-by-step solution:

### Step 1: Understand the Energy Conservation Principle
When the sphere rolls down the incline from a height \( h \), its potential energy at the top is converted into kinetic energy at the bottom. The potential energy (PE) at the 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 2: Write the Kinetic Energy Expression
When the sphere rolls without slipping, its kinetic energy (KE) consists of two parts: translational kinetic energy and rotational kinetic energy. The total kinetic energy is given by:

\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]

For a solid sphere, the moment of inertia \( I \) is:

\[ I = \frac{2}{5} m r^2 \]

where \( r \) is the radius of the sphere. The angular velocity \( \omega \) is related to the linear velocity \( v \) by the equation:

\[ \omega = \frac{v}{r} \]

### Step 3: Substitute the Moment of Inertia and Angular Velocity
Substituting \( I \) and \( \omega \) into the kinetic energy equation gives:

\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \]

This simplifies to:

\[ KE = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \left(\frac{1}{2} + \frac{1}{5}\right) mv^2 \]

### Step 4: Combine the Kinetic Energy Terms
Finding a common denominator (which is 10) for the kinetic energy terms:

\[ KE = \left(\frac{5}{10} + \frac{2}{10}\right) mv^2 = \frac{7}{10} mv^2 \]

### Step 5: Set Potential Energy Equal to Kinetic Energy
According to the conservation of energy:

\[ mgh = \frac{7}{10} mv^2 \]

### Step 6: Cancel Out Mass and Solve for \( v^2 \)
Canceling \( m \) from both sides gives:

\[ gh = \frac{7}{10} v^2 \]

Rearranging this equation to solve for \( v^2 \):

\[ v^2 = \frac{10gh}{7} \]

### Step 7: Take the Square Root to Find \( v \)
Taking the square root of both sides gives:

\[ v = \sqrt{\frac{10gh}{7}} \]

### Final Answer
Thus, the speed of the homogeneous solid sphere after rolling down the inclined plane is:

\[ v = \sqrt{\frac{10gh}{7}} \]