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 gh)/(7))`
B. `sqrt((6gh)/(5))`
C. `sqrt((4 gh)/(5))`
D. `sqrt(2gh)`

Explanation

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 follow these steps:

### Step 1: Understand the Energy Conservation Principle
Initially, the sphere is at rest at height \( h \). The potential energy (PE) at this height is given by:
\[ PE = mgh \]
where \( m \) is the mass of the sphere and \( g \) is the acceleration due to gravity.

### Step 2: Determine the Final Energy State
When the sphere reaches the bottom of the incline, all the potential energy will be converted into kinetic energy (KE). The kinetic energy consists of two components: translational kinetic energy and rotational kinetic energy.

The translational kinetic energy (KE_trans) is given by:
\[ KE_{\text{trans}} = \frac{1}{2} mv_{\text{cm}}^2 \]
where \( v_{\text{cm}} \) is the velocity of the center of mass.

The rotational kinetic energy (KE_rot) for a solid sphere is given by:
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a solid sphere, the moment of inertia \( I \) is:
\[ I = \frac{2}{5} m r^2 \]
and for rolling without slipping, the relationship between translational and rotational motion is:
\[ v_{\text{cm}} = r \omega \implies \omega = \frac{v_{\text{cm}}}{r} \]

### Step 3: Substitute for Rotational Kinetic Energy
Substituting \( \omega \) into the equation for rotational kinetic energy:
\[ KE_{\text{rot}} = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v_{\text{cm}}}{r}\right)^2 = \frac{1}{2} \cdot \frac{2}{5} m r^2 \cdot \frac{v_{\text{cm}}^2}{r^2} = \frac{1}{5} mv_{\text{cm}}^2 \]

### Step 4: Total Kinetic Energy
The total kinetic energy at the bottom of the incline is:
\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = \frac{1}{2} mv_{\text{cm}}^2 + \frac{1}{5} mv_{\text{cm}}^2 \]
Factoring out \( mv_{\text{cm}}^2 \):
\[ KE_{\text{total}} = m v_{\text{cm}}^2 \left(\frac{1}{2} + \frac{1}{5}\right) = m v_{\text{cm}}^2 \left(\frac{5}{10} + \frac{2}{10}\right) = m v_{\text{cm}}^2 \cdot \frac{7}{10} \]

### Step 5: Apply Conservation of Energy
Setting the initial potential energy equal to the total kinetic energy at the bottom:
\[ mgh = m v_{\text{cm}}^2 \cdot \frac{7}{10} \]
Canceling \( m \) from both sides:
\[ gh = v_{\text{cm}}^2 \cdot \frac{7}{10} \]

### Step 6: Solve for \( v_{\text{cm}} \)
Rearranging gives:
\[ v_{\text{cm}}^2 = \frac{10gh}{7} \]
Taking the square root:
\[ v_{\text{cm}} = \sqrt{\frac{10gh}{7}} \]

### Conclusion
Thus, the speed of the homogeneous solid sphere after rolling down the inclined plane is:
\[ v_{\text{cm}} = \sqrt{\frac{10gh}{7}} \]