The speed of a homogenous solid sphere after rolling down an inclined plane of vertical height h, from rest without sliding is

Options

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

Explanation

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

### Step 1: Identify Initial Energy
When the sphere is at height \( h \), it has potential energy and no kinetic energy since it starts from rest. The potential energy (PE) at height \( h \) is given by:
\[ PE = mgh \]
where \( m \) is the mass of the sphere and \( g \) is the acceleration due to gravity.

### Step 2: Identify Final Energy
When the sphere reaches the bottom of the incline, its potential energy is converted into kinetic energy. The total kinetic energy (KE) consists of translational kinetic energy and rotational kinetic energy:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
For a solid sphere, the moment of inertia \( I \) about its center of mass is:
\[ I = \frac{2}{5} m r^2 \]
Since the sphere rolls without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ v = r \omega \quad \Rightarrow \quad \omega = \frac{v}{r} \]

### Step 3: Substitute Angular Velocity
Substituting \( \omega \) in the kinetic energy equation:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \left(\frac{1}{2} + \frac{1}{5}\right) mv^2 \]
Finding a common denominator (which is 10):
\[ KE = \left(\frac{5}{10} + \frac{2}{10}\right) mv^2 = \frac{7}{10} mv^2 \]

### Step 4: Apply Conservation of Energy
According to the conservation of energy, the initial potential energy equals the final kinetic energy:
\[ mgh = \frac{7}{10} mv^2 \]
Cancelling \( m \) from both sides:
\[ gh = \frac{7}{10} v^2 \]

### Step 5: Solve for \( v \)
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 speed of the homogeneous solid sphere after rolling down the inclined plane is:
\[ v = \sqrt{\frac{10}{7} gh} \]