The speed of a uniform spherical shell after rolling down an inclined plane of vertical height h from rest, is

Options

A. `sqrt((10" gh")/(7))`
B. `sqrt((6" gh")/(5))`
C. `sqrt((4" gh")/(5))`
D. `sqrt(2" gh")`

Explanation

To find the speed of a uniform spherical shell after rolling down an inclined plane of vertical height \( h \) from rest, we can use the principle of conservation of energy. Here’s the step-by-step solution:

### Step 1: Understand the Energy Conservation Principle
The total mechanical energy at the top of the incline (potential energy) will be equal to the total mechanical energy at the bottom of the incline (kinetic energy).

### Step 2: Write the Potential Energy at the Top
At the top of the incline, the potential energy (PE) is given by:
\[ PE = mgh \]
where \( m \) is the mass of the spherical shell, \( g \) is the acceleration due to gravity, and \( h \) is the vertical height.

### Step 3: Write the Kinetic Energy at the Bottom
At the bottom of the incline, the kinetic energy (KE) consists of both translational and rotational kinetic energy:
\[ KE = KE_{translational} + KE_{rotational} \]
The translational kinetic energy is:
\[ KE_{translational} = \frac{1}{2} mv^2 \]
The rotational kinetic energy for a uniform spherical shell is:
\[ KE_{rotational} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a uniform spherical shell, the moment of inertia \( I \) is:
\[ I = \frac{2}{3} mr^2 \]
Using the relationship between linear velocity \( v \) and angular velocity \( \omega \) (where \( v = r\omega \)), we can express \( \omega \) as:
\[ \omega = \frac{v}{r} \]
Thus, substituting \( \omega \) into the rotational kinetic energy gives:
\[ KE_{rotational} = \frac{1}{2} \left(\frac{2}{3} mr^2\right) \left(\frac{v^2}{r^2}\right) = \frac{1}{3} mv^2 \]

### Step 4: Combine the Kinetic Energies
Now, we can combine both kinetic energies:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{3} mv^2 = \left(\frac{3}{6} + \frac{2}{6}\right) mv^2 = \frac{5}{6} mv^2 \]

### Step 5: Set Potential Energy Equal to Kinetic Energy
Now, we set the potential energy equal to the total kinetic energy:
\[ mgh = \frac{5}{6} mv^2 \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{5}{6} v^2 \]

### Step 6: Solve for \( v^2 \)
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{6gh}{5} \]

### Step 7: Solve for \( v \)
Taking the square root of both sides gives:
\[ v = \sqrt{\frac{6gh}{5}} \]

### Final Answer
Thus, the speed of the uniform spherical shell after rolling down the inclined plane is:
\[ v = \sqrt{\frac{6gh}{5}} \]