Two objects, a ring and a spherical shell of same mass and radius are released from the top of two Identical inclined plane. If they are rolling without slipping, then ratio of speed of center of mass of the two objects when they will reach the bottom of the inclined plane is

Options

A. `sqrt 5 : sqrt 3 `
B. `sqrt 5 : sqrt 6 `
C. `sqrt 5 : sqrt 2 `
D. `sqrt 5 : 1 `

Answer

To find the ratio of the speeds of the center of mass of a ring and a spherical shell when they reach the bottom of an identical inclined plane, we can follow these steps:

### Step 1: Understand the Problem
We have two objects: a ring and a spherical shell, both with the same mass \( m \) and radius \( r \). They are released from the same height \( h \) on identical inclined planes and roll without slipping.

### Step 2: Identify the Moment of Inertia
The moment of inertia \( I \) for each object is given by:
- For the ring:
\[ I_{\text{ring}} = mr^2 \]
- For the spherical shell:
\[ I_{\text{shell}} = \frac{2}{3}mr^2 \]

### Step 3: Apply Conservation of Mechanical Energy
The total mechanical energy at the top (potential energy) is equal to the total mechanical energy at the bottom (kinetic energy).

At the top:
- Potential Energy (PE) = \( mgh \)
- Kinetic Energy (KE) = 0 (since they are at rest)

At the bottom:
- Potential Energy = 0
- Kinetic Energy = \( \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \)

Since they are rolling without slipping, we have the relation:
\[ \omega = \frac{v}{r} \]

### Step 4: Write the Energy Conservation Equation
Using the conservation of energy:
\[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \]
Substituting \( \omega = \frac{v}{r} \):
\[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{r}\right)^2 \]

### Step 5: Substitute Moment of Inertia
For the ring:
\[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2 \]
Thus:
\[ gh = v^2 \implies v = \sqrt{gh} \]

For the spherical shell:
\[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{3}mr^2\right)\left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ mgh = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = \frac{5}{6}mv^2 \]
Thus:
\[ gh = \frac{5}{6}v^2 \implies v^2 = \frac{6gh}{5} \implies v = \sqrt{\frac{6gh}{5}} \]

### Step 6: Calculate the Ratio of Speeds
Now we find the ratio of the speeds of the center of mass:
\[ \frac{v_{\text{ring}}}{v_{\text{shell}}} = \frac{\sqrt{gh}}{\sqrt{\frac{6gh}{5}}} = \sqrt{\frac{5}{6}} \]

### Final Answer
The ratio of the speeds of the center of mass of the ring to the spherical shell when they reach the bottom of the inclined plane is:
\[ \frac{v_{\text{ring}}}{v_{\text{shell}}} = \sqrt{\frac{5}{6}} \]