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 solve the problem of finding the ratio of the speed of the center of mass of a ring and a spherical shell when they reach the bottom of an 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 rolling down identical inclined planes without slipping. We need to find the ratio of their speeds at the bottom of the incline.

### Step 2: Identify the moment of inertia
The moment of inertia (I) is crucial for rolling objects. For the two objects, we have:
- For the ring: \( I_{ring} = mR^2 \)
- For the spherical shell: \( I_{shell} = \frac{2}{3} mR^2 \)

### Step 3: Use the rolling motion equations
When an object rolls without slipping, the total mechanical energy at the top is converted into translational and rotational kinetic energy at the bottom. The equation for the speed \( v \) of the center of mass when rolling down an incline can be expressed as:

\[ v = \sqrt{\frac{2gh}{1 + \frac{I}{mR^2}}} \]

where \( h \) is the height of the incline.

### Step 4: Calculate the speed for the ring
For the ring, substituting the moment of inertia into the equation:

\[ v_{ring} = \sqrt{\frac{2gh}{1 + \frac{mR^2}{mR^2}}} = \sqrt{\frac{2gh}{1 + 1}} = \sqrt{\frac{2gh}{2}} = \sqrt{gh} \]

### Step 5: Calculate the speed for the spherical shell
For the spherical shell, substituting the moment of inertia:

\[ v_{shell} = \sqrt{\frac{2gh}{1 + \frac{\frac{2}{3}mR^2}{mR^2}}} = \sqrt{\frac{2gh}{1 + \frac{2}{3}}} = \sqrt{\frac{2gh}{\frac{5}{3}}} = \sqrt{\frac{6gh}{5}} \]

### Step 6: Find the ratio of speeds
Now, we find the ratio of the speeds of the center of mass of the ring and the spherical shell:

\[ \text{Ratio} = \frac{v_{ring}}{v_{shell}} = \frac{\sqrt{gh}}{\sqrt{\frac{6gh}{5}}} = \frac{\sqrt{gh}}{\sqrt{6gh/5}} = \frac{\sqrt{5}}{\sqrt{6}} = \sqrt{\frac{5}{6}} \]

### Conclusion
Thus, the ratio of the speeds of the center of mass of the ring to the spherical shell when they reach the bottom of the incline is:

\[ \text{Ratio} = \sqrt{5} : \sqrt{6} \]