In pure rolling, fraction of its total energy associated with rotation is `alpha` for a ring and `beta` for a solid sphere. Then

Options

A. `alpha=(1)/(2)`
B. `alpha=(1)/(4)`
C. `beta=(2)/(5)`
D. `beta=(2)/(7)`

Answer

To solve the problem, we need to find the fractions of total energy associated with rotation for a ring and a solid sphere during pure rolling.

### Step-by-Step Solution:

#### Step 1: Analyze the Total Kinetic Energy of the Ring
The total kinetic energy (KE) of a ring in pure rolling motion can be expressed as the sum of translational kinetic energy (TKE) and rotational kinetic energy (RKE):

\[ KE_{\text{ring}} = KE_{\text{trans}} + KE_{\text{rot}} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]

For a ring, the moment of inertia \(I\) is given by \(I = mr^2\). Since it is in pure rolling motion, we have \(\omega = \frac{v}{r}\). Substituting these values, we get:

\[ KE_{\text{ring}} = \frac{1}{2} mv^2 + \frac{1}{2} (mr^2) \left(\frac{v}{r}\right)^2 \]

This simplifies to:

\[ KE_{\text{ring}} = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2 \]

#### Step 2: Calculate the Fraction of Energy Associated with Rotation for the Ring
The rotational kinetic energy for the ring is:

\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} (mr^2) \left(\frac{v}{r}\right)^2 = \frac{1}{2} mv^2 \]

Now, the fraction of total energy associated with rotation (\(\alpha\)) is:

\[ \alpha = \frac{KE_{\text{rot}}}{KE_{\text{total}}} = \frac{\frac{1}{2} mv^2}{mv^2} = \frac{1}{2} \]

#### Step 3: Analyze the Total Kinetic Energy of the Solid Sphere
For a solid sphere, the total kinetic energy in pure rolling motion is:

\[ KE_{\text{sphere}} = KE_{\text{trans}} + KE_{\text{rot}} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]

The moment of inertia \(I\) for a solid sphere is \(I = \frac{2}{5} mr^2\). Substituting this into the equation gives:

\[ KE_{\text{sphere}} = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} mr^2\right) \left(\frac{v}{r}\right)^2 \]

This simplifies to:

\[ KE_{\text{sphere}} = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \left(\frac{5}{10} + \frac{2}{10}\right) mv^2 = \frac{7}{10} mv^2 \]

#### Step 4: Calculate the Fraction of Energy Associated with Rotation for the Solid Sphere
The rotational kinetic energy for the solid sphere is:

\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{2}{5} mr^2\right) \left(\frac{v}{r}\right)^2 = \frac{1}{5} mv^2 \]

Now, the fraction of total energy associated with rotation (\(\beta\)) is:

\[ \beta = \frac{KE_{\text{rot}}}{KE_{\text{total}}} = \frac{\frac{1}{5} mv^2}{\frac{7}{10} mv^2} = \frac{2}{7} \]

### Final Results
- The fraction of total energy associated with rotation for the ring (\(\alpha\)) is \(\frac{1}{2}\).
- The fraction of total energy associated with rotation for the solid sphere (\(\beta\)) is \(\frac{2}{7}\).

### Conclusion
The values of \(\alpha\) and \(\beta\) are:
- \(\alpha = \frac{1}{2}\)
- \(\beta = \frac{2}{7}\)