When a sphere rolls without slipping the ratio of its kinetic energy of translation to its total kinetic energy is.

Options

A. `5//7`
B. `7//10`
C. `2//7`
D. none of these

Explanation

To find the ratio of the kinetic energy of translation to the total kinetic energy of a sphere rolling without slipping, we can follow these steps:

### Step 1: Define the Kinetic Energies
1. **Translational Kinetic Energy (TKE)**: The translational kinetic energy of the sphere is given by the formula:
\[ TKE = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the sphere and \( v \) is its translational velocity.

2. **Rotational Kinetic Energy (RKE)**: The rotational kinetic energy of the sphere is given by:
\[ RKE = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

### Step 2: Moment of Inertia for a Sphere
The moment of inertia \( I \) for a solid sphere is:
\[ I = \frac{2}{5} m r^2 \]
where \( r \) is the radius of the sphere.

### Step 3: Relate Angular Velocity to Translational Velocity
For a sphere rolling without slipping, the relationship between angular velocity \( \omega \) and translational velocity \( v \) is:
\[ \omega = \frac{v}{r} \]
Substituting this into the formula for rotational kinetic energy gives:
\[ RKE = \frac{1}{2} I \left(\frac{v}{r}\right)^2 = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) \]
This simplifies to:
\[ RKE = \frac{1}{5} m v^2 \]

### Step 4: Calculate Total Kinetic Energy
The total kinetic energy \( TKE_{total} \) is the sum of translational and rotational kinetic energies:
\[ TKE_{total} = TKE + RKE = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 \]
To combine these, we need a common denominator:
\[ TKE_{total} = \frac{5}{10} m v^2 + \frac{2}{10} m v^2 = \frac{7}{10} m v^2 \]

### Step 5: Find the Ratio
Now, we can find the ratio of the translational kinetic energy to the total kinetic energy:
\[ \text{Ratio} = \frac{TKE}{TKE_{total}} = \frac{\frac{1}{2} m v^2}{\frac{7}{10} m v^2} \]
The \( m v^2 \) terms cancel out:
\[ \text{Ratio} = \frac{\frac{1}{2}}{\frac{7}{10}} = \frac{1}{2} \times \frac{10}{7} = \frac{5}{7} \]

### Final Answer
Thus, the ratio of the kinetic energy of translation to the total kinetic energy when a sphere rolls without slipping is:
\[ \frac{5}{7} \]

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