A solid spherical ball is rolling without slipping down an inclined plane. The fraction of its total kinetic energy associated with rotation is

Options

A. `(2)/(5)`
B. `(2)/(7)`
C. `(3)/(5)`
D. `(3)/(7)`

Answer

To find the fraction of the total kinetic energy associated with rotation for a solid spherical ball rolling without slipping down an inclined plane, we can follow these steps:

### Step 1: Understand the Kinetic Energy Components
The total kinetic energy (K_total) of the rolling ball consists of two parts:
1. Translational Kinetic Energy (K_trans)
2. Rotational Kinetic Energy (K_rot)

The formulas for these energies are:
- Translational Kinetic Energy: \( K_{trans} = \frac{1}{2} mv^2 \)
- Rotational Kinetic Energy: \( K_{rot} = \frac{1}{2} I \omega^2 \)

### Step 2: Relate Linear and Angular Velocity
For a ball rolling without slipping, the relationship between linear velocity (v) and angular velocity (ω) is given by:
\[ v = r \omega \]
where r is the radius of the ball.

### Step 3: Moment of Inertia for a Solid Sphere
The moment of inertia (I) for a solid sphere about its center is:
\[ I = \frac{2}{5} m r^2 \]

### Step 4: Substitute for K_rot
We can express the rotational kinetic energy in terms of v:
\[ K_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \omega^2 \]
Substituting \( \omega = \frac{v}{r} \):
\[ K_{rot} = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \]
\[ K_{rot} = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \frac{v^2}{r^2} = \frac{1}{5} mv^2 \]

### Step 5: Substitute for K_total
Now, we can express the total kinetic energy:
\[ K_{total} = K_{trans} + K_{rot} = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 \]
Finding a common denominator:
\[ K_{total} = \frac{5}{10} mv^2 + \frac{2}{10} mv^2 = \frac{7}{10} mv^2 \]

### Step 6: Calculate the Fraction of K_rot to K_total
Now we can find the fraction of the total kinetic energy that is associated with rotation:
\[ \text{Fraction} = \frac{K_{rot}}{K_{total}} = \frac{\frac{1}{5} mv^2}{\frac{7}{10} mv^2} \]
The \( mv^2 \) terms cancel out:
\[ \text{Fraction} = \frac{1/5}{7/10} = \frac{1}{5} \cdot \frac{10}{7} = \frac{2}{7} \]

### Final Answer
Thus, the fraction of the total kinetic energy associated with rotation is:
\[ \frac{2}{7} \]

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