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 solve the problem, we need to determine the fraction of the total kinetic energy of a solid spherical ball that is associated with its rotation when it rolls without slipping down an inclined plane.

### Step 1: Understand the motion of the ball
The ball is rolling down the inclined plane without slipping. This means that there is a relationship between the linear velocity (v) of the center of mass of the ball and its angular velocity (ω):
\[ v = \omega r \]
where \( r \) is the radius of the ball.

### Step 2: Write the expressions for kinetic energy
The total kinetic energy (KE_total) of the ball consists of two parts: translational kinetic energy (KE_trans) and rotational kinetic energy (KE_rot).
- Translational Kinetic Energy:
\[ KE_{trans} = \frac{1}{2} mv^2 \]
- Rotational Kinetic Energy:
\[ KE_{rot} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the ball.

### Step 3: Calculate the moment of inertia for a solid sphere
For a solid sphere, the moment of inertia about its center of mass is given by:
\[ I = \frac{2}{5} m r^2 \]

### Step 4: Substitute the moment of inertia into the rotational kinetic energy formula
Now substituting the moment of inertia into the rotational kinetic energy formula:
\[ KE_{rot} = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \omega^2 \]
\[ KE_{rot} = \frac{1}{5} m r^2 \omega^2 \]

### Step 5: Substitute the relationship between v and ω
Using the relationship \( v = \omega r \), we can express \( \omega \) in terms of \( v \):
\[ \omega = \frac{v}{r} \]
Now substituting this into the rotational kinetic energy:
\[ KE_{rot} = \frac{1}{5} m r^2 \left(\frac{v}{r}\right)^2 \]
\[ KE_{rot} = \frac{1}{5} m r^2 \frac{v^2}{r^2} \]
\[ KE_{rot} = \frac{1}{5} mv^2 \]

### Step 6: Calculate the total kinetic energy
Now, we can calculate the total kinetic energy:
\[ KE_{total} = KE_{trans} + KE_{rot} \]
\[ KE_{total} = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 \]
To add these, we need a common denominator:
\[ KE_{total} = \frac{5}{10} mv^2 + \frac{2}{10} mv^2 \]
\[ KE_{total} = \frac{7}{10} mv^2 \]

### Step 7: Find the fraction of kinetic energy associated with rotation
Now, we can find the fraction of the total kinetic energy that is associated with rotation:
\[ \text{Fraction} = \frac{KE_{rot}}{KE_{total}} \]
\[ \text{Fraction} = \frac{\frac{1}{5} mv^2}{\frac{7}{10} mv^2} \]
The \( mv^2 \) cancels out:
\[ \text{Fraction} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7} \]

### Final Answer
The fraction of the total kinetic energy associated with rotation is \( \frac{2}{7} \).

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