To find the total kinetic energy of a uniform sphere rolling without slipping, we need to consider both its translational and rotational kinetic energy. Here’s the step-by-step solution:
### Step 1: Identify the mass and velocity
The mass of the sphere is given as \( m = 500 \, \text{g} = 0.5 \, \text{kg} \) (since \( 1 \, \text{kg} = 1000 \, \text{g} \)). The velocity of the center of mass is given as \( v = 0.02 \, \text{m/s} \).
### Step 2: Write the formula for total kinetic energy
The total kinetic energy \( K \) of a rolling object is the sum of its translational kinetic energy \( K_t \) and rotational kinetic energy \( K_r \):
\[
K = K_t + K_r
\]
Where:
- Translational kinetic energy \( K_t = \frac{1}{2} m v^2 \)
- Rotational kinetic energy \( K_r = \frac{1}{2} I \omega^2 \)
### Step 3: Calculate translational kinetic energy
Substituting the values into the translational kinetic energy formula:
\[
K_t = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.5 \, \text{kg} \times (0.02 \, \text{m/s})^2
\]
Calculating \( v^2 \):
\[
(0.02)^2 = 0.0004 \, \text{m}^2/\text{s}^2
\]
Now substituting back:
\[
K_t = \frac{1}{2} \times 0.5 \times 0.0004 = 0.0001 \, \text{J}
\]
### Step 4: Calculate rotational kinetic energy
For a solid sphere, the moment of inertia \( I \) about its center is given by:
\[
I = \frac{2}{5} m r^2
\]
Using the relationship for rolling without slipping, \( v = r \omega \), we can express \( \omega \) as:
\[
\omega = \frac{v}{r}
\]
Substituting \( \omega \) into the rotational kinetic energy formula:
\[
K_r = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right)
\]
This simplifies to:
\[
K_r = \frac{1}{5} m v^2
\]
Now substituting the values:
\[
K_r = \frac{1}{5} \times 0.5 \, \text{kg} \times 0.0004 = 0.00004 \, \text{J}
\]
### Step 5: Calculate total kinetic energy
Now, we can add the translational and rotational kinetic energy:
\[
K = K_t + K_r = 0.0001 \, \text{J} + 0.00004 \, \text{J} = 0.00014 \, \text{J}
\]
### Step 6: Final answer
Thus, the total kinetic energy of the rolling sphere is:
\[
K = 1.4 \times 10^{-4} \, \text{J}
\]
