A uniform hollow sphere of mass 200 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is:

Options

A. `4.17xx10^(-4)J`
B. `8.72xx10^(-3)J`
C. `6.25xx10^(-4)J`
D. `3.61xx10^(-4)J`

Explanation

To calculate the kinetic energy of a uniform hollow sphere rolling without slipping, we need to consider both its translational and rotational kinetic energy. Here’s a step-by-step solution:

### Step 1: Identify the mass and speed
The mass \( m \) of the hollow sphere is given as 200 g, which we convert to kilograms:
\[ m = 200 \, \text{g} = 0.2 \, \text{kg} \]
The speed \( v_c \) of the center of mass is given as 5.00 cm/s, which we convert to meters per second:
\[ v_c = 5.00 \, \text{cm/s} = 0.05 \, \text{m/s} \]

### Step 2: Write the formula for kinetic energy
The total kinetic energy \( KE \) of the rolling sphere is the sum of its translational kinetic energy \( KE_{trans} \) and its rotational kinetic energy \( KE_{rot} \):
\[ KE = KE_{trans} + KE_{rot} \]
Where:
\[ KE_{trans} = \frac{1}{2} m v_c^2 \]
\[ KE_{rot} = \frac{1}{2} I \omega^2 \]

### Step 3: Calculate translational kinetic energy
Substituting the values into the translational kinetic energy formula:
\[ KE_{trans} = \frac{1}{2} \times 0.2 \, \text{kg} \times (0.05 \, \text{m/s})^2 \]
Calculating:
\[ KE_{trans} = \frac{1}{2} \times 0.2 \times 0.0025 = 0.00025 \, \text{J} \]

### Step 4: Calculate the moment of inertia for a hollow sphere
The moment of inertia \( I \) of a hollow sphere is given by:
\[ I = \frac{2}{3} m r^2 \]
Since we do not have the radius \( r \) directly, we will express \( \omega \) in terms of \( v_c \):
\[ \omega = \frac{v_c}{r} \]

### Step 5: Substitute \( \omega \) into the rotational kinetic energy formula
Now substituting \( \omega \) into the rotational kinetic energy formula:
\[ KE_{rot} = \frac{1}{2} I \left(\frac{v_c}{r}\right)^2 \]
Substituting \( I \):
\[ KE_{rot} = \frac{1}{2} \left(\frac{2}{3} m r^2\right) \left(\frac{v_c^2}{r^2}\right) = \frac{1}{3} m v_c^2 \]

### Step 6: Calculate rotational kinetic energy
Now we can calculate \( KE_{rot} \):
\[ KE_{rot} = \frac{1}{3} \times 0.2 \, \text{kg} \times (0.05 \, \text{m/s})^2 \]
Calculating:
\[ KE_{rot} = \frac{1}{3} \times 0.2 \times 0.0025 = \frac{0.0005}{3} \approx 0.0001667 \, \text{J} \]

### Step 7: Calculate total kinetic energy
Now we can sum the translational and rotational kinetic energies:
\[ KE = KE_{trans} + KE_{rot} = 0.00025 \, \text{J} + 0.0001667 \, \text{J} \approx 0.0004167 \, \text{J} \]

### Final Result
The total kinetic energy of the hollow sphere is approximately:
\[ KE \approx 4.17 \times 10^{-4} \, \text{J} \]