A unifrom sphere of mass 500 g rolls without slipping on a plants horizontal surface with its center moving at a speed of `5.00cm//s` .It kinetic energy is :

Options

A. `6.25xx 10 ^(-4)J`
B. `8.75xx 10^(-4)J`
C. `8.75 xx 10^(-3)J`
D. `1.13xx 10^(-3)J`

Answer

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 a step-by-step solution:

### Step 1: Identify the mass and velocity
The mass \( m \) of the sphere is given as 500 g, which we convert to kilograms:
\[ m = 500 \, \text{g} = 0.5 \, \text{kg} \]
The velocity \( v \) of the center of the sphere is given as 5.00 cm/s, which we convert to meters per second:
\[ v = 5.00 \, \text{cm/s} = 5.00 \times 10^{-2} \, \text{m/s} \]

### Step 2: Calculate translational kinetic energy
The translational kinetic energy \( KE_{trans} \) is given by the formula:
\[ KE_{trans} = \frac{1}{2} mv^2 \]
Substituting the values:
\[ KE_{trans} = \frac{1}{2} \times 0.5 \, \text{kg} \times (5.00 \times 10^{-2} \, \text{m/s})^2 \]
Calculating \( (5.00 \times 10^{-2})^2 \):
\[ (5.00 \times 10^{-2})^2 = 25 \times 10^{-4} = 2.5 \times 10^{-3} \]
Now substituting this back into the equation:
\[ KE_{trans} = \frac{1}{2} \times 0.5 \times 2.5 \times 10^{-3} = 0.25 \times 2.5 \times 10^{-3} = 0.625 \times 10^{-3} \, \text{J} \]

### Step 3: Calculate rotational kinetic energy
For a sphere, the moment of inertia \( I \) is given by:
\[ I = \frac{2}{5} m r^2 \]
The angular velocity \( \omega \) can be related to the linear velocity \( v \) by:
\[ \omega = \frac{v}{r} \]
Thus, the rotational kinetic energy \( KE_{rot} \) is given by:
\[ KE_{rot} = \frac{1}{2} I \omega^2 \]
Substituting \( I \):
\[ KE_{rot} = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ KE_{rot} = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) = \frac{1}{5} mv^2 \]
Now substituting the values:
\[ KE_{rot} = \frac{1}{5} \times 0.5 \, \text{kg} \times (5.00 \times 10^{-2} \, \text{m/s})^2 \]
Using \( (5.00 \times 10^{-2})^2 = 2.5 \times 10^{-3} \):
\[ KE_{rot} = \frac{1}{5} \times 0.5 \times 2.5 \times 10^{-3} = 0.1 \times 2.5 \times 10^{-3} = 0.25 \times 10^{-3} \, \text{J} \]

### Step 4: Calculate total kinetic energy
Now, we can find the total kinetic energy \( KE_{total} \):
\[ KE_{total} = KE_{trans} + KE_{rot} \]
Substituting the values:
\[ KE_{total} = 0.625 \times 10^{-3} + 0.25 \times 10^{-3} = 0.875 \times 10^{-3} \, \text{J} \]
This can be expressed as:
\[ KE_{total} = 8.75 \times 10^{-4} \, \text{J} \]

### Final Answer
The total kinetic energy of the sphere is:
\[ \boxed{8.75 \times 10^{-4} \, \text{J}} \]