A thin hollow sphere of mass m is completely filled with a liquid of mass m. When the sphere rolls with a velocity v kinetic energy of the system is (neglect friction)

Options

A. `(1)/(2)mv^(2)`
B. `mv^(2)`
C. `(4)/(3)mv^(2)`
D. `(4)/(5) mv^(2)`

Explanation

To solve the problem of finding the total kinetic energy of a thin hollow sphere filled with liquid when it rolls with a velocity \( v \), we will consider both the translational and rotational kinetic energies of the system. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the components of kinetic energy
The total kinetic energy \( K \) of the system can be expressed as the sum of translational kinetic energy \( K_t \) and rotational kinetic energy \( K_r \):
\[ K = K_t + K_r \]

### Step 2: Calculate the translational kinetic energy
The translational kinetic energy of the system is given by the formula:
\[ K_t = \frac{1}{2} M v^2 \]
where \( M \) is the total mass of the system. In this case, the mass of the hollow sphere is \( m \) and the mass of the liquid inside is also \( m \), so:
\[ M = m + m = 2m \]
Thus, the translational kinetic energy becomes:
\[ K_t = \frac{1}{2} (2m) v^2 = mv^2 \]

### Step 3: Calculate the rotational kinetic energy
For a thin hollow sphere, the moment of inertia \( I \) is given by:
\[ I = \frac{2}{3} m r^2 \]
where \( r \) is the radius of the sphere. The angular velocity \( \omega \) of the sphere can be related to the linear velocity \( v \) by the equation:
\[ v = \omega r \quad \Rightarrow \quad \omega = \frac{v}{r} \]
The rotational kinetic energy is given by:
\[ K_r = \frac{1}{2} I \omega^2 \]
Substituting the expressions for \( I \) and \( \omega \):
\[ K_r = \frac{1}{2} \left(\frac{2}{3} m r^2\right) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ K_r = \frac{1}{2} \cdot \frac{2}{3} m r^2 \cdot \frac{v^2}{r^2} = \frac{1}{3} m v^2 \]

### Step 4: Combine the kinetic energies
Now, we can combine the translational and rotational kinetic energies:
\[ K = K_t + K_r = mv^2 + \frac{1}{3} mv^2 \]
Factoring out \( mv^2 \):
\[ K = mv^2 \left(1 + \frac{1}{3}\right) = mv^2 \left(\frac{3}{3} + \frac{1}{3}\right) = mv^2 \cdot \frac{4}{3} \]

### Step 5: Final result
Thus, the total kinetic energy of the system is:
\[ K = \frac{4}{3} mv^2 \]

### Conclusion
The correct answer is option 3: \( \frac{4}{3} mv^2 \).

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