The rotational kinetic energy of a body is E. In the absence of external torque, the mass of the body is halved and its radius of gyration is doubled. Its rotational kinetic energy is

Options

A. (a)0.5E
B. (b)0.25E
C. (c)E
D. (d)2E

Answer

To solve the problem, we need to analyze how the rotational kinetic energy changes when the mass of the body is halved and its radius of gyration is doubled. Let's go through the solution step by step.

### Step 1: Understand the formula for rotational kinetic energy
The rotational kinetic energy (E) of a body is given by the formula:
\[ E = \frac{1}{2} I \omega^2 \]
where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

### Step 2: Relate moment of inertia to mass and radius of gyration
The moment of inertia \(I\) can be expressed in terms of the mass \(m\) and the radius of gyration \(k\) as:
\[ I = m k^2 \]
Substituting this into the kinetic energy formula gives:
\[ E = \frac{1}{2} (m k^2) \omega^2 \]

### Step 3: Analyze the changes in mass and radius of gyration
According to the problem:
- The mass of the body is halved: \(m' = \frac{m}{2}\)
- The radius of gyration is doubled: \(k' = 2k\)

### Step 4: Calculate the new moment of inertia
Using the new mass and radius of gyration, the new moment of inertia \(I'\) becomes:
\[ I' = m' (k')^2 = \left(\frac{m}{2}\right) (2k)^2 = \frac{m}{2} \cdot 4k^2 = 2m k^2 \]

### Step 5: Determine the new rotational kinetic energy
The new rotational kinetic energy \(E'\) can be expressed as:
\[ E' = \frac{1}{2} I' \omega'^2 \]
Since angular momentum \(L = I \omega\) is conserved (no external torque), we have:
\[ L = I \omega = I' \omega' \]
This means:
\[ I \omega = I' \omega' \implies \omega' = \frac{I}{I'} \omega \]
Substituting \(I = mk^2\) and \(I' = 2mk^2\):
\[ \omega' = \frac{mk^2}{2mk^2} \omega = \frac{1}{2} \omega \]

### Step 6: Substitute \(\omega'\) into the new kinetic energy formula
Now substituting \(\omega' = \frac{1}{2} \omega\) into the expression for \(E'\):
\[ E' = \frac{1}{2} (2mk^2) \left(\frac{1}{2} \omega\right)^2 \]
\[ E' = \frac{1}{2} (2mk^2) \left(\frac{1}{4} \omega^2\right) = \frac{1}{2} \cdot 2mk^2 \cdot \frac{1}{4} \omega^2 = \frac{1}{4} mk^2 \omega^2 \]
Since \(E = \frac{1}{2} mk^2 \omega^2\), we can relate \(E'\) to \(E\):
\[ E' = \frac{1}{4} \cdot 2E = \frac{1}{2} E \]

### Final Answer
Thus, the new rotational kinetic energy \(E'\) is:
\[ E' = \frac{1}{2} E \]