Question

The moment of inertia of a body about the given axis, rotating with angular velocity $1 \text{ rad/s}$ is numerically equal to 'P' times its rotational kinetic energy. The value of 'P' is

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $2$
• D. $1$

Hint:

For any problem where a variable equals $1$, plug it in immediately to simplify your primary equations before trying to manipulate ratios or systems of equations. This keeps the algebra straightforward.

Answer 1

The Correct Option is C

Step 1: Read what is being asked.
A body spins about a fixed axis with angular speed $\omega = 1\ \text{rad/s}$. We are told its moment of inertia $I$ equals $P$ times its rotational kinetic energy $E_k$. Our job is to find the pure number $P$.
Step 2: Write the kinetic-energy formula.
For any rigid body rotating about an axis, the rotational kinetic energy is $E_k = \tfrac{1}{2} I \omega^2$. This is the only physics we need here.
Step 3: Put in the given angular speed.
Substitute $\omega = 1\ \text{rad/s}$, so $\omega^2 = 1$. Then $E_k = \tfrac{1}{2} I (1)^2 = \tfrac{1}{2} I$. So numerically the kinetic energy is just half the moment of inertia.
Step 4: Turn the words into an equation.
The statement $I = P \cdot E_k$ becomes $I = P \cdot \tfrac{1}{2} I$, because we just found $E_k = \tfrac{1}{2} I$.
Step 5: Solve for $P$.
Since the body is actually rotating, $I \neq 0$, so we can cancel $I$ from both sides: $1 = \tfrac{P}{2}$. Multiplying across gives $P = 2$.
Step 6: Sanity check.
When $\omega = 1$, the factor $\tfrac{1}{2}\omega^2$ becomes $\tfrac{1}{2}$, so $E_k$ is half of $I$, which means $I$ must be twice $E_k$. That confirms $P = 2$.
\[ \boxed{P = 2} \]