Question

A body of mass 'm' and radius of gyration 'K' has an angular momentum 'L'. Then its angular velocity is

• A. $\frac{L}{mK^2}$
• B. $\frac{mK^2}{L}$
• C. $\frac{K^2}{mL}$
• D. $\frac{mK^2}{L^2}$

Hint:

Think of this as a direct rotational analog to linear motion variables! Linear momentum is $p = mv \rightarrow v = \frac{p}{m}$. In rotation, replace momentum $p$ with $L$, and replace mass $m$ with its inertial counterpart $mK^2$ to instantly get $\omega = \frac{L}{mK^2}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks for an expression for the angular velocity ($\omega$) of a rigid rotating body in terms of its mass ($m$), radius of gyration ($K$), and total angular momentum ($L$).

Step 2: Key Formula or Approach:
The angular momentum $L$ of a rigid body rotating about a fixed axis with angular velocity $\omega$ is related to its moment of inertia $I$ by:
$$L = I\omega$$ The moment of inertia $I$ can also be expressed in terms of the body's mass and its radius of gyration $K$:
$$I = mK^2$$

Step 3: Detailed Explanation:
Substitute the radius of gyration expression for the moment of inertia ($I = mK^2$) directly into the angular momentum equation:
$$L = (mK^2)\omega$$ To isolate the angular velocity $\omega$, divide both sides of the equation by the entire mass-distribution term $(mK^2)$:
$$\omega = \frac{L}{mK^2}$$

Step 4: Final Answer:
The expression for angular velocity is $\frac{L}{mK^2}$, which perfectly matches option (A).