Question

A body of mass $m$ and radius of gyration $K$ has an angular momentum $L$. Its angular velocity is

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

Hint:

Dimensional analysis is exceptionally powerful for simple rotational expressions. The unit of angular momentum is $\text{kg}\cdot\text{m}^2/\text{s}$ and the unit of moment of inertia ($mK^2$) is $\text{kg}\cdot\text{m}^2$. Dividing them cancels out the mass and length dimensions completely, leaving $\text{s}^{-1}$ (radians per second), which confirms the correct layout.

Answer 1

The Correct Option is C

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

Step 2: Key Formula or Approach:
1. The angular momentum $L$ of a body spinning with an angular velocity $\omega$ about a fixed axis is related to its moment of inertia $I$ by:
$$L = I\omega \implies \omega = \frac{L}{I}$$ 2. The moment of inertia $I$ of any body can be expressed in terms of its total mass $m$ and its radius of gyration $K$ as:
$$I = mK^2$$

Step 3: Detailed Explanation:
Substitute the fundamental expression for the moment of inertia ($I = mK^2$) directly into the angular momentum formula:
$$L = (mK^2)\omega$$ To solve for the angular velocity $\omega$, divide both sides of the equation by the term $mK^2$:
$$\omega = \frac{L}{mK^2}$$

Step 4: Final Answer:
The angular velocity of the body is $\frac{L}{mK^2}$, which maps perfectly to option (C).