Question 497: Centre of Mass and Rotational Motion

Question

A body is rotating about its own axis. Its rotational kinetic energy is ' x ' and its angular momentum is ' $y$ '. Hence its moment of inertia about its own axis is

Accepted Answer

We are asked about rotational kinetic energy $x$, angular momentum $y$, and the moment of inertia $I$.
Step 1: Recall formulas
Rotational kinetic energy:
$ K = \frac{1}{2} I \omega^2 $
Angular momentum:
$ L = I \omega $
Here:
$ K = x, \quad L = y $
Step 2: Eliminate $\omega$
From angular momentum:
$ \omega = \frac{L}{I} = \frac{y}{I} $
Substitute into the kinetic energy:
$ x = 	frac{1}{2} I \left(\frac{y}{I}ight)^2 $
$ x = \frac{1}{2} I \cdot \frac{y^2}{I^2} $
$ x = \frac{1}{2} \frac{y^2}{I} $
Step 3: Solve for $I$
$ I = \frac{y^2}{2x} $
Final Answer:
The moment of inertia is
$ \boxed{\frac{y^2}{2x}} $
Ã¢ÂÂ This corresponds to
Option B
.

Answer

$\frac{x^2}{2 y}$

Answer

$\frac{y^2}{2 x}$

Answer

$\frac{x}{2 y}$

Answer

$\frac{y}{2 x}$

Centre of Mass and Rotational Motion

A body is rotating about its own axis. Its rotational kinetic energy is ' x ' and its angular momentum is ' y '. Hence its moment of inertia about its own axis is

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