Question 476: Centre of Mass and Rotational Motion

Question

The angular momentum of a rotating body is ' $L$ '. When the frequency of rotating body is tripled and its kinetic energy is made one-third, the new angular momentum becomes

Accepted Answer

Given:
Angular momentum of a rotating body initially = $L$.
Frequency of rotation is tripled $(f 	o 3f)$.
Kinetic energy becomes one-third of its initial value.
We want: new angular momentum in terms of $L$.
Step 1: General relations
Angular velocity: $\omega = 2\pi f$.
Rotational KE: $K = 	frac{1}{2} I \omega^2$.
Angular momentum: $L = I\omega$.
Step 2: Initial conditions
Initial angular momentum: $L = I\omega$.
Initial KE: $K = 	frac{1}{2} I \omega^2$.
Step 3: After changes
New angular velocity: $\omega' = 3\omega$.
Now required KE is one third the original:
$ K' = 	frac{1}{3}K = 	frac{1}{3}\cdot 	frac{1}{2} I \omega^2. $
But also,
$ K' = 	frac{1}{2} I' (\omega')^2 = 	frac{1}{2} I'(3\omega)^2 = 	frac{9}{2} I' \omega^2. $
Equating,
$ 	frac{9}{2} I' \omega^2 = 	frac{1}{3}\cdot 	frac{1}{2} I \omega^2. $
Cancel $	frac{1}{2}\omega^2$:
$ 9I' = 	frac{1}{3} I, $
$ I' = 	frac{I}{27}. $
So, the moment of inertia must reduce.
Step 4: New angular momentum
$ L' = I' \omega' = \left(	frac{I}{27}ight)(3\omega) = 	frac{I\omega}{9}. $
But $I\omega = L$.
So,
$ L' = 	frac{L}{9}. $
Ã¢ÂÂ Final Answer
$ \boxed{	frac{1}{9}L} $
Correct Option: A

Answer

$\frac{1}{9} \mathrm{~L}$

Answer

$\frac{1}{3} \mathrm{~L}$

Answer

6 L

Answer

9 L

Centre of Mass and Rotational Motion

The angular momentum of a rotating body is ' L '. When the frequency of rotating body is tripled and its kinetic energy is made one-third, the new angular momentum becomes

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