Question

The angular momentum of wheel changes from 2L to 5Lin 3sec. What is the magnitude of the troque acting on it ​

Answer 1

Since the direction of torque is opposite to rotation, it opposes the initial rotation.

As such, the magnitude of angular momentum changing from 2L to 3L implies:

Initial momentum = 2L.

Final momentum = -3L (it has reversed direction due to torque.)

So, net change in momentum = (-3L) - 2L = -5L.

Average Torque = Constant Torque = -5L/5 sec = -L/sec.

Answer 2

The magnitude of the torque acting on it is equal to L.

Explanation:

It is given that,

Initial value of angular momentum of wheel, $L_i=2\ L$

Final value of angular momentum of wheel, $L_f=5\ L$

Time, t = 3 s

We know that the mathematical form of law of conservation of angular momentum is given by :

$\tau=\dfrac{dL}{dt}$

$\tau=\dfrac{L_f-L_i}{t}$

$\tau=\dfrac{5L-2L}{3}$

$\tau=L$

So, the magnitude of the torque acting on it is equal to L. Hence, this is the required solution.

Learn more,

Angular momentum

https://brainly.in/question/12849196