The angular momentum of a rotating body changes from A0 to 4A0 in 4 min. The torque acting on the body is:

Question

The angular momentum of a rotating body changes from A0 to 4A0 in 4 min. The torque acting on the body is: (A)  (3/4)A0  (B)  4A0  (C)  3A0  (D)  (3/2)A0

Tardigrade Admin · Accepted Answer

The rate of change of angular momentum (dJ) of a body is equal to the external torque

(τ)

acting upon the body.
Torque = rate of change of angular momentum or

τ = \frac{dJ}{d t}

Given,

dJ = 4 A_{0} - A_{0}, d t = 4 min

∴

τ = \frac{3}{4} A_{0}

This formula

\frac{dJ}{d t}

is similar to the formula for linear motion.

Q. The angular momentum of a rotating body changes from $A_{0}$ to $4 A_{0}$ in $4$ min. The torque acting on the body is:

$\frac{3}{4} A_{0}$

$4 A_{0}$

$3 A_{0}$

$\frac{3}{2} A_{0}$

Q. The angular momentum of a rotating body changes from A0​ to 4A0​ in 4 min. The torque acting on the body is:

43​A0​

4A0​

3A0​

23​A0​

The rate of change of angular momentum (dJ) of a body is equal to the external torque (τ) acting upon the body. Torque = rate of change of angular momentum or τ=dtdJ​ Given, dJ=4A0​−A0​,dt=4min ∴ τ=43​A0​ This formula dtdJ​ is similar to the formula for linear motion.

Q. The angular momentum of a rotating body changes from $A_{0}$ to $4 A_{0}$ in $4$ min. The torque acting on the body is:

$\frac{3}{4} A_{0}$

$4 A_{0}$

$3 A_{0}$

$\frac{3}{2} A_{0}$