Question

obtain an expression relating the torque with
anguler acceleration for a rigid body( 12 th new syllabus 2020)
​

Answer 1

A system rotating with an angular momentum in presence of a torque suffers a change in the angular momentum and the rate of change of angular momentum is directly proportional to the torque acting on it.

If I be the moment of inertia of the system and ω be the angular velocity then angular momentum $\large\rm { L = I \omega}$

Rate of change of angular momentum provided the shape of system does not change is :

$\large\rm { \tau = \frac{d}{dt} ( I \omega)}$

$\large\rm { \ }$

$\large\rm { \:\:\:\:\;\:\:\:\:\:\;\:\:\:\: = I \frac{d\omega}{dt}}$

$\large\rm { \ }$

$\large\rm { \:\:\:\:\;\:\:\:\:\:\;\:\:\:\: = I \alpha}$

where alpha is the angular acceleration and τ is the torque

Additional Information

dimension of torque : $\large\rm { [ML^{2}T^{-2}]}$

Unit- Newton-metre ( N•m)

Answer 2

Answer:

A system rotating with an angular momentum in presence of a torque suffers a change in the angular momentum and the rate of change of angular momentum is directly proportional to the torque acting on it.

If I be the moment of inertia of the system and ω be the angular velocity then angular momentum