Question

An object's moment of inertia is 2.0kg?m2 . Its angular velocity is increasing at the rate of 3.8rad/s per second.

What is the net torque on the object?

Answer 1

we know,

torque= momentInertia*angularacceleration

=> torque= 2.0 x 3.8

=>torque =7.6 N-m

Answer 2

$Rate\,of\,change\,of\,angular\,velocity,\,i.e.\,angular\,acceleration=\frac{\mathrm{d} \omega}{\mathrm{d} t}= \alpha\\ \alpha =3.8\,rad/s^2\\ Moment\,of\,intertia \,,I=2\,kg\,m^2\\ Torque\,,\tau\,=I\,\alpha=7.6\,N\,m$

Answer 3

$\tau = I\alpha =I \frac{dw}{dt} = (2)(3.8) = 7.6 \ N.m$

Answer 4

here,

angular acceleration , a = 3.8rad/s^2

then

Torque = I * a = 2 * 3.8 = 7.6 N.m