Question

A 10 kg wheel is given acceleration of 20 rad / sec- by an applied torque 2Nm. Calculate
it's - a) Moment of inertia
b) Radius of gyration​

Answer 1

Explanation:

torque = to moment of inertia into ang acceleration

Answer 2

a) Moment of inertia , $I=0.1\ kg-m^2$

b) Radius of gyration​, r = 0.1 m

Explanation:

It is given that,

Mass of the wheel, m = 10 kg

Angular acceleration of the wheel, $\alpha =20\ rad/s$

Torque, $\tau=2\ N-m$

(a) The torque of wheel is given by the below formula as :

$\tau=I\times \alpha$

I is the moment of inertia

$I=\dfrac{\tau}{\alpha }$

$I=\dfrac{2}{20}$

$I=0.1\ kg-m^2$

(b) Let r is the radius of gyration. It is given by :

$I=mr^2$

$r=\sqrt{\dfrac{I}{m}}$

$r=\sqrt{\dfrac{0.1}{10}}$

r = 0.1 m

hence, this is the required solution.

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