Question

A circular disc of mass 10 kg and radius 0.2m is set into rotation about an axis passing through its centre and perpendicular to its plane by applying torque 10 Nm. Calculate angular velocity of the disc that it will attain at the end of 6 s from the rest. (Ans: 300 rad/s)

Answer 1

initial angular velocity,$\omega_0$=0
Let final angular velocity is $\omega$
we know, torque is rate of change in angular momentum,
so, $\tau=\frac{dL}{dt}$

$\tau dt = I(\omega-\omega_0)$

$\tau (t - 0) = I\omega$

$\tau t =\frac{1}{2}mr^2\omega$ [ as moment of inertia of disc about an axis passing through centre and perpendicular to its plane. ]

now, Put $\tau=10Nm$
m = 10kg, r = 0.2 m and t = 6s

10 × 6 = 1/2 × 10 × (0.2)² × $\omega$

$\omega=300 rad/s$

Answer 2

Answer:300 rad /s

Explanation:use the relation I* angular acceleration = torque .

Get angular acceleration and using that substitute in omega= initial angular velocity + alpha *time ....you will get omega that is angular velocity

Answer 3

Answer:300 rad /s

Explanation:use the relation I* angular acceleration = torque .

Get angular acceleration and using that substitute in omega= initial angular velocity + alpha *time ....you will get omega that is angular velocity