Question

A disc on a frictionless axle, starting from rest (0 rpm) can spin up to a rotation rate of 3820 rpm in a period of 2 seconds. (This is equivalent to angular acceleration of 200 rad/s.) Moment of inertia of disc = 5 kg-m. a) How much net torque was applied to the disc during the 2-s period? Answer: b) How much net torque would be needed to change the angular acceleration to 400 rad/s"? Answer: c) If the angular acceleration is 400 rad/s, how long will it take to spin up from 0 to 3820 rpm? Answer:

Answer 1

Answer:

1000 Nm

2000 Nm

1.00007 seconds

Explanation:

I = Moment of inertia = 5 kgm²

$\alpha$ = Angular acceleration

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

t = Time taken

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=5\times 200\\\Rightarrow \tau=1000\ Nm$

The torque of the disc would be 1000 Nm

If $\alpha=400\ rad/s^2$

$\tau=I\alpha\\\Rightarrow \tau=5\times 400\\\Rightarrow \tau=2000\ Nm$

The torque of the disc would be 2000 Nm

From equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{3820\times \frac{2\pi}{60}-0}{400}\\\Rightarrow t=1.00007\ s$

It would take 1.00007 seconds to reach 3820 rpm