Question

A 19 kg solid disk of radius0.44 m is rotated about an

axisthrough its center. If the disk accelerates from rest to an
angularspeed of 2.5 rad/s while rotating1.0 revolutions, what net
torque isrequired?

Answer 1

0.915 Nm

Step-by-step explanation:

1 revolution = 2π rad

We can use the following equation of motion to find out the acceleration acting on the disk

`\omega^2 - \omega_0^2 = 2\alpha\Delta \theta`

where
`\omega v = 2.5 rad/s is the final angular velocity of the disk, [tex]\omega_0` = 0 rad/s is the initial velocity of the can when it starts from rest,
`\Delta \theta` is the angular distance traveled,
`\alpha` is the angular acceleration of the disk, which we care looking for:

`2.5^2 - 0 = 2*\alpha*2\pi`

`\alpha = (2.5^2)/(2*2\pi) \approx 0.5 rad/s^2`

The moment of inertia of the solid disk is:

`I = (1)/(2)mR^2 = (1)/(2)19*0.44^2 = 1.8392 kgm^2`

where m is the mass and R is the radius of the disk

The net torque applied is

`T = \alpha*I = 0.5 * 1.8392 = 0.915 Nm`