Question

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

Answer 1

Answer:

0.915 Nm

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:

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

$\alpha = \frac{2.5^2}{2*2\pi} \approx 0.5 rad/s^2$

The moment of inertia of the solid disk is:

$I = \frac{1}{2}mR^2 = \frac{1}{2}19*0.44^2 = 1.8392 kgm^2$

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

The net torque applied is

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