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
[tex]\omega^2 - \omega_0^2 = 2\alpha\Delta \theta[/tex]
where [tex]\omega v = 2.5 rad/s is the final angular velocity of the disk, [tex]\omega_0[/tex] = 0 rad/s is the initial velocity of the can when it starts from rest, [tex]\Delta \theta[/tex] is the angular distance traveled, [tex]\alpha[/tex] is the angular acceleration of the disk, which we care looking for:
[tex]2.5^2 - 0 = 2*\alpha*2\pi[/tex]
[tex]\alpha = \frac{2.5^2}{2*2\pi} \approx 0.5 rad/s^2[/tex]
The moment of inertia of the solid disk is:
[tex]I = \frac{1}{2}mR^2 = \frac{1}{2}19*0.44^2 = 1.8392 kgm^2[/tex]
where m is the mass and R is the radius of the disk
The net torque applied is
[tex] T = \alpha*I = 0.5 * 1.8392 = 0.915 Nm[/tex]
