The rate of rotation of a solid disk with a radius of 2 m and mass of 2 kg constantly changes from 5 Hz to 9 Hz. If the change in rotational frequency occurs over 4 s, what torque was applied to the disk?

1 Answer
Jan 21, 2018

8pi mN ~~25 mN

Explanation:

The applied torque is
vectau=vecRxxvecF
The reaction to this torque is
vectau = I vec alpha

The magnitude of the torque is:

tau = I alpha

where I=1/2MR^2 and alpha= (omega_2- omega_1)/(Deltat)

because omega=2pi f

tau = I alpha= 1/cancel2MR^2*(cancel2pi(f_2-f_1))/(Deltat)

Substitute M=2 kg, R= 2m, f_2=9 Hz, f_1= 5Hz; Deltat =4s

tau = 2kg*(2m)^2*pi*((9-5)Hz)/(4s)= 8pi m*(kgm)/s^2~~25mN

To generate this toque, you need a force tau/R=(25mN)/(2m)= 12.5N