Question

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?

Answer 1

`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