Question

A flywheel in the form of a uniformly thick disk of radius 1.63 m, has a mass of 90.6 kg and spins counterclockwise at 291 rpm. What is the constant torque required to stop it in 4.25 min?

Answer 1

The torque is `=-14.44Nm`

Explanation:

The torque is the rate of change of angular momentum

`tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt=Ialpha`

`alpha` is the acceleration or deceleration

where `I` is the moment of inertia

For a solid disc, `I=(mr^2)/2`

The mass of the flywheel is `m=90.6 kg`

Thwe radius of the disc is `r=1.63m`

So, the moment of inertia is

`I=90.6*(1.63)^2/2=120.36kgm^2`

The initial angular velocity is `omega_i=2pi*n/60=2pi*291/60=30.47rads^-1`

The final angular velocity is `omega_f=0ms^-1`

Apply the equation to calculate `alpha`

`omega_f=omega_i+alphat`

`t=4.25*60=255s`

`alpha=-30.47/255=-0.12rads^-2`

Therefore,

The torque is

`tau=Ialpha=-120.36*0.12=-14.44Nm`