Question

An object with a mass of `2 kg` is traveling in a circular path of a radius of `4 m`. If the object's angular velocity changes from `1 Hz` to `5 Hz` in `2 s`, what torque was applied to the object?

Answer 1

The torque is `=402.1Nm`

Explanation:

The torque is the rate of change of angular momentum

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

where `I` is the moment of inertia

The mass of the object is `m=2kg`

The radius of the path is `r=4m`

For the object, `I=(mr^2)`

So, `I=2*(4)^2=32kgm^2`

The initial angular velocity is

`omega_0=2pif_0=2*pi*1=2pirads^-1`

The final angular velocity is

`omega_1=2pif_1=2*pi*5=10pirads^-1`

The time is `t=2s`

The angular acceleration is

`alpha=(omega_1-omega_0)/t=(10pi-2pi)/2=4pirads^-2`

So,

The torque is `tau= 32*4pi=402.1Nm`