Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `15 m`. If the object's angular velocity changes from `4 Hz` to `34 Hz` in `9 s`, what torque was applied to the object?

Answer 1

The torque is `=7068.6Nm`

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

For the object, `I=mr^2`

The mass is `m=3 kg`

The radius of the path is `r=15m`

So the moment of inertia is

,`I=3*(15)^2=675kgm^2`

The angular acceleration is

`alpha=(2pif_1-2pif_0)/t=(34-4)pi/9=10.47rads^-2`

So,

The torque is

`tau=Ialpha=675*10.47=7068.6Nm`