Question

An object with a mass of `7 kg` is traveling in a circular path of a radius of `9 m`. If the object's angular velocity changes from `5 Hz` to `6 Hz` in `8 s`, what torque was applied to the object?

Answer 1

442.26 Nm

Explanation:

Torque acting on a particle is given as, `tau` = I`prop`
(Where, I is the moment if inertia and `prop` is angular acceleration)
Now,`I = m(r^2)`
Or, I= 567 SI unit
`prop` = `(domega) / dt`
= (`omega` final -`omega` initial)/`t` where `omega` is angular velocity
= `(2pi/t)` (n final - n initial) (as omega=2`pi`n) where n is frequency
= 0.78 `(rad)/(s^2)`
Hence, `tau` = `(567*0.78)`N.m
Or,442.26 N.m

Answer 2

The torque is `=445.3Nm`

Explanation:

The torque is the rate of change of angular momentum

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

where `I` is the moment of inertia

The mass of the object is `m=7kg`

The radius of the path is `r=9m`

For the object, the moment of inertia is `I=mr^2`

So, `I=7*(9)^2=567kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(f_2-f_1)/t*2pi=(6-5)/8*2pi`

`=(1/4pi)rads^(-2)`

So,

The torque is `tau= 567*1/4pi=445.3Nm`