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 `4 Hz` to `5 Hz` in `3 s`, what torque was applied to the object?

Answer 1

We will need to know the angular acceleration that existed during those 3 s. If this was a linear motion problem we could use the suvat formula
`v = u + a*t`
The equivalent formula in angular motion is
`omega_f = omega_i + alpha*t`

Using that last formula,
`5*(2*pi " radians")/s = 4*(2*pi " radians")/s + alpha*3 s`

Solving for `alpha`,
`alpha = ((2*pi " radians")/s)/(3 s) = 2.09 " radians"/s^2`

Next we need the angular motion equivalent of F=m*a (Newton's 2nd Law). The equivalent is

`tau = I*alpha`

I will assume the object can be considered a point mass. This requires that the dimensions of the object be much smaller than the 4 m radius. With that assumption, the rotational inertia, `I`, of the rotating mass is
`I = m*r^2 = 2 kg*(4 m)^2 = 32 kg*m^2`

So now we can calculate the torque, `tau`.

`tau = 32 kg*m^2 * 2.09 " radians"/s^2 = 67 kg*m^2/s^2`
`tau = 67 N*m`

You might ask how the combination of units `kg*m^2/s^2` simplified into `N*m`. Note that you can group `kg*m/s^2` apart from the other m. And `kg*m/s^2` is the definition of the Newton.

Torque is the product of force and the length of the lever arm. So, we have our answer.

I hope this helps,
Steve