Question

An object with a mass of `3 kg` is traveling in a circular path of a radius of `2 m`. If the object's angular velocity changes from `2 Hz` to `16 Hz` in `5 s`, what torque was applied to the object?

Answer 1

`tau=211"Nm"`

Explanation:

I will assume that you meant that the frequency changes from `2"Hz"` to `16"Hz"`.

Torque can be expressed as the rate of change of angular momentum.

`(1)" "color(darkblue)(tau=(dL)/(dt))`

Angular momentum can be described by the equation:

`L=Iomega`

where `I` is the moment of inertia and `omega` is the angular velocity

Substituting this into equation `(1)`, we have:

`(2)" "tau=(d(Iomega))/(dt)`

The angular velocity an be expressed by the equation:

`omega=2pif`

Substituting this into equation `3`, we get:

`(3)" "tau=(d(I*2pif))/(dt)`

Because we are given that the frequency of the motion is changing we can move `I` and `2pi` outside of the differential as constants and rewrite equation `(3)` as:

`(4)" "tau=2piI*(df)/(dt)`

Finally, for a (`~~`point) mass traveling a circular path about a center axis, the moment of inertia is given by `I=mr^2`.

Additionally, we will use the average rate of change, meaning that

`(df)/(dt)=(Deltaf)/(Deltat)=(f_f-f_i)/(t_f-t_i)`.

Substituting both of the above into equation `(4)`, we get:

`(5)" "color(darkblue)(tau=2pimr^2*(f_f-f_i)/(t_f-t_i)`

We are given the following information:

• `|->m=3"kg"`
• `|->r=2"m"`
• `|->t=5"s"`
• `|->f_i=2"s"^-1`
• `|->f_f=16"s"^-1`

Substituting these values into equation `(5)`, we get:

`tau=2pi(3"kg")(2"m")^2*(16"s"^-1-2"s"^-1)/(5"s"-0"s")`

`=>=24pi"kgm"^2*14/5s^-2`

`=>211.12"Nm"`

`:.tau~~211"Nm"`