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?

1 Answer
Aug 11, 2017

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"