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
Explanation:
I will assume that you meant that the frequency changes from
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 andomega is the angular velocity
Substituting this into equation
(2)" "tau=(d(Iomega))/(dt)
The angular velocity an be expressed by the equation:
omega=2pif
Substituting this into equation
(3)" "tau=(d(I*2pif))/(dt)
Because we are given that the frequency of the motion is changing we can move
(4)" "tau=2piI*(df)/(dt)
Finally, for a (
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
(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
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"