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 and#omega# 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"#