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

Answer:

#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"#