The rate of rotation of a solid disk with a radius of #6 m# and mass of #5 kg# constantly changes from #2 Hz# to #12 Hz#. If the change in rotational frequency occurs over #5 s#, what torque was applied to the disk?

1 Answer
Aug 9, 2017

Answer:

#tau~~1131"Nm"# (counterclockwise)

Explanation:

Torque can be expressed by the following equation:

#color(skyblue)(tau=Ialpha)#

where #I# is the moment of inertia of the object and #alpha# is its angular acceleration

Since a specific #I# is not provided for the disk, it will be assumed that it is a uniform disk which rotates about a center axis. Therefore, we can assume #color(darkblue)(I=1/2MR^2)#.

HyperPhysics

The average angular acceleration is given by:

#color(red)(alpha_(avg)=(Deltaomega)/(Deltat))#

where #omega# is the angular velocity of the disk and #Deltat# is the period over which the change occurs

Since #omega=2pif#, we can write the angular acceleration in terms of the frequency as:

#color(red)(alpha_(avg)=(2pi(f_f-f_i))/(Deltat))#

Putting all of the above together, we have the equation:

#color(darkblue)(tau=1/2MR^2*(2pi(f_f-f_i))/(Deltat))#

We are given the following information:

  • #|->"M"=5"kg"#
  • #|->"R"=6"m"#
  • #|->"f"_i=2"s"^-1#
  • #|->"f"_f=12"s"^-1#
  • #|->Deltat=5"s"#

Substituting these values into the equation we derived above, we have:

#tau=1/2(5"kg")(6"m")^2*(2pi(12"s"^-1-2"s"^-1)/(5"s"))#

#=1130.973"Nm"#

#~~1131"Nm"#

Note this occurs in the counterclockwise direction as the convention is that counterclockwise is positive.