# 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

#### Answer:

#### Explanation:

**Torque** can be expressed by the following equation:

#color(skyblue)(tau=Ialpha)# where

#I# is themoment of inertiaof the object and#alpha# is itsangularacceleration

Since a specific

The *average* angular acceleration is given by:

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

#omega# is theangularvelocity of the disk and#Deltat# is the period over which the change occurs

Since

#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.