# 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 themoment of inertiaand#omega# is theangular 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"#