An object with a mass of # 3 kg# is traveling in a circular path of a radius of #7 m#. If the object's angular velocity changes from # 3 Hz# to # 29 Hz# in #3 s#, what torque was applied to the object?

1 Answer
Mar 4, 2016

Answer:

Use the basics of rotation around a fixed axis. Remember to use #rads# for the angle.

#τ=2548π(kg*m^2)/s^2=8004,78(kg*m^2)/s^2#

Explanation:

The torque is equal to:

#τ=I*a_(θ)#

Where #I# is the moment of inertia and #a_(θ)# is the angular acceleration.

The moment of inertia:

#I=m*r^2#

#I=3kg*7^2m^2#

#I=147kg*m^2#

The angular acceleration:

#a_(θ)=(dω)/dt#

#a_(θ)=(d2πf)/dt#

#a_(θ)=2π(df)/dt#

#a_(θ)=2π(29-3)/3((rad)/s)/s#

#a_(θ)=52/3π(rad)/s^2#

Therefore:

#τ=147*52/3πkg*m^2*1/s^2#

#τ=2548π(kg*m^2)/s^2=8004,78(kg*m^2)/s^2#