An object with a mass of # 2 kg# is traveling in a circular path of a radius of #2 m#. If the object's angular velocity changes from # 3 Hz# to # 9 Hz# in # 1 s#, what torque was applied to the object?

1 Answer

Answer:

# 96pi Nm#

Explanation:

Comparing linear motion and Rotational motion for understanding

For Linear motion #-- #For rotational motion,

mass #-> # moment of Inertial

Force #-> # Torque

velocity #-> # Angular velocity

acceleration #-> # ANgular acceleration

So,

#F= ma# #-> # #-> # # tau = I alpha#

Here,

#alpha = (omega _2 -omega _1)/(Delta t) =(2pixxn_2-2pixxn_1)/(Deltat)=(2pi)xx( (9-3))/1 s^(-2) = 12pis^(-2) #

and
#I = mr^2 = 2kg*2^2 m^2 = 8 kgm^2#

So #tau = 8 kgm^2 * 12pis^(-2) = 96pi Nm#