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 # 3 Hz# to # 6 Hz# in # 5 s#, what torque was applied to the object?

Question

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 # 3 Hz# to #  6 Hz# in # 5 s#, what torque was applied to the object?

Isabella Alvarez · Accepted Answer

#tau=72pi/5# #tau=I* theta " torque"# #theta=(Delta omega)/(Delta t)# #theta=(2*pi*6-2.pi*3)# #theta=1(2 pi-6 pi)/5# #theta=(6 pi)/5# #I=m*r^2=3*2^2=12# #tau=12*(6 pi)/5# #tau=72pi/5#

Julia Aguilar · Answer

undefined To calculate the torque applied to the object, you can use the equation:

Torque = Moment of inertia * Angular acceleration

First, calculate the moment of inertia (I) using the formula for a point mass rotating about an axis at a distance (r):

I = m * r^2

where:
m = mass of the object (3 kg)
r = radius of the circular path (2 m)

I = 3 kg * (2 m)^2 = 12 kg*m^2

Next, calculate the angular acceleration (α) using the change in angular velocity (ω) and the time interval (t):

Angular acceleration (α) = (final angular velocity - initial angular velocity) / time

α = (6 Hz - 3 Hz) / 5 s = 0.6 Hz/s

Now, plug the values of moment of inertia and angular acceleration into the torque formula:

Torque = 12 kg*m^2 * 0.6 Hz/s = 7.2 N*m