Determining revolution using Angular Motion

  • Thread starter petereater
  • Start date
  • #1

Homework Statement



The combination of an applied force and a friction force produces a constant total torque of 35.0 N · m on a wheel rotating about a fixed axis. The applied force acts for 5.90 s. During this time, the angular speed of the wheel increases from 0 to 10.1 rad/s. The applied force is then removed, and the wheel comes to rest in 60.4 s.

Find the total number of revolutions of the wheel during the entire interval of 66.3 s.

Homework Equations


t=66.3, angular acceleration= (10.1/66.3)
theta[final]=theta[initial] +( omega [initial]* t) + (.5*angular acceleration*t^2)

The Attempt at a Solution


I simple plugged and solved the relevant equation and converted theta[final] into revolutions by multiplying 2pi. There was no theta initial and omega initial is zero. Unfortunately, I didn't get the correct answer. I think my problem is that the angular acceleration maybe incorrect. If it is, then I don't understand why it isn't.
 

Answers and Replies

  • #2
23
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your angular acceleration looks like you took the second angular velocity and divided it by the entire time of motion.

Realize that angular acceleration can only occur if there is torque about that axis. The problem here is that you have two sources of torque: 1. applied force 2. friction. During the time interval, one of these forces is removed (which is why the system is able to come to a stop)

So you should have two angular accelerations: 1. one before the one of the forces is removed, and 2. the angular acceleration after the force is removed.

The only thing you need to decide is which force to remove for you calculation.
 

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