Angular Acceleration problem

  • Thread starter rlmurra2
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  • #1
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A uniform rod (mass=2.0 kg, length=.60m) is free to rotate about a frictionless pivot at one end. The rod is released from rests in the horizontal position. What is the magnitude of the angular acceleration of the rod at the instant it is 60 degrees below the horizontal?

Please help! I really need this one!
 

Answers and Replies

  • #2
OlderDan
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rlmurra2 said:
A uniform rod (mass=2.0 kg, length=.60m) is free to rotate about a frictionless pivot at one end. The rod is released from rests in the horizontal position. What is the magnitude of the angular acceleration of the rod at the instant it is 60 degrees below the horizontal?

Please help! I really need this one!
You should show us your attempt to solve this problem, then ask for help

https://www.physicsforums.com/showthread.php?t=4825

In case you have no idea where to get started, you need to calculate the moment of inertia of a rod pivoting at one end. You might find that by looking it up, or by using the parallel axis theorem after looking up the moment of inertia of a rod rotating about its center. Then you will need to calculate the torque acting on the rod about the pivot point. This torque comes from the force of gravity acting at the center of the rod. Then you will use "Newton's second law for rotation" to find the angular acceleration as the ratio of the torque to the moment of inertia.
 
  • #3
Astronuc
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Please show some work, such as the appropriate formula for calculating angular momentum as a function of the position (angle) of a rod pivoting at one end.
 

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