Angular Acceleration of A rod with friction. Dynamics (General Plane)

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Discussion Overview

The discussion revolves around the dynamics of a slender rod being lifted at an angle, specifically focusing on calculating the angular acceleration of the rod when it is released. The context includes considerations of forces, moments, and friction acting on the rod.

Discussion Character

  • Homework-related, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant presents a scenario involving a 100 kg rod lifted to a 30-degree angle, detailing the forces and moments acting on it.
  • The participant expresses uncertainty about their calculations, particularly regarding the acceleration of the center of gravity and the resulting angular acceleration.
  • Another participant questions the clarity of the actual question being posed, indicating a need for more specificity.
  • A later reply suggests that the center of mass will experience acceleration in both the y-direction and the x-direction, implying a more complex motion than initially considered.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the initial calculations or the interpretation of the problem. Multiple viewpoints are presented regarding the dynamics of the rod and the nature of the forces involved.

Contextual Notes

Participants have not fully resolved the assumptions about the motion of the center of mass and the implications of friction on the angular acceleration calculation.

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Homework Statement


A 100 kg slender rod is lifted on the left end until the angle between the ground and the right end is 30 degrees. The right end is still in contact with the ground. The coefficient of friction between the rod and the ground is .5 and the length of the rod is 2.1 m.


Homework Equations



∑fx=m(ag)x
∑fy=m(ag)y
∑Mg=Ig∂
Ff=.5Fn


The Attempt at a Solution




I summed the forces in the x direction and got -Ff=m(ag)x
In the y direction I got -981+Fn=m(ag)y but i think the y acceleration of the center of gravity may be zero because of the fact that the bar can't translate up or down due to the contact with the ground. It can only rotate or slip horizontally on the ground. Finally i summed the moments about the center of gravity and got that fn(1.05cos(30))+Ff(1.05sin(30))=36.75∂

I got a final answer of 17.26 rad/s^2 counter clockwise, but i don't think it is correct. Any suggestions?
 
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What is the actual question?
 
paisiello2 said:
What is the actual question?


Find the angular acceleration at the instant it is released. So the angle is 30 degrees
 
The center of mass will accelerate in the y-direction as well as the x-direction.
 

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