Calculate the Initial Calue of the Force at the Bearing

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SUMMARY

The discussion focuses on calculating the initial force exerted by a bearing on a uniform slender bar pivoted at one end and released from a horizontal position. The participant utilized the equation ƩMO = IGα + maGd, leading to the derivation of angular acceleration α = -mg(0.5L)/IG. By applying a normal and tangential coordinate system, the final force exerted by the bearing was determined to be 88.29 N.

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  • Understanding of dynamics and rotational motion
  • Familiarity with the equations of motion for rigid bodies
  • Knowledge of moment of inertia (IG) calculations
  • Ability to apply normal and tangential coordinate systems in physics problems
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Northbysouth
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Homework Statement



The uniform slender bar is pivoted at and swings freely in the vertical plane. If the bar is released from rest in the horizontal position, calculate the initial value of the force exerted by the bearing on the bar an instant after release.

I have attached an image of the question


Homework Equations





The Attempt at a Solution



I'm honestly not sure how to start this question. I'm pretty sure that this is a straightforward question

I think I may need to use the equation:

ƩMO = IGα + maGd

Any advice would be appreciated. Thank you
 

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Ahh, wait I've figure it out now.

I used the fact that:

ƩMO: IGα = -mg(L/2)

Solving for α = -mg(0.5L)/IG

where L is the length of the bar

Then I applied a normal and tangential coordinate system, with t pointing downwards and n pointing towards O.

ƩFt: mrα = Oy - mg

Plugging in the formula that I solved for α above I get

Oy = mr[-mg(0.5L)/Ig] + mg

My final answer was 88.29 N
 

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