What are the Rotational Dynamics of a Thin Rod Held by a Pivot Point and String?

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SUMMARY

The discussion focuses on the rotational dynamics of a thin rod of length l and mass m, fixed at one end by a pivot and supported by a string attached at 3/4l. The moment of inertia for the rod is given as ml²/3. Key calculations include determining the pivot point's force on the rod, the angular acceleration immediately after the string is cut, and the transverse accelerations of both the center of mass and the far end of the rod. The torque exerted by the pivot point is calculated as 0.25mgl, although there is confusion regarding the pivot's ability to exert torque.

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  • Study the derivation of moment of inertia for various shapes, focusing on thin rods
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rotational dynamics rod--please help!

Homework Statement


A thin rod of length l and mass m is fixed at a far ends by a free pivot and held from rotating with a string attached at 3/4l and fixed toa ceiling. The moment of inertia of a thin rod about a far end is ml2/3

a) what is the pivot point exerting on the rod

The rope is cut at t =0 seconds. at t =0 seconds find:
b) the angular acceleration of the rod.
c) the transverse acceleration of the rod's center of mass.
d) the transverse aceleration of point A (the far end of the rod)
e) the force exerted by the pivot point

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You have to at least try the problem and explain your thoughts before asking for help.
 


First, total downward torque = upward torque.
.5mgl + Torque of pivot point = .75mgl
torque of pivot point = .25mgl.
This doesn't make sense because the pivot point can't exert a torque.
Thats all I got to...and I couldn't figure out anthing else lol
I have all the answers, I just need to understand hwo to do it.
 

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