- #1

noob314

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## Homework Statement

A uniform disk of mass M and radius R is pivoted about a point on its edge. The disk is released from rest when its center of mass is at the same height as the pivot. What is the reaction force exerted on the disk by the pivot the initial instant the disk is released? The instant the disk is at its lowest point?

## Homework Equations

[tex]I_{of disk, CM}= \frac{1}{2}MR^{2}[/tex]

## The Attempt at a Solution

Instant disk is released

I'm guessing [tex]\Sigma \vec{F}_{x} = 0[/tex] and since there are no horizontal forces acting on the disk, the reaction force of x ([tex]R_{x}[/tex]) = 0.

It looks like there are forces acting on y, so [tex]\Sigma \vec{F}_{y } = Mg + R_{y}[/tex]. Since it is not in a static equilibrium,[tex]\Sigma \vec{F}_{y }[/tex] is not 0. That's where I'm stuck. My textbook doesn't have anything about reaction forces, and my lecture only covered reaction forces in static equilibriums.

Instant disk is at its lowest point

I don't know how to start on this one. Obviously there is some force going towards the left. There is probably radial acceleration at the bottom, but I'm not sure about tangential. Since the tangential acceleration is going towards the left before it reaches the lowest point, would the tangential acceleration be 0 at the lowest point and then towards the right once it passes it?

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