# Reaction force on disk by pivot

by noob314
Tags: disk, force, pivot, reaction
 P: 19 1. The problem statement, all variables and given/known data 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? 2. Relevant equations $$I_{of disk, CM}= \frac{1}{2}MR^{2}$$ 3. The attempt at a solution Instant disk is released I'm guessing $$\Sigma \vec{F}_{x} = 0$$ and since there are no horizontal forces acting on the disk, the reaction force of x ($$R_{x}$$) = 0. It looks like there are forces acting on y, so $$\Sigma \vec{F}_{y } = Mg + R_{y}$$. Since it is not in a static equilibrium,$$\Sigma \vec{F}_{y }$$ 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? Attached Thumbnails
 P: 19 I'm starting to get frustrated. This was on an exam, but my professor refuses to give out the solutions or even the answers, so I can't even check if I'm close to the answer. This looks like it's suppose to be simple, but I'm just not getting it. For the instant the disk is released $$\Sigma \vec{F}_{x} = Ma_{x} = 0$$ with $$\vec{F}_{r, x} = 0$$ I'm assuming $$a_{x}$$ would be the radial acceleration, which would be 0 because the object would still be at rest, correct? $$\Sigma \vec{F}_{y} = Mg + R_{y} = Ma_{y}$$ So in this case, would $$a_{y}$$ be the tangential acceleration? For the instant the disk is at lowest point Would the radial acceleration be 3g/2? $$a_{r} = R\omega^{2}$$ $$U_{i} = K_{f}$$ $$= MgR = \frac{1}{2} I\omega^{2}$$ Because the pivot point is not at the center of mass, the parallel axis theorem would be used. $$= MgR = \frac{1}{2}(\frac{1}{3}MR^{2}+MR^{2})\omega^{2}$$ $$= MgR = \frac{2}{3}MR^{2}\omega^{2}$$ $$= \frac{3g}{2} = R\omega^{2}$$ $$a_{r} = R\omega^{2} = \frac{3g}{2}$$ I'm still confused as to whether the angular acceleration or net torque would be equal to 0. In order to find the angular acceleration, I need to find the net torque and vice versa, but I don't know how to find either at the lowest point.