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Reaction force on disk by pivot 
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#1
Dec1009, 08:51 PM

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 [tex]I_{of disk, CM}= \frac{1}{2}MR^{2}[/tex] 3. 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? 


#2
Dec1109, 05:16 PM

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I am sure your textbook mentions somewhere (in one form or another) that action is equal and opposite to reaction and calls this Newton's Third law. So the reaction force exerted by the disk on the pivot is equal and opposite to the action force exerted by the pivot on the disk. That force has two components, call them F_{x} and F_{y}. Draw a standard free body diagram and say that
F_{net,x} = ma_{x} F_{net,y} = ma_{y} τ_{Net}=Iα Obviously, torques and the moment of inertia are to computed with respect to the pivot. The tangential acceleration is zero when the net torque is zero. The radial acceleration is zero when the center of mass moves in a straight line. 


#3
Dec1109, 09:32 PM

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 [tex]\Sigma \vec{F}_{x} = Ma_{x} = 0[/tex] with [tex]\vec{F}_{r, x} = 0[/tex] I'm assuming [tex]a_{x}[/tex] would be the radial acceleration, which would be 0 because the object would still be at rest, correct? [tex]\Sigma \vec{F}_{y} = Mg + R_{y} = Ma_{y}[/tex] So in this case, would [tex]a_{y}[/tex] be the tangential acceleration? For the instant the disk is at lowest point Would the radial acceleration be 3g/2? [tex]a_{r} = R\omega^{2}[/tex] [tex]U_{i} = K_{f}[/tex] [tex]= MgR = \frac{1}{2} I\omega^{2}[/tex] Because the pivot point is not at the center of mass, the parallel axis theorem would be used. [tex]= MgR = \frac{1}{2}(\frac{1}{3}MR^{2}+MR^{2})\omega^{2}[/tex] [tex]= MgR = \frac{2}{3}MR^{2}\omega^{2}[/tex] [tex]= \frac{3g}{2} = R\omega^{2} [/tex] [tex]a_{r} = R\omega^{2} = \frac{3g}{2}[/tex] 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. 


#4
Dec1209, 09:35 AM

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Reaction force on disk by pivot
As soon as the disk is released, the forces at the pivot generate no torque about the pivot. The only torque is generated by gravity. To find this torque, you need to find where it is applied with respect to the pivot. Can you tell where the external force of gravity acts on the disk?



#5
Dec1209, 11:05 AM

P: 19

Well, the torque from the external force of gravity would act on the center of mass of the disk, which is distance R from the pivot. That would give me to net torque on the object which would allow me to find the angular acceleration, and then the tangential acceleration. But I still don't see how that will help me find the reaction force or the radial acceleration.



#6
Dec1209, 09:31 PM

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Read posting #2. You have three equations and three unknowns, the acceleration and the two reaction forces. The radial acceleration is zero instantaneously as soon as the disk is released but the tangential acceleration is not. You can find the tangential acceleration from the angular acceleration. Once you know that, you can find the vertical reaction force. The horizontal force is (instantaneously) zero as soon as the disk is released.
You are on the right track for part (b) except that the moment of inertia of a disk about its CM is not (1/3)MR^{2}. 


#7
Dec1209, 10:10 PM

P: 19

Thanks, I understand it now.



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