Reaction force on disk by pivot

In summary, 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. There are two components to that force, Fx and Fy. The radial acceleration is zero when the net torque is zero and the tangential acceleration is zero when the net torque is zero.
  • #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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  • #2
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 Fx and Fy. Draw a standard free body diagram and say that

Fnet,x = max
Fnet,y = may
τNet=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
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
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
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
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)MR2.
 
  • #7
Thanks, I understand it now.
 

Related to Reaction force on disk by pivot

1. What is the definition of reaction force on a disk by pivot?

The reaction force on a disk by pivot is the force exerted by the pivot on the disk in response to the force applied by the disk on the pivot. It is equal in magnitude and opposite in direction to the force applied by the disk and is responsible for keeping the disk in rotational equilibrium.

2. How is the reaction force on a disk by pivot calculated?

The reaction force on a disk by pivot can be calculated using the principle of moments, which states that the sum of all the clockwise moments is equal to the sum of all the anticlockwise moments. In this case, the moment due to the applied force is equal to the moment due to the reaction force, allowing us to calculate the magnitude of the reaction force.

3. What factors affect the reaction force on a disk by pivot?

The reaction force on a disk by pivot is affected by several factors, including the magnitude and direction of the applied force, the distance between the force and the pivot, and the weight and position of the disk. Additionally, the friction between the pivot and the disk can also affect the reaction force.

4. Why is the reaction force on a disk by pivot important in rotational equilibrium?

The reaction force on a disk by pivot is essential in maintaining rotational equilibrium because without it, the disk would either accelerate or decelerate in its rotation. This force ensures that the net torque acting on the disk is equal to zero, allowing it to remain in a state of static equilibrium.

5. Are there any real-life applications of the reaction force on a disk by pivot?

Yes, there are many real-life applications of the reaction force on a disk by pivot. One example is the use of a pivot in a seesaw or balance scale. The reaction force on the pivot helps to keep the seesaw or scale in balance, allowing for accurate measurements. Another application is in the steering mechanism of a car, where the reaction force on the pivot helps to turn the wheels and maintain the car's direction of motion.

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