Dynamics: disc released horizontally, with pin fixed to edge (rotation)

In summary, the conversation discusses a first year dynamics question involving torque and force equations. The use of the parallel axis theorem is suggested and the steps to calculate the angular acceleration and acceleration of the disk's center of mass are outlined. The conversation then moves on to finding the vertical reaction of the pin and clarifies the concept of net force acting on the disk. It is concluded that the net force, including the reaction force at the pin, acts on the disk and does not necessarily have to be at the center.
  • #1
Aralox
5
0
Hi guys, I am really stumped by this 1st yr Dynamics question, and would appreciate any help on how to approach the question.

Homework Statement


q4KaJ.png



Homework Equations


Torque equations and Force equations
Moment of inertia for disc rotating around centre: (mr^2)/2
After using parallel axis theorem, the I about the edge is I = (3/2)mr^2

Thank you,
Aralox
 
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  • #2
Welcome to PF.
Aralox said:

Homework Equations


Torque equations and Force equations
Moment of inertia for disc rotating around centre: (mr^2)/2
After using parallel axis theorem, the I about the edge is I = (3/2)mr^2

Thank you,
Aralox
Using the parallel axis theorem is a good start. Can you do the following:

1. Given I, and the torque about the pivot point, can you calculate the angular acceleration of the disk?

2. From the angular acceleration about the pivot, you should be able to figure out the acceleration of the disk's center of mass.
 
  • #3
Thanks for the reply!

1. Given I, and the torque about the pivot point, can you calculate the angular acceleration of the disk?
I used Net Torque = mgr, as initially the weight is perp. to radius to pivot.
Since net torque = Iα (α is angular acc.),
mgr = Iα.
Since I = (3/2) mr^2, I simplified the previous equation to

α = 2g/3r, which happens to be a constant, so it looks like i have the initial angular acceleration of the disc about the pivot at the edge.


2. From the angular acceleration about the pivot, you should be able to figure out the acceleration of the disk's center of mass.
To find this, i used the equation a = αr,
so acceleration of disc's center: a = (2g/3r) * r = (2/3)g

How would i find the vertical reaction of the pin from this though?
Just brainstorming, could i do something where i took the disc to be not moving and the world rotating around it, meaning i would have the edge moving at a = (2/3)g, which i could get some kind of reaction force from?
This interpretation/application step is what I am having the most trouble with.


Thanks,
Aralox

PS: Just a test to figure out this 'latex reference' thing
[itex]\sum Torque[/itex]
 
  • #4
Looks good so far :approve:
Aralox said:
.
.
.
...so acceleration of disc's center: a = (2g/3r) * r = (2/3)g

How would i find the vertical reaction of the pin from this though?
Just brainstorming, could i do something where i took the disc to be not moving and the world rotating around it, meaning i would have the edge moving at a = (2/3)g, which i could get some kind of reaction force from?
This interpretation/application step is what I am having the most trouble with.
Well, now that you know the acceleration of the disk's center of mass, you pretty much know what the net force (vector sum of all forces) acting on the disk must be...
 
  • #5
Alrighty, so if i had:

[itex]\sum F[/itex] = N - mg = ma,
which leads to N = (5/3)mg

would N be the 'vertical component of reaction at the pin'? Isnt N the force (reaction force of some sort) vector originating at the centre of the disc, since I am taking net force at the centre?
 
  • #6
Just a little bump :)
 
  • #7
Sorry, somehow I missed your earlier post.
Aralox said:
Alrighty, so if i had:

[itex]\sum F[/itex] = N - mg = ma,
which leads to N = (5/3)mg
Well, a is downward, right? And you're taking the downward direction to be negative.

would N be the 'vertical component of reaction at the pin'?
Yes.
Isnt N the force (reaction force of some sort) vector originating at the centre of the disc, since I am taking net force at the centre?
No, N is acting at the pin. You're taking the net force acting on the disk, meaning the vector sum of all forces acting on the disk, whether they act at the center or not.
 
  • #8
Ohhhhh now i understand! I've always had this idea in my head that when taking the net force they must all act at the same point, but now i think about it, that can't be the case.

Thank you for all the help Redbelly98! :)
 

What is dynamics?

Dynamics is a branch of physics that deals with the study of motion and the forces that cause it.

What is a disc released horizontally with a pin fixed to the edge?

A disc released horizontally with a pin fixed to the edge refers to a scenario in which a disc-shaped object is released with one of its edges attached to a fixed pin, causing it to rotate as it falls.

What causes the disc to rotate in this scenario?

The disc rotates due to the force of gravity acting on its center of mass and the constraint of the fixed pin at one edge, causing it to pivot and rotate as it falls.

How does the rotation of the disc change over time?

As the disc falls, its rotational speed increases due to the torque produced by the force of gravity and the fixed pin. However, as it reaches the bottom of its fall, the rotational speed decreases due to the conversion of kinetic energy into potential energy.

What factors can affect the dynamics of this scenario?

The dynamics of this scenario can be affected by factors such as the mass and shape of the disc, the distance between the center of mass and the fixed pin, and the strength and direction of the force of gravity.

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