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Forces Concerning a Rectangular Prism

by hyperddude
Tags: forces, prism, rectangular
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hyperddude
#1
Jan3-13, 08:30 PM
P: 15
1. The problem statement, all variables and given/known data
I'm wondering this for any object with moment of inertia I, but I'll ask this question for a rectangle for simplicity and I'm sure I can extend it to general objects.

Say we have a rectangular object (with mass m, height h, and width L) that is attached to a wall by some sort of point support exactly at its corner. If we hold it horizontally then let it go and swing, what is the force that the support exerts on the ruler from right after we let go as it swings? Here's a picture:




2. Relevant equations
Moment of inertia of this rectangle: [itex]\frac{m(h^2 + w^2)}{12} + m((h/2)^2+(w/2)^2) = \frac{m(h^2+w^2)}{3}[/itex]


3. The attempt at a solution
I was thinking about concentrating the mass into a point mass at the location of the rectangle's center of mass. This would be like a simple pendulum with starting position at angle [itex]θ[/itex] below the horizontal, where [itex]θ=\tan^{-1}(h/w)[/itex]. There would be a force of mg downward, which has a component of mg*sin(θ) in the direction of swinging and a component of mg*cos(θ) pulling on the "string" of the pendulum. The force that the support supplies, in return, would be opposite to and equal in magnitude to this pull. I'm not so sure this is correct though...
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Nugatory
#2
Jan3-13, 11:47 PM
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Try calculating the torque around the pivot point.
hyperddude
#3
Jan3-13, 11:53 PM
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Quote Quote by Nugatory View Post
Try calculating the torque around the pivot point.
To do this, I'm thinking that the "lever arm" is the the line segment from the pivot point to the center of mass. The force due to gravity perpendicular to this lever arm is mg*sin(θ), as stated in my first post. So, the torque should be [itex]L(mg\sin{θ})[/itex] where [itex]L = \sqrt{h^2 + w^2}[/itex] and [itex]θ=\tan^{-1}(h/w)[/itex].

How does this help me find the force exerted by the support at the pivot point though?

Studiot
#4
Jan4-13, 06:59 AM
P: 5,462
Forces Concerning a Rectangular Prism

Say we have a rectangular object (with mass m, height h, and width L) that is attached to a wall by some sort of point support exactly at its corner.
Won't the edge foul the wall?

You say it has moment of inertia I so do you understand the calculation of moments of inertia?

Nugatory has offered a method that uses the same approach.
hyperddude
#5
Jan4-13, 12:11 PM
P: 15
Woops, [itex]θ=\tan^{-1}(w/h)[/itex], not [itex]h/w[/itex].

With the moment of inertia, I can figure out the motion of this rectangle. How does that relate to the force that the pivot supplies? Does it supply a force equal to the component of the force of gravity on the rectangle, that is perpendicular to the component in the direction of the swinging motion?
haruspex
#6
Jan4-13, 03:56 PM
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Quote Quote by hyperddude View Post
How does that relate to the force that the pivot supplies? Does it supply a force equal to the component of the force of gravity on the rectangle, that is perpendicular to the component in the direction of the swinging motion?
A component of the pivot's force will certainly do that. What we know about the initial motion of the centre of mass of the ruler is that it will maintain a constant distance from the pivot. The net force in that radial (diagonal) direction is therefore zero. But the force from the pivot will also have a component perpendicular to that radial line. Together with the corresponding component from gravity this will provide the torque necessary for the angular acceleration. Relating the angular acceleration to the linear one should give you enough equations to solve it.
Calculating the torque around the pivot point also works. That gives you the angular acceleration, and the linear acceleration to add to that must be that which holds the pivot point still. Then you can figure out what force at the pivot point when combined with gravity produces that linear acceleration. The two methods are equivalent.
haruspex
#7
Jan4-13, 03:58 PM
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Quote Quote by Studiot View Post
Won't the edge foul the wall?
Presumably the wall is behind the ruler, not next to it.
hyperddude
#8
Jan4-13, 11:59 PM
P: 15
Quote Quote by haruspex View Post
A component of the pivot's force will certainly do that. What we know about the initial motion of the centre of mass of the ruler is that it will maintain a constant distance from the pivot. The net force in that radial (diagonal) direction is therefore zero. But the force from the pivot will also have a component perpendicular to that radial line. Together with the corresponding component from gravity this will provide the torque necessary for the angular acceleration. Relating the angular acceleration to the linear one should give you enough equations to solve it.
Calculating the torque around the pivot point also works. That gives you the angular acceleration, and the linear acceleration to add to that must be that which holds the pivot point still. Then you can figure out what force at the pivot point when combined with gravity produces that linear acceleration. The two methods are equivalent.
So does the linear acceleration (from the pivot, I assume) contribute to the angular acceleration from gravity? I don't quite see how they are related.
haruspex
#9
Jan5-13, 12:58 AM
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Quote Quote by hyperddude View Post
So does the linear acceleration (from the pivot, I assume) contribute to the angular acceleration from gravity? I don't quite see how they are related.
There isn't "linear acceleration from the pivot". There's a force from the pivot. The ruler as a whole undergoes some angular acceleration α. The centre also undergoes a linear acceleration a. The instantaneous linear acceleration of any other point on the ruler is then given by the vector expression a+αr, where r is the vector position relative to the centre. At the pivot, this must equate to 0.
The two aspects of movement result from two forces, gravity and the force at the pivot, acting together.
I didn't really understand your question, so it's hard to guess whether I've answered it.
hyperddude
#10
Jan5-13, 01:17 AM
P: 15
Quote Quote by haruspex View Post
There isn't "linear acceleration from the pivot". There's a force from the pivot. The ruler as a whole undergoes some angular acceleration α. The centre also undergoes a linear acceleration a. The instantaneous linear acceleration of any other point on the ruler is then given by the vector expression a+αr, where r is the vector position relative to the centre. At the pivot, this must equate to 0.
The two aspects of movement result from two forces, gravity and the force at the pivot, acting together.
I didn't really understand your question, so it's hard to guess whether I've answered it.
My question was what the force from the pivot was. You mentioned in a previous post that there would be a component acting in the radial direction (thus making the net force in the radial direction 0). What about perpendicular to the radius?

The instantaneous linear acceleration of any other point on the ruler is then given by the vector expression a+αr, where r is the vector position relative to the centre. At the pivot, this must equate to 0.
Does that mean the component of the pivot's force perpendicular to the radius is 0?
Basic_Physics
#11
Jan5-13, 09:27 AM
P: 358
You are basically considering a physical pendulum here. You do have circular motion of the cm about the pivot point. It seems you are particularly interested in the radial acceleration of the pendulum which will then enable you to evaluate the force.
hyperddude
#12
Jan5-13, 12:44 PM
P: 15
Quote Quote by Basic_Physics View Post
You are basically considering a physical pendulum here. You do have circular motion of the cm about the pivot point. It seems you are particularly interested in the radial acceleration of the pendulum which will then enable you to evaluate the force.
By radial, you mean in the direction of the radius, right? Is that the only thing I need to consider in finding the force by the pivot?
haruspex
#13
Jan5-13, 02:46 PM
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Quote Quote by hyperddude View Post
My question was what the force from the pivot was. You mentioned in a previous post that there would be a component acting in the radial direction (thus making the net force in the radial direction 0). What about perpendicular to the radius?
As I wrote previously
the force from the pivot will also have a component perpendicular to that radial line. Together with the corresponding component from gravity this will provide the torque necessary for the angular acceleration.
I feel it will become clearer if you just go ahead and develop the usual equations.
hyperddude
#14
Jan5-13, 04:02 PM
P: 15
Quote Quote by haruspex View Post
I feel it will become clearer if you just go ahead and develop the usual equations.
Alright.

The torque due to the component of gravity perpendicular to the radius is [itex]mgL\sin{θ}[/itex]. Since [itex]\sin{θ}=\frac{w}{\sqrt{w^2+h^2}}[/itex] and [itex]L=\frac{\sqrt{w^2+h^2}}{2}[/itex], the torque is therefore [itex]\frac{mgw}{2}[/itex].

This torque is equal to [itex]Iα[/itex] and [itex]I=\frac{m(w^2+h^2)}{3}[/itex]. So, [itex]α=\frac{3gw}{2(w^2+h^2)}[/itex].

[itex]a = αL = \frac{3gw}{2(w^2+h^2)}*\frac{\sqrt{w^2+h^2}}{2}[/itex]

[itex]=\frac{3gw}{4\sqrt{w^2+h^2}}[/itex]

The acceleration from gravity, [itex]g\sin{θ}=\frac{gw}{\sqrt{w^2+h^2}}[/itex] is larger than [itex]a[/itex] found just above! So, I'm thinking that the pivot supplies a force in the opposite direction to make the net acceleration [itex]α=\frac{3gw}{4(w^2+h^2)}[/itex]

This force is [itex]m(\frac{gw}{\sqrt{w^2+h^2}}-\frac{3gw}{4(w^2+h^2)})=\frac{mgw}{4\sqrt{w^2+h^2}}[/itex]

So, my conclusion is that the force from the pivot is in two components: the first is in the radial direction (which makes the radial net force 0). This force has a magnitude of [itex]\frac{mgh}{\sqrt{h^2+w^2}}[/itex], which is the radial component of force due to gravity.

The other component is perpendicular to the previous component. This was found to be [itex]\frac{mgw}{4\sqrt{w^2+h^2}}[/itex].

Is my thinking correct?
haruspex
#15
Jan5-13, 04:27 PM
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Quote Quote by hyperddude View Post
The torque due to the component of gravity perpendicular to the radius is [itex]mgL\sin{θ}[/itex]. Since [itex]\sin{θ}=\frac{w}{\sqrt{w^2+h^2}}[/itex] and [itex]L=\frac{\sqrt{w^2+h^2}}{4}[/itex], the torque is therefore [itex]\frac{mgw}{4}[/itex].
A divisor of 2 has become 4 in there somewhere.
It's easier just to say the line of action of mg is distance w/2 from the pivot, so the moment is mgw/2.
This torque is equal to [itex]Iα[/itex] and [itex]I=\frac{m(w^2+h^2)}{3}[/itex]. So, [itex]α=\frac{3gw}{16(w^2+h^2)}[/itex].
2, not 16.
[itex]a = αL = \frac{3gw}{4(w^2+h^2)}*\frac{\sqrt{w^2+h^2}}{4}[/itex]

[itex]=\frac{3gw}{16\sqrt{w^2+h^2}}[/itex]
4, not 16.
The acceleration from gravity, [itex]g\sin{θ}=\frac{gw}{\sqrt{w^2+h^2}}[/itex] is larger than [itex]a[/itex] found just above! So, I'm thinking that the pivot supplies a force in the opposite direction to make the net acceleration [itex]α=\frac{3gw}{16(w^2+h^2)}[/itex]

This force is [itex]m(\frac{gw}{\sqrt{w^2+h^2}}-\frac{3gw}{16(w^2+h^2)})=\frac{13gw}{16\sqrt{w^2+h^2}}[/itex]

So, my conclusion is that the force from the pivot is in two components: the first is in the radial direction (which makes the radial net force 0). This force has a magnitude of [itex]\frac{mgh}{\sqrt{h^2+w^2}}[/itex], which is the radial component of force due to gravity.

The other component is perpendicular to the previous component. This was found to be [itex]\frac{13mgw}{16\sqrt{w^2+h^2}}[/itex].

Is my thinking correct?
The noted numerical errors aside, yes.
hyperddude
#16
Jan5-13, 05:10 PM
P: 15
Oh, good catch. No wonder the rest of my problem wasn't working out. Thanks!


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