How do i determine the final position?

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  • #26
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I didn't use the term compressible, just compressive. Even if the idealized rod is incompressable, that doesn't mean it can't have internal pressures.
Whether it is "compressible" or "compressive", it is still about "compression" which means making it smaller to take less space. Although the action itself is compression, is still best to avoid the term.

I just wasn't sure if gravity can be considered as a vector force applied at the mid-point of the rod.
Well, center of mass/gravity 101. In a uniform, symmetric and rigid object, the center is always at the midpoint.

Assuming that there is no initial horizontal component of velocity, then since there are no external horizontal forces, then the center of mass only moves and accelerates vertically. You could attach the end of the rod to massless pin that could slide freely in a horizontal slot, the rod could swing or rotate and the center of mass would not move horizontally, since there are no external horizontal forces.
I am not sure how you can attach a pin that slides; and yet allow the end of the rod to move when there is a pin attached to it. Maybe you can draw me a picture to clarify.

You should end up with two differential equations related to vertical acceleration of the center of mass with respect to the angle of the rod, one relating the upwards force from the frictionless surface at point A to vertical acceleration of center of mass, and the other relating acceleration of the center of mass to the angular acceleration of the rod due to the torque related to the upwards force (frictionless surface) at point A and downwards force (gravity) at the center of mass, divided by the angular inertia of a rod rotating about one end.
What equations did you use? How did you end up with differential equations? Can you show me your calculations?
 
  • #27
rcgldr
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You could attach the end of the rod to massless pin that could slide freely in a horizontal slot, the rod could swing or rotate and the center of mass would not move horizontally, since there are no external horizontal forces.

I am not sure how you can attach a pin that slides; and yet allow the end of the rod to move when there is a pin attached to it.
As I mentioned, imagine the massless pin is bound on both sides by frictionless horizontal slots. Vertical motion of the pin is prevented, as well as motion in the direction of the pin, but there is no resistance (force) to horiztonal motion of the pin and the end of the rod the pin is attached to. It's bascially a pendulum with the pivot point free to move along a horiztonal line that is perpendicular to the axis of rotation of the pendulum. In this case the intial state is of an inverted pendulum.

You should end up with two differential equations related to vertical acceleration of the center of mass with respect to the angle of the rod, one relating the upwards force from the frictionless surface at point A to vertical acceleration of center of mass, and the other relating acceleration of the center of mass to the angular acceleration of the rod due to the torque related to the upwards force (frictionless surface) at point A and downwards force (gravity) at the center of mass, divided by the angular inertia of a rod rotating about one end.

What equations did you use? How did you end up with differential equations? Can you show me your calculations?
I didn't derive the equations, I just described them. Eventually you'll have equations that relate the acceleration of the center of mass versus angle of the rod. You'll need to integrate these equations (if they can be integrated) to calculate acceleration, velocity, position, and angle versus time.
 
  • #28
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As I mentioned, imagine the massless pin is bound on both sides by frictionless horizontal slots. Vertical motion of the pin is prevented, as well as motion in the direction of the pin, but there is no resistance (force) to horiztonal motion of the pin and the end of the rod the pin is attached to. It's bascially a pendulum with the pivot point free to move along a horiztonal line that is perpendicular to the axis of rotation of the pendulum. In this case the intial state is of an inverted pendulum.
A picture is worth a thousand words. You are confusing me more now that you are relating it to a pendulum. Just draw me a picture please.

I didn't derive the equations, I just described them. Eventually you'll have equations that relate the acceleration of the center of mass versus angle of the rod. You'll need to integrate these equations (if they can be integrated) to calculate acceleration, velocity, position, and angle versus time.
How can you be sure what the outcome of the equations will be if you didn't derive them? Sorry but baseless proof is meaningless for me.
 
  • #29
Philip Wood
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As threatened earlier, here's the energy treatment of the problem. Much easier to formulate the equation to be solved.

I did differentiate it wrt t to check that it agreed with my forces and torques treatment (thumbnail in post 20) and it does – except for a slipped sign in that thumbnail.
 

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  • #30
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As threatened earlier, here's the energy treatment of the problem. Much easier to formulate the equation to be solved.
I love your work this time because i am able to understand 90% of your calculations. I still need some clarifications though. Please help me to understand.

It took me some time to understand but it seems that you love the dot derivative notation to indicate velocity and acceleration. So now that is out of the way, i wonder how you calculate your radius.

In the transnational kinetic energy part, you seemed to have chosen a variable radius. This radius you have chosen is always the x-projection between point A and the center of mass. This is why you got:[tex]\frac{Lcos\theta}{2}[/tex]
But then the radius you have chosen for the rotational kinetic energy is different. And i don't know how you came up with:[tex]\frac{L}{\sqrt{12}}[/tex]
Which also makes me wonder, what is your pivot point? Why there are two different radius? Shouldn't the pivot point be point A and the radius just L/2?
 
  • #31
Philip Wood
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Don't think about it in terms of radius and pivot point.

You agree that y = [itex]\frac{L}{2}[/itex]sin[itex]\theta[/itex] ?

In that case [itex]\dot{y}[/itex] = [itex]\frac{L}{2}[/itex]cos[itex]\theta[/itex] [itex]\dot{θ}[/itex].

The rotational KE must be calculated about the centre of mass (CM). The formula for rotational KE is [itex]\frac{1}{2}[/itex]Iω2. I, the moment of inertia of the rod about its centre, is [itex]\frac{1}{12}[/itex]mL2.

But ω = [itex]\dot{θ}[/itex], the rate of change of the angle θ shown on the diagram.

Hope it all makes sense now.
 
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  • #32
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You agree that y = [itex]\frac{L}{2}[/itex]sin[itex]\theta[/itex] ?

In that case [itex]\dot{y}[/itex] = [itex]\frac{L}{2}[/itex]cos[itex]\theta[/itex] [itex]\dot{θ}[/itex].
Interesting. You derived velocity by calculating the derivative of y. I did it differently. I substituted the relationship between velocity, radius and angular velocity like this: [itex]\dot{y} = r \dot{\theta}[/itex]. Then you can compare both results and see that that [itex]\frac{L}{2}[/itex]cos[itex]\theta[/itex] happens to be your radius for angular the velocity. And according to the physical geometry, it is the projection between point A and the center of mass in the x-axis.

The rotational KE must be calculated about the centre of mass (CM). The formula for rotational KE is [itex]\frac{L}{2}[/itex]Iω2.
I think it was a typo and you meant that the rotational KE is [itex]\frac{1}{2}[/itex]Iω2

I, the moment of inertia of the rod about its centre, is [itex]\frac{1}{12}[/itex]mL2.
Let's define moment of inertia again: [itex]I = mr^2[/itex].
So, if [itex]I = m(\frac{L^2}{12})[/itex], means your radius is [itex]\frac{L}{\sqrt{12}}[/itex]. How come?
 
  • #33
Philip Wood
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Sorry about the typo.

[itex]\frac{L}{\sqrt{12}}[/itex] is what's called the 'radius of gyration' of the rod about its centre. The interpretation is this: if all the rod's mass were to be gathered together at one point, then if that point were to be at the radius of gyration from the rod's centre, then the rod would have the same moment of inertia about its centre as the original rod.

Do you, I wonder, need to learn up about moment of inertia: (a) its rôle in mechanics, and (b) how to calculate moments of inertia of a few simple bodies, like cylinders and rods? Sorry if this seems patronising, but you do seem very keen to get to the bottom of this stuff.
 
  • #34
rcgldr
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angular inertia of a rod rotating about one end.
That should have been rod rotating by it's center, since the end of the rod is not constrained horizontally (frictionless surface) and the center of mass moves and acceleates only vertically. Philip Wood already did the math using angular inertia for a rod rotating about it's center.

The 'radius of gyration' of the rod about its centre. The interpretation is this: if all the rod's mass were to be gathered together at one point ...
or all the mass located in a hollow cylinder at that radius of gyration, or any collection of point masses, as long as all of them are located at the radius of gyration.
 
  • #35
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Sorry if this seems patronising, but you do seem very keen to get to the bottom of this stuff.
Of course i am. I am sure you didn't pick that value for radius of gyration randomly. I want to be able to understand the calculations myself.

[itex]\frac{L}{\sqrt{12}}[/itex] is what's called the 'radius of gyration' of the rod about its centre.
I found how the number was obtained. It seems that i was just picking a radius away from my chosen pivot point; which is wrong? It seems that we have to calculate this radius by integrating the stick.
[itex]dm = \frac{M}{L}dr[/itex]
[itex]I = \int_{\frac{-L}{2}}^{\frac{L}{2}} r^2 dm = \frac{M}{L} \int_{\frac{-L}{2}}^{\frac{L}{2}} r^2dr[/itex]
[itex]I = \frac{M}{L} \frac{r^3}{3} |_{\frac{-L}{2}}^{\frac{L}{2}} = \frac{M}{3L} (\frac{L^3}{8} + \frac{L^3}{8}) = M \frac{L^2}{12}[/itex]

However, i question that if finding the radius of gyration this way is really correct? Here is a simple example i am going to use. Assume that our problem, point A is a fixed pivot point. And i am going to try to find the torque of the stick when gravity is the only relevant force.
[itex]\tau = \alpha I = \frac{a_t}{r} mr^2 = F_t r[/itex]

-First, let's calculate the inertia and radius of gyration; with the fixed pivot point A. So the points of integration are from 0 to L.
[itex]I = \int_{0}^{L} r^2 dm = \frac{M}{L} \int_{0}^{L} r^2dr = \frac{M}{L} \frac{r^3}{3} |_{0}^{L} = \frac{M}{3L} (L^3 - 0) = M \frac{L^2}{3}[/itex]
-Therefore the radius of gyration when the pivot is at the end of the stick is: [itex]\frac{L}{\sqrt{3}}[/itex]
-The tangent acceleration that can be obtained by the geometry is equals: [itex]a_t = -gcos(\theta)[/itex]
-Then the torque equals: [itex]\tau = \alpha I = \frac{-gcos(\theta)}{\frac{L}{\sqrt{3}}}M \frac{L^2}{3} = \frac{-\sqrt{3}LMgcos(\theta)}{3}[/itex]
-But the torque is also equals: [itex]\tau = F_t r = -Mgcos(\theta) \frac{L}{2}[/itex]
-Or even if i use the radius of gyration instead of a distance between the pivot and the center of mass: [itex]\tau = F_t r = -Mgcos(\theta) \frac{L}{\sqrt{3}}[/itex]
-Even if i do a mix and match on the torque that contains moment of inertia: [itex]\tau = \alpha I = \frac{-gcos(\theta)}{\frac{L}{2}}M \frac{L^2}{3} = \frac{-2LMgcos(\theta)}{3}[/itex]

One last thing, it took me to try it with a different notation to actually find out that this is wrong:
You agree that y = [itex]\frac{L}{2}[/itex]sin[itex]\theta[/itex] ?

In that case [itex]\dot{y}[/itex] = [itex]\frac{L}{2}[/itex]cos[itex]\theta[/itex] [itex]\dot{θ}[/itex].
This is what you are doing:
[itex]x = y^2[/itex]
[itex]\frac{dx}{dz} = \frac{d}{dz}(y^2) = 2y \frac{dy}{dz}[/itex] which is wrong
 
  • #36
Philip Wood
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Agree with your moment of inertia calculations. Can't (with only a few minutes to spare at present) follow your dynamics.

Don't see any connection between your stuff with y2 at the bottom of your post and my differentiation of sinθ wrt t, using the chain rule.

Shall be off PF for some 48 hours starting in a few minutes.
 
  • #37
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Don't see any connection between your stuff with y2 at the bottom of your post and my differentiation of sinθ wrt t, using the chain rule.

Chain rule can't be applied such way because sinθ is not differentiable with respect to t. It is due to the fact that θ does not contain the time variable. Not even with the radius relation: [itex]\theta = \frac{d}{r}[/itex]

Here is what your correct chain rule looks like:[tex]\frac{dy}{dt} = \frac{d}{dt}sin\theta[/tex][tex]\frac{dy}{dt} = \frac{d}{dt}(sin\theta)\frac{d}{dt}(\theta)[/tex][tex]\dot{y} = \dot{\theta}\frac{d}{dt}(sin\theta)[/tex]Meaning that you can't get rid of the sinθ and suddenly change it to cosθ.
 
  • #38
Philip Wood
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Chain rule for this case :

[itex]\frac{dy}{dt}[/itex] = [itex]\frac{dy}{d\theta}[/itex] [itex]\times[/itex] [itex]\frac{d\theta}{dt}[/itex].
 
  • #39
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You are right.
[itex]y = sin\theta[/itex]
[itex]\frac{dy}{dt} = \frac{d}{d\theta}(sin\theta) \frac{d\theta}{dt}[/itex]
[itex]\dot{y} = cos\theta \dot{\theta}[/itex]
And it can be confirmed by integration:
[itex]\frac{dy}{dt} = cos\theta \frac{d\theta}{dt}[/itex]
[itex]\int{dy} = \int cos\theta \frac{d\theta}{dt}dt[/itex]
[itex]y = sin\theta[/itex]

Which is somewhat awkward because it means that my first example is also correct:
[itex]\frac{dx}{dz} = \frac{df(y)}{dy} \frac{dy}{dz}[/itex]
[itex]x = y^2[/itex]
[itex]\frac{dx}{dz} = \frac{y^2}{dy} \frac{dy}{dz}[/itex]
[itex]\frac{dx}{dz} = 2y \frac{dy}{dz}[/itex]
And confirmed by integration:
[itex]\int{dx} = \int 2y \frac{dy}{dz} dz[/itex]
[itex]x = y^2[/itex]
 
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