Inertia Problem

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  • #1
nabaa
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PHYSICS HELP:


I solved a using parallel axis theorem:

ML^2/12 + M(L/2)^2 = ML^2/3

other than that i'm lost..

all my equations on inertia:

torque = rFsintheta = Iangularaccel.


b. I tried using the equation Fsinthetat but it didn't work
 
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  • #2
nabaa
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some more relevant equations:

E_r=1/2Iw^2

L=Iw

Iend= Icm + Md^2
 
  • #3
cepheid
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τ = Iα is indeed the starting point for determining the equation of motion. The magnitude of the torque is equal to the component of the force that is perpendicular to the lever arm, multiplied by the length of that lever arm. So, in this case, the force is the component of the weight that is perpendicular to the rod (not the component that acts parallel to the rod, which doesn't contribute to the torque). The lever arm is the distance between the point at which the force acts and the centre of rotation.

So, τ will depend on the angular position (theta). α also depends on theta. It is the second derivative of it with respect to time. Btw, do you know differential calculus?
 
  • #4
nabaa
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i'm right now taking double variable calculus and so am familiar with it, yes

and thank you!

so I got that (ML^2/12)(d^2A/dt)=M(L/2-d)sintheta?
 
  • #5
nabaa
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by using (Irod)(second derivative of A versus t) = Mass(L at center - distance from center)sinangle
 
  • #6
cepheid
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I thought that the rod was pivoting around its end. So why did you use the equation for the moment of inertia of a rod pivoting around its centre?

The equation for α should be:

[tex] \alpha = \frac{d^2\theta}{dt^2} [/tex]

A is a constant. It is the initial angle i.e.:

[tex] \theta(t=0) = A [/tex]

I'll get back to you regarding the right hand side of your equation.
 
  • #7
cepheid
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Okay for the right hand side -- I don't see a force. Let's start with that. Answer these questions one by one:

What is weight of the rod?

What is the component of the rod's weight that is in the direction perp. to the rod? (This is the component that produces a torque about the pivot point).

Now, let's focus on r (the lever arm):

At what point can the weight (gravitational force) be considered to be applied?

How far from the pivot point is this application point of the force? This distance is the lever arm (r).
 
  • #8
nabaa
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oh, that's right it would be the end of the rod, sorry

and also, maybe other relevant equations?

angle = s/r where s is arc length and r is distance till

w = dangle/dt

changeinangle = angle2 - angle1

angular accel = dw/dt
 
  • #9
nabaa
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the weight of the rod is M

the component that it's perpendicular to is at L, yes? at the end of the rod

when is gravity not taken into account? isn't it always, or is it only after L/2 or until it's rotated at an angle not 0 degrees

isn't it a whole L away from the pivot point?
 
  • #10
cepheid
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the weight of the rod is M

NO. That is the mass of the rod.

the component that it's perpendicular to is at L, yes? at the end of the rod

What I meant was, decompose the force vector (which is vertical) into components that are parallel and perpendicular to the rod, and use trigonometry to find the magnitudes of those components.

when is gravity not taken into account?

Never. Gravity is always present.

isn't it a whole L away from the pivot point?

NO. For an extended body, the weight vector can be considered to act at the centre of mass.
 
  • #11
nabaa
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oh so the weight of the rod is

(9.81M)

vectors:

(9.81M)sinangle = vertical component
(9.81M)cosangle = horizontal component

so...

(9.81M)(L/2) is weight vector at com
 
  • #12
cepheid
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oh so the weight of the rod is

(9.81M)

Yeah

(9.81M)sinangle = vertical component
(9.81M)cosangle = horizontal component

Not exactly. The weight Mg IS vertical. The components are neither horizontal nor vertical. It should be:

(9.81M)sinangle = component perpendicular to rod
(9.81M)cosangle = component parallel to rod

(9.81M)(L/2) is weight vector at com

Is this supposed to be the torque? If so, you're missing something. The length is now correct. However, the force is wrong. Remember, only the perpendicular component contributes to the torque. That's why we calculated it in the first place.
 
  • #13
nabaa
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(9.81M)sinangle = component perpendicular to rod
(9.81M)cosangle = component parallel to rod

oh yea, when I drew a picture this made a lot more sense

so torque = (9.81Msintheta)(L/2)
 
  • #14
cepheid
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oh yea, when I drew a picture this made a lot more sense

so torque = (9.81Msintheta)(L/2)

Perfect! So now you have both sides of your equation, right?

[tex] I\alpha = \tau [/tex]

[tex] \left(\frac{ML^2}{3}\right)\frac{d^2\theta}{dt^2} = \frac{MgL}{2}\sin \theta [/tex]

Now, this is a differential equation that you don't know how to solve (and neither do I), because it is non-linear. In order to make it solvable, you have to use the fact that the equation said the total angular displacement is very small. What is an approximation to sin[itex]\theta[/itex] when [itex]\theta[/itex] is very small?
 
  • #15
nabaa
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when sin is very small it equals about 0, so does that mean that you'll have to set d^2angle/dt^2 = 0?
 
  • #16
cepheid
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  • #18
cepheid
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What is x???

EDIT: What I mean is, you should use your symbols consistently. If you want to use "angle" to represent the angle, then it should be:

sin(angle) ~ angle

and if you want to use "x" for the angle, it should be

sin(x) ~ x

Pick one and stick to it consistently.
 
  • #19
nabaa
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oops x should be angle, right?

meaning:

MgLangle/2 = (ML^2/3)(d^2angle/dt^2)

(MgL/2)/(ML^2/3) = (d^2angle/dt^2)/angle

3g/2L = (d^2angle/dt^2)/angle

how do i isolate angle, then?
 
  • #20
cepheid
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oops x should be angle, right?

meaning:

MgLangle/2 = (ML^2/3)(d^2angle/dt^2)

(MgL/2)/(ML^2/3) = (d^2angle/dt^2)/angle

3g/2L = (d^2angle/dt^2)/angle

Yes, exactly! :smile:

how do i isolate angle, then?

It's a differential equation. The second derivative of the angle is proportional to the negative of the angle itself (there should be a negative sign on one side, we just missed it). Do you know how to solve this kind of differential equation?
 
  • #21
nabaa
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well usually in this instance i'd use integration but i don't know how to do so with a double derivative, or are you saying we don't need integration?
 
  • #22
cepheid
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Ah I see. Okay. It's clear that you haven't done differential equations before. To answer your question, for a second derivative, you can just integrate twice, IF you know what the function is. In this case, however, we CAN"T integrate because we don't know what the function, [itex]\theta (t) [/itex] IS. In fact, that's what we are trying to solve for. ALL we know is that:

[tex] \frac{d^2 \theta (t)}{dt^2} \propto -\theta(t) [/tex]

This type of equation, in which a function is related to one or more of its own derivatives is called a differential equation. When you use this information to find out what the function is, that is called "solving" the differential equation. In this case, it's called a "second-order" differential equation because the highest-order derivative in the equation is a second derivative. There are formal methods for solving differential equations. The good news is that you don't need to know these methods in order to solve this problem. In this problem, we can just GUESS the solution:

Can you think of a function whose second derivative is proportional to the negative of itself?

Hint: the function should describe the motion of the rod with time? What does your intuition tell you will happen if you displace the rod by a small angle and then let it go? What kind of [itex] \theta (t) [/itex] might describe that result?
 
  • #23
nabaa
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oh..

sin(X)

or cos(x)
 
  • #24
cepheid
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oh..

sin(X)

or cos(x)

Right! Except I think you mean sin(t) or cos(t). :wink: Time is the independent variable here. Theta is changing with time. If you displace the rod by an angle A, and then let it go, it will oscillate with time (swing back and forth). That's how we know that the solution must be a periodic function (which is the "hint" I was getting at with my previous post).

Question 1: How do you know whether it is sin or cos? Hint: what is the angular displacement at t = 0? It's equal to the maximum displacment (the amplitude of the oscillation) at t = 0, right? Which function can be made consistent with that?

Quesiton 2: sin(t) or cos(t) cannot be the full answer. Remember that we have:

[tex]
\frac{d^2 \theta (t)}{dt^2} \propto -\theta(t)
[/tex]

which means that:


[tex]
\frac{d^2 \theta (t)}{dt^2} = -C\theta(t)
[/tex]

where "C" is some constant. In this case, you worked out the constant three posts ago. How must you modify the solution in order to get this constant out front after differentiating twice?
 
  • #25
nabaa
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C = 3g/2L

so the equation is

d^2angle(t)/dt^2 = −(3g/2L)(sin(t))

and angle(t) =(3g/2L)(sin(t))
 
  • #26
nabaa
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is that right?
 
  • #27
cepheid
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C = 3g/2L

Yes

so the equation is

d^2angle(t)/dt^2 = −(3g/2L)(sin(t))

No, for 2 reasons:

reason 1: the equation is actually d^2angle(t)/dt^2 = −(3g/2L)(angle(t))

reason 2: Is sin(t) the right choice?. Again, consider the system at t = 0 (initial conditions)

and angle(t) =(3g/2L)(sin(t))

No, due to reason 1 above.

EDIT: Hint: If the solution were really just a constant times sin(t) or cos(t), then this would mean that the rod goes through a full swing cycle in 2*pi seconds. But there is no reason why it should be that in particular. The swing period should depend on the parameters of the rod.
 
  • #28
nabaa
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ohhh so it would end up being (3g/2L)(cost), and the negatives would still cancel (integrate) out
 
  • #29
cepheid
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ohhh so it would end up being (3g/2L)(cost), and the negatives would still cancel (integrate) out

No, that's still wrong. Look very closely at reason 1 above. You have to somehow differentiate a function twice and get a result that is -(3g/2L) times the original function. NEITHER -(3g/2L)sin(t) NOR -(3g/2L)cos(t) can satisfy that requirement (try it). BUT if you modify them appropriately, it can be done. See my hint that I added to my previous post as an edit.
 
  • #30
nabaa
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to be honest I don't understand what's wrong with them? isn't the double derivative of 3g/2Lcost = −3g/2Lcost
 
  • #31
cepheid
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to be honest I don't understand what's wrong with them? isn't the double derivative of 3g/2Lcost = −3g/2Lcost

Yes, but this is not equal to -3g/2L times the original function. If the original function is (3g/2L)cos(t), then -3g/2L*(original function) would be -(9g2/4L2)cos(t), which is NOT what you get when you differentiate twice.

So, your proposed function does NOT satisfy your differential equation.

Think about this for a second. What do we know? SOME SORT of sinusoidal function will satisfy the differential equation, but we don't know its exact form. So, in order to figure it out, we need to take into account everything that we DO know:

- we know that at t = 0, angle(t) = A, which means that the function has to be of the type that is non-zero when you give it a zero argument (this helps you decide between sine and cosine)

- we know that A is the maximum displacement, which means that the amplitude of the oscillation is A, which means that the rod goes back and forth between an angular position of +A and -A. This tells you that A has to be multiplying the sinusoidal function in front (and nothing else).

- we know that the frequency has to be in there somewhere and should be set by the physical properties of the rod and its environment. We also know that it should be consistent with the differential equation, which contains that constant C that appears out front AFTER twice differentiating (but maybe now I am giving away too much...)
 
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  • #32
nabaa
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cosine because at angle = 0 cosine is 1

A in front, okay


Acost/(3g/2L)??

??
 
  • #33
cepheid
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cosine because at angle = 0 cosine is 1

A in front, okay


Acost/(3g/2L)??

??

Mean answer: Why are you asking me?! Differentiate it twice and see if it works or not.

Nice answer: Working "backwards" is difficult, so I will give you a BIG hint. What is the physical significance of the parameter [itex]\omega[/itex] in the expression:

[tex] \theta (t) = A\cos(\omega t) [/tex]

What do you get when you differentiate it once?

[tex] \frac{d\theta (t)}{dt} = \frac{d}{dt}[A\cos(\omega t)] =~? [/tex]

What do you get when you differentiate it again?

[tex] \frac{d^2 \theta (t)}{dt^2}= \frac{d^2}{dt^2}[A\cos(\omega t)] =~? [/tex]

Assuming you STARTED with a function of this form, what must be true in order for the result of the above to match your differential equation?
 
  • #34
nabaa
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differentiating once would yield -wAsin(wt)

twice would yield

-w^2Acos(wt)

so w^2 must = 3g/2L

w = sqrt(3g/2L)

function = Acos(sqrt(3g/2L)t), yes?
 
  • #35
cepheid
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differentiating once would yield -wAsin(wt)

twice would yield

-w^2Acos(wt)

so w^2 must = 3g/2L

w = sqrt(3g/2L)

function = Acos(sqrt(3g/2L)t), yes?

Beautiful! You have just answered BOTH part (b) and part (c) since T = 1/f and w = 2*pi*f is the angular frequency. Notice how similar the expression for T is to the expression for the period of a *simple* pendulum. The latter is derived in a similar way, by setting up the equation of motion. Now you know where that comes from! It is ridiculously late in my time zone, so, good night!
 

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