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Inertia Problem

  • Thread starter 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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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))
 

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