Mechanics: A plank on a spring causing two cylinders to undergo harmonic motion (rotation)

  • #1
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Homework Statement:
Rolling without slip. The masses, R and dX are given. I need to find the period.
Relevant Equations:
$$T =2π \sqrt{\frac{m}{k}}$$

1589274523120.png

Here is the picture on the system.
I have to find the period (T). The masses, R and dX is given. The systam at first is at rest, then at t = 0 we pull the plank to dX distance from its originial position.
In the thread (https://www.physicsforums.com/threa...-cylinders-to-accelerate.988793/#post-6340092) I learnt how to calculate the acceleration with a constant force pulling the plank.
Now I can think on two options. The system is going to oscillate forever, due to the lack of friction. Or the system will eventually stop, because the the plank losing energy with rotating the cylinders. If the second case is true, then I need the period as a function of time? How should I start this problem?
Any help is appreciated!
 
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Answers and Replies

  • #2
BvU
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If the second case is true,
It is not: there is no mention of any loss of mechanical energy. When the cylinders roll past the equilibrium point, they simply transfer their kinetic energy back to the plank.
 
  • #3
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Thank you! Wild guess: "cylinders roll past the equilibrium point"
They leave equilibrium when they start rolling? So due action-reaction the kinetic energy is in balance between the plank and cylinders. My guess is based on the observation that the string accelerates and decelerates the plank and cylinders (through the plank) at the same time.
 
  • #4
BvU
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we pull the plank to dX distance from its originial position.
Problem statement does not mention it, but I suppose we then let go of the plank :biggrin: and call that t=0 and x = dX
They leave equilibrium when they start rolling?
At t=0 they start rolling; they have already left equilibrium (which was at x = 0).
 
  • #5
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It was too hard (and I have no time) to calculate the period with Newton's laws. Instead I tried Lagrangian mechanics. Can you help me with this?
Here is how I calculated:
$$ 0 = \frac {d} {dt} (\frac {∂L}{∂ Ẋ}) - \frac {∂L}{∂x} $$
$$ L = KE-PE$$
$$PE = \frac{1}{2} ⋅ k ⋅ x^2$$
$$KE = KE_1+KE_2+KE_3$$
$$KE = \frac{1}{2} ⋅ m_1 ⋅ Ẋ^2 + (\frac{1}{2} ⋅ m_2 ⋅ ( \frac{Ẋ}{2})^2 + \frac{1}{2} ⋅ I_2 ⋅ ω^2)+(\frac{1}{2} ⋅ m_3 ⋅ ( \frac{Ẋ}{2})^2 + \frac{1}{2} ⋅ I_3 ⋅ ω^2)$$

$$KE = \frac{1}{2} ⋅ m_1 ⋅ Ẋ^2 + (\frac{1}{2} ⋅ m_2 ⋅ ( \frac{Ẋ}{2})^2 + \frac{1}{2} ⋅ m_2 ⋅ R^2⋅ (\frac{Ẋ}{2R})^2)+(\frac{1}{2} ⋅ m_3 ⋅ ( \frac{Ẋ}{2})^2 + \frac{1}{2} ⋅ \frac{1}{2} ⋅m_3 ⋅R^2⋅ (\frac{Ẋ}{2R})^2)$$

$$KE= \frac{1}{2} ⋅ m_1 ⋅ Ẋ^2+\frac{1}{4} ⋅ m_2 ⋅ Ẋ^2+\frac{3}{16} ⋅ m_3⋅ Ẋ^2$$

$$L = \frac{1}{2} ⋅ m_1 ⋅ Ẋ^2+\frac{1}{4} ⋅ m_2 ⋅ Ẋ^2+\frac{3}{16} ⋅ m_3⋅ Ẋ^2-\frac{1}{2} ⋅ k ⋅ x^2$$

$$0=\frac{d}{dt}(\frac {∂L}{∂Ẋ})-\frac {∂L}{∂x}=\frac{1}{2}⋅m_1⋅Ẍ+\frac{1}{} ⋅m_2⋅Ẍ+\frac{3}{8}⋅m_3⋅Ẍ+k⋅x$$

With some forming. Is this correct?:
$$Ẍ+\frac{k}{m_1+\frac{1}{2}m_2+\frac{3}{8}m_3}⋅X=0$$

And putting into the period equation (I think it's wrong):

$$T=2⋅π\sqrt{\frac{m_1+\frac{1}{2}m_2+\frac{3}{8}m_3}{k}}$$

How can I calculate the period from this information?
 
  • #6
BvU
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##T## is the period !?

Didn't check the math -- that's your responsibility

Slightly surprised the R falls out, but it's possible. Nice exercise !
 
  • #7
kuruman
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##T## is the period !?

Didn't check the math -- that's your responsibility

Slightly surprised the R falls out, but it's possible. Nice exercise !
The spring doesn't care what is attached to its end. The rolling items can be replaced by their effective masses given by ##m_{2,eff}=I_2/ R_2^2## and ##m_{3,eff}=I_3 /R_3^2##. That's why the radii drop out. Note: The moments of inertia are about the point of contact, not about the center of the circles because there is kinetic energy of the center of mass and kinetic energy about the center of mass.

@Hohen: To quote Goldstein, using a Lagrangian here is like killing a fly with a sledgehammer. Many things can go wrong.
 
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  • #8
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T is the period !?
yes
The spring doesn't care what is attached to its end
Okay, but why the amplitude isn’t in the equation? It doesnt matter for the frequency/time period how long the initial distance was when the system started moving?
@Hohen: To quote Goldstein, using a Lagrangian here is like killing a fly with a sledgehammer. Many things can go wrong.
The second object in this problem was to write the Lagrangian equation of the system. That’s why I calculated out.
 
  • #9
kuruman
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yes

Okay, but why the amplitude isn’t in the equation? It doesnt matter for the frequency/time period how long the initial distance was when the system started moving?
If by "equation" you mean the expression for the period, in simple harmonic motion the period is independent of the amplitude. You should have encountered this by now.
The second object in this problem was to write the Lagrangian equation of the system. That’s why I calculated out.
OK. Next time please post the entire statement of the problem. A few words of caution: Before you write down the Lagrangian, you need to consider what you are going to use for generalized coordinates so that the position of each object can be uniquely specified. Writing the kinetic energy of the rolling objects as ##\frac{1}{2}I\omega^2## won't do because ##\omega## is not a generalized coordinate. For a cylinder of mass ##m_2## I would use ##\theta_2## for the angle by which in turns and ##x_2## for the position of its center. Same thing for ##m_3##. Of course there are constraints to figure out. Write them down, don't apply them in your head. Finally, if I put a box around the plank and the cylinders, all I will see is this box executing simple harmonic motion just like a horizontal spring-mass system. This means that there are enough constraints to apply so that the Lagrangian will generate only one equation of motion.
 
  • #10
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For a cylinder of mass ##m_2## I would use ##\theta_2## for the angle by which in turns and ##x_2## for the position of its center.
Like this?
##I_2⋅\omega_2^2=(m_2⋅R^2)⋅(\frac{v}{2R})^2=p_2⋅\dot \theta_2^2=(m_2⋅x _2^2)⋅(\frac{\dot x_2}{2x_2})^2##
The velocities are devided with ##2## because ##v## is the velocity of the plank (so it equals vrot of the cylinders)
 
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  • #11
kuruman
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Like this?
##I_2⋅\omega_2^2=(m_2⋅R^2)⋅(\frac{v}{2R})^2=p_2⋅\dot \theta_2^2=(m_2⋅x _2^2)⋅(\frac{\dot x_2}{2x_2})^2##
The velocities are devided with ##2## because ##v## is the velocity of the plank (so it equals vrot of the cylinders)
That's not how it works. Let ##x_2## be the position of ##m_2## and ##\theta_2## the angle by which it has turned. Its kinetic energy has two terms: ##T_2=\frac{1}{2}m_2\dot x_2^2+\frac{1}{2}I_2\dot\theta_2^2##. The first term is the kinetic energy of the CM and the second term the kinetic energy about the CM.

What constraint relates ##x_2## and ##\theta_2## and why? What constraint relates ##x_2## and ##x_1##, the position of the plank, and why? Repeat with ##x_3## and ##\theta_3##. Your Lagrangian will have 5 generalized coordinates subject to 4 constraints. You can eliminate any four of the coordinates in favor of the fifth. You have chosen ##x_1## (what you call ##X##) and that's an excellent choice. From my experience, the formal procedure of first considering separate generalized coordinates for each object and then applying the constraints algebraically leaves less room for error especially when something that looks "obvious" actually is not.
 
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  • #12
kuruman
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Actually your original calculation for the period is not wrong. I got around to doing it and the effective mass of the oscillator is ##M_{eff}=m_1+\frac{1}{2}m_2+\frac{3}{8}m_3## which gives a period$$T=2\pi\sqrt{\frac{M_{eff}}{k}}$$as you have already posted.
 
  • #13
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That means my calculation with the Lagrangain is correct, I just used wrong generalized coordinates?
 
  • #14
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What constraint relates ##x_2## and ##\theta_2## and why? What constraint relates ##x_2## and ##x_1##, the position of the plank, and why?
I'm stuck with this. The only relation I found between ##x_2## and ##\theta_2## is ##x_2 = \theta_2 ⋅ R##
So with derivation ##\frac{d(x_2)}{dt} = \dot \theta_2 ⋅ R##
About ##x_1## and ##x_2## all I can think of that ##x_2## travels half a distance as ##x_1##.
 
  • #15
kuruman
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I'm stuck with this. The only relation I found between ##x_2## and ##\theta_2## is ##x_2 = \theta_2 ⋅ R##
So with derivation ##\frac{d(x_2)}{dt} = \dot \theta_2 ⋅ R##
That is correct for rolling without slipping. If there were slipping, that would not be correct. Just so that you see that good old ##v=\omega R## may or may not be applicable.
About ##x_1## and ##x_2## all I can think of that ##x_2## travels half a distance as ##x_1##.
That is also correct but why? Hint: What is the speed of the top point of contact relative to the center and what is the speed of the center relative to the bottom point of contact?
 
  • #16
kuruman
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That means my calculation with the Lagrangain is correct, I just used wrong generalized coordinates?
There are no right or wrong generalized coordinates as long as they uniquely specify the state of all the components of a system. The way you got the Lagrangian in terms of one coordinate might fail you if the problem becomes more complicated, for example having a third smaller cylinder rolling inside the hollow one. The strategy of first using enough coordinates to specify each component separately and then using constraints to whittle down the number of coordinates and hence the equations of motion, will see you through. This strategy is particularly useful when you have relative motion between two or more components (not the case here).
 
  • #17
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That is also correct but why? Hint: What is the speed of the top point of contact relative to the center and what is the speed of the center relative to the bottom point of contact?
I think I get it. The speed of ##P_{bottom}## relative to the ground is 0.
The speed of ##P_{center}## relative to the ##P_{bottom}## is ##R⋅\omega##
##P_{top}##'s speed relative to the ##P_{center}## is ##R⋅\omega##
So ##P_{center}##'s speed relative to the ground is ##2⋅R⋅\omega##
Another attempt to generalize the coordinates again:

##KE_2=KE_T+KE_R=\frac{1}{2}⋅m_2 ⋅\dot x_2^2+\frac{1}{2}⋅I_2⋅\omega_2^2=##
##=\frac{1}{2}⋅m_2⋅\dot x_2^2+\frac{1}{2}⋅(m_2⋅R^2) ⋅(\frac{\dot x_2}{R})^2##

R here is a so called arbitrary constant? And now it's safe to put ##\dot x_2 = \frac{\dot x}{2}## into the equation? If this isn't correct… another relation came into my mind between ##x_1## and ##x_2##. A wave function. The distance between them. At ##x=0## the distance is ##0## and at the ##\pm A_{max}## the distance is at max. Now as I wrote it down this can't be applied because the plank isn't a point mass (or its attached to 2 objects. So at the other object I have to use another coordinate for the position of the plank)?
 
  • #18
kuruman
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I think I get it. The speed of ##P_{bottom}## relative to the ground is 0.
The speed of ##P_{center}## relative to the ##P_{bottom}## is ##R⋅\omega##
##P_{top}##'s speed relative to the ##P_{center}## is ##R⋅\omega##
So ##P_{center}##'s speed relative to the ground is ##2⋅R⋅\omega##
You got it! :partytime:

Another attempt to generalize the coordinates again:
##KE_2=KE_T+KE_R=\frac{1}{2}⋅m_2 ⋅\dot x_2^2+\frac{1}{2}⋅I_2⋅\omega_2^2=##
##=\frac{1}{2}⋅m_2⋅\dot x_2^2+\frac{1}{2}⋅(m_2⋅R^2) ⋅(\frac{\dot x_2}{R})^2##
This is correct so far.
R here is a so called arbitrary constant?
No. ##R## is the radius of the cylinders as shown in your drawing. It is a given constant, just as the masses are given.
And now it's safe to put ##\dot x_2 = \frac{\dot x}{2}## into the equation?
If this isn't correct…
You need to have a clear mental picture of what distance is what and what affects what. Let's put the origin to where the left end of the plank is when he spring is relaxed. Then certainly ##\frac{1}{2}kx_1^2## is the potential energy term in the Lagrangian. Could ##x_1## denote the position of the plank? Sure, if you specify that it is defined as the distance of the left tip of the plank from the origin. Because the plank is a rigid body, once you know the position of its left tip, you know the position of any other point on it that is always at a constant distance away from it. Now for the cylinder ##m_2##. When the spring is relaxed, (##x_1##=0) its center is at distance ##x_2## from the origin. That distance can be anything and you cannot write down a relation between ##x_1## and ##x_2##, but you don't have to because ##x_2## appears nowhere in the Lagrangian. All you need is a relation between ##\dot x_1## and ##\dot x_2## which you already have. Similar considerations apply for the other cylinder.

Of course to specify the solution completely you need the initial conditions at ##t=0##, namely
##x_1(0),~\dot x_1(0),~x_2(0)## and ##x_3(0)##. The rest of the initial conditions, ##\dot x_2(0),~\dot \theta_2(0),~\dot x_3(0),~\dot \theta_3(0)## can be obtained from the constraints.
 

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