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A rod falling on a frictionless surface

  1. Jan 23, 2016 #1
    1. The problem statement, all variables and given/known data

    Consider a massless rod of length $L$ with a small mass $m1$ attached on one end, and $m2$ attached on the other end. The rod is initially in the vertical position at rest on a frictionless surface, with $m1$ on bottom and $m2$ on top. A small impulse is applied to the top of the rod in the horizontal direction, and it begins to fall.

    Find the speeds of $m1$ and $m2$ when $m2$ is about to hit the ground.

    2. Relevant equations

    I.Conservation of energy: $m_2gL=1/2m_1v_1^2+1/2m_2v_2^2$
    II.Conservation of momentum in the horizontal direction: $m_1v_1=m_2v_{2_x}$
    III.Center of mass is located at $d_{cm}=(m_1*0+m_2*L)/(m_1+m_2)$ along the rod (with $m1$ at the origin).

    3. The attempt at a solution

    Since the only force acting on the system is gravity and the normal force on $m1$, which are both in the vertical direction, linear momentum is conserved in the horizontal direction, and therefore the center of mass of the rod does not change its position in the horizontal direction.

    The net force on the system is $(m_1+m_2)g-N$ downwards, but this isn't very helpful since the normal force $N$ changes with time.

    I had an idea to consider angular momentum about the mass $m1$ or maybe about the center of mass of the rod, but I'm not really sure how to proceed.

    The problem is solved if we can find the horizontal component $v_{2_x}$ in terms of $v_2$, because then we can use equations I and II above to solve for $v1$ and $v2$. Equivalently, the problem is solved if we find $v_{2_y}$. But $v_{2_y}$ is related to the vertical speed of the center of mass of the system, $v_{cm}$ by a constant factor (using similar triangles). Therefore it suffices to find $v_{cm}$ just before $m_2$ hits the ground.

    I am tempted to say, by conservation of energy, $(m_1+m_2)gd_{cm}=1/2(m_1+m_2)v_{cm}^2$, or $\sqrt{2gd_{cm}}=v_{cm}$. For some reason I don't think I can apply conservation of energy to the center of mass, but let's go on anyway.

    By similar triangles, the height $h_cm$ of the center of mass above the ground is related to the height $h_{m_2}$ of $m_2$ above the ground, by: $h_{m_2}=h_{cm}*L/d_{cm}$. Thus $v_{2_y }=v_{cm}*L/d_{cm}$. I continued like this and I got an immediate contradiction. So I know I did something wrong, not with the algebra, but with the assumption that you can apply conservation of energy to the center of mass.

    So now I'm stuck.
     
  2. jcsd
  3. Jan 23, 2016 #2

    haruspex

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    An awkward aspect of such a question is whether the lower mass might lose contact with the ground before the upper mass lands. If the upper mass were given a sufficient horizontal impulse at the start, this could certainly happen.

    Putting that aside for now, think about energy and linear momentum.
    Angular momentum will not be conserved, no matter what axis you choose. The weight of the upper mass and the normal reaction create an unknown torque. You cannot eliminate the normal reaction by taking moments about the lower mass. You have to use an axis which is either stationary or passes through the system mass centre, nothing else gives the right answer. (Not quite true, but near enough.)

    Not sure whether working with the common mass centre helps. It can be done either way. I would set a variable, x, as the horizontal displacement of the lower mass and theta as the angle to the vertical. You can then represent the horizontal and vertical displacements of the upper mass in terms of those, and find the velocities by differentiating. Finally, plug those into the energy and horizontal momentum equations.
     
  4. Jan 23, 2016 #3
    I don't think that method will work, because it only gives you the rate of change of the horizontal and vertical displacements of each mass in terms of the angle to the vertical, not in terms time (i.e you can only find dx/dtheta, not dx/dt).

    The problem is relatively simple, so I'm thinking there should be a simple observation which solves it without too much algebra.
     
  5. Jan 23, 2016 #4

    haruspex

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    Why do you need to know the time? You can find rates of change as a function of position without knowing them as a function of time.
    (I obtained the solution before posting.)
     
  6. Jan 23, 2016 #5
    Well how can you plug them into the energy and momentum equations, which involve velocities (rate of change as a function of time)? Could you show me the first steps of your solution, and I'll try to complete it from there.
     
  7. Jan 23, 2016 #6

    haruspex

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    First, represent the position of each mass in terms of the two variables I mentioned, x and theta.
    Then differentiate as necessary to obtain velocities. Plug those into the two conservation equations.
    You should be able to find the angular velocity as a function of the angle.
     
  8. Jan 24, 2016 #7
    Yes taking the center of mass of the system to have x-coordinate zero, I found the position vector of m1 as <dsin(theta),0> and for m2, <(d-L)sin(theta), Lcos(theta)>, where d is the distance along the rod measured from m1 where the center of mass is located. And, yes, I can differentiate these with respect to theta, but my point is that the derivative does not represent the velocity vector. The velocities in the momentum and conservation equations are of the form distance/time. The "velocities" we have here are of the form distance/angle. I don't understand how you can substitute things for each-other if they have different units...
     
  9. Jan 24, 2016 #8

    haruspex

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    That's because you are differentiating with respect to angle when you should be differentiating with respect to time. Use the chain rule: ##\frac d{dt}f(\theta)=\frac {df}{d\theta}\frac {d\theta}{dt}##
     
  10. Jan 24, 2016 #9
    Ok that was the second thing I was going to say. I did use the chain rule, but I still think it doesn't work.

    You get the velocity vector for m1 is $<dsin(theta)*dtheta/dt,0>$and the velocity vector for m2 is $<(d-L)sin(theta)*(dtheta/dt), Lcos(theta)*(dtheta/dt)>$. Notice that the conservation of linear momentum equation is useless, because if you plug in $v1=dsin(theta)*dtheta/dt$ and $v_2_x=(d-L)sin(theta)*(dtheta/dt)$, you get a formula in terms of the masses, the distance d, and L, which is the same as the center of mass formula. So we only have the energy conservation formula left, but that doesn't give us enough information (and you can't get rid of dtheta/dt)...

    Can you please show me the steps I'm missing?

    I'm pretty sure there is a nicer way to do the problem. Did you get the final result to be sqrt{2gl}?
     
  11. Jan 24, 2016 #10

    haruspex

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    If working with mass centre, yes, you do not need to look at linear momentum. You should end up with an equation relating ##\dot{\theta}## to ##\theta##. Then you plug in ##\theta=\pi/2## and get the angular velocity when the upper mass hits the ground. I can't comment further without seeing your working.
    I ended up with ##m_2L\dot {\theta}^2=2g(m_1+m_2)##.
     
  12. Jan 24, 2016 #11
    I am trying to proceed using your hints, but I must say I am quite confused by what you are saying.

    Ok so first we are looking at different theta's. My theta is the angle of the rod to the horizontal, and if I understand correctly, your theta is the angle of the rod to the vertical?

    If we are using your theta, then r2=<(L-d)sin(theta), Lcos(theta)>, v2=<(L-d)cos(theta)*theta', -Lsin(theta)*theta'>. So when theta=pi/2, v2=<0, -L*theta'>. So, if I understand correctly, m2 has zero velocity in the horizontal direction when it is about to hit the ground. And according to you, the answer should be

    |v2|=L*theta'=sqrt(2gL*(m1+m2)/m2)

    This doesn't seem right.
     
  13. Jan 24, 2016 #12

    haruspex

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    You're right, I dropped a term. I now get ##L\dot \theta^2(m_1+m_2\sin^2(\theta))=2g(1-\cos(\theta))(m_1+m_2)##. On the ground, that reduces to ##L\dot\theta^2=2g##.
    Yes, by conservation of momentum, neither mass will have any horizontal velocity when the rod is horizontal. So I could have avoided all the algebra and just observed that the top mass will hit the ground at the same speed as if in free fall.
    We are left with the issue of proving the system does not become airborne before that.
     
  14. Jan 24, 2016 #13
    Actually, the horizontal impulse given at the top is assumed to be very small, so that it's initial angular velocity is zero. So you don't have to worry about the system becoming airborne before m2 hits the ground.

    Now, I still don't think your answer is right, because according to you, v2=sqrt(2gL), and this does not depend on the masses m1 or m2 at all. But clearly they matter...

    I would really appreciate if you could explain to me exactly how you are getting that equation for theta'. I got v1 and v2 in terms of theta and theta'. Did you simply plug them into the conservation of energy equation?

    Also, there is something weird going on here. As I said in the post above yours, m2 has zero horizontal velocity when m2 is about to hit the ground. Therefore m1 has zero horizontal velocity, and zero total velocity. So v1=0. But then the conservation of energy equation directly gives v2=sqrt(2gL).
     
  15. Jan 24, 2016 #14
    Are you sure about this? I got the same result myself but I thought I must have done something wrong because it is too simple. The masses have to matter, surely,...
     
  16. Jan 24, 2016 #15

    haruspex

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    One might expect that the mass ratio matters, but it does not have to be so.
    You agree that there are no horizontal velocities when the rod is horizontal, so if the lower mass is still one the ground all the energy has gone into the vertical velocity of the upper mass. It follows immediately that ##v_2=\sqrt{2gL}##, matching the answer I got by finding the velocity as a function of angle.
    Yes, the horizontal impulse at the top is small, but that that does not make it clear that the lower mass stays grounded. I note that the problem implies the upper mass is substantially the greater. Why does it do that?
     
  17. Jan 24, 2016 #16
    Upon further reflection, this makes sense. The motion of m2 can be thought of as rotation about the center of mass plus translation of the center of mass. When the rod is horizontal, the velocity vector due to rotational motion is perpendicular to the ground. But the translation of the center of mass is always vertically downwards, so indeed when m2 is about to hit the ground, it's velocity vector should point directly downwards. As you said, it immediately follows that v1=0 from conservation of momentum, and the result is immediate.

    As far as the mass m1 being smaller, that is a mistake. I did not mean to write "a small mass".
     
  18. Jan 24, 2016 #17

    haruspex

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    Good.
    I checked for becoming airborne on the basis that if and when that happens there will be no angular acceleration. (Even though the equations we're using are as though normal force could go negative, it should still be the case that angular acceleration would pass through zero.)
    Applying that to my general equation I found the mass ratio would have to equal ##-(1-\cos(\theta))^2##. Clearly that never happens, but comes closest to being true when the lower mass is tiny and the rod is near horizontal. So even a small initial nudge leaves open the possibility of becoming airborne if the lower mass is very much the smaller.
    Of course, if it does become airborne the problem gets very much harder!
     
  19. Jan 25, 2016 #18

    Suraj M

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    1 small question, you aren't taking the centre of mass to be on the axis of rotation?
     
  20. Jan 25, 2016 #19

    haruspex

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    No, the mass centre can't be on the axis of rotation, except right at the start. As the rod falls, the axis of rotation will describe an arc of a circle. At first it will move out horizontally, and finish moving vertically, ending up where the lower mass finishes. The centre of the circle of which it is a quadrant will be at the point on the ground where the rod initially stood.
     
  21. Jan 25, 2016 #20

    Suraj M

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    Oh that makes everything pretty simple.
     
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