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Stick on table

  1. Apr 26, 2005 #1
    A thin uniform stick of mass [tex]m[/tex] with its bottom end resting on a frictionless table is released from rest at an angle [tex]A[/tex] to the vertical. Find the force exerted by the table on the stick at the moment of release. Express your answer in terms of [tex]m, g[/tex] and [tex]A[/tex].

    Ive been reliably informed (from the lecturer) that angular momentum is to be used but I don’t really understand how, or how it would effect the force exerted by the table.

    Any help is appreciated

    picture is here

    http://picturerack.com/users/includes/flopt.asp?todo=3&inp=/richiec/
     
  2. jcsd
  3. Apr 26, 2005 #2

    Andrew Mason

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    I don't see how angular momentum can be a factor since it is 0 at the moment of release. But torque (dL/dt) is definitely a factor.

    What is the normal force at the moment of release and what is the rate of angular momentum change at that moment? Think of the torque as acting on the centre of mass of the rod.

    AM
     
  4. Apr 26, 2005 #3
    AM,

    I saw this problem and it looks tricky. When you say torque acts at the center of the rod, do you mean "about" the center? or are you talking about the force of gravity acting at the center? the torque depends on the refrence point right?
     
  5. Apr 26, 2005 #4
    NoName,

    This is a pretty good problem!

    Since the table is frictionless, what will the path of the center of mass of the stick be as it falls? So where will the center of rotation be?
     
  6. Apr 26, 2005 #5

    Andrew Mason

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    What I should have said that gravity acts at the centre of mass which is at the centre of the rod.

    The torque is a little tricky. You can look at it a couple of ways. You could look at it as the gravitational force on the centre of mass about the sliding end. Or you could look at it as the normal force on the sliding end about the centre of mass. It works out to be the same.

    AM
     
  7. Apr 26, 2005 #6
    Do u mean the torque is the same measured both ways? How can you know that without knowing the normal force, which is what the question asks for?
     
  8. Apr 26, 2005 #7

    siddharth

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    Think of the forces acting on the rod. Draw the free body diagram of the rod. So which way will the center of mass of the rod fall? .

    The instantaneous center of roation is a point about which the rod will appear to purely rotate for that instant.

    Remember, the instantaneous center of rotation need not lie on the rod itself and will also be at different points as the rod falls. Once you find that point, since you know that the rod appears to purley rotate about that point, find the torque due to the various forces. Also find a relation between the acceleration of the center of mass and the angular acceleration about the center of rotation and you should be able to calculate the Normal force.
     
  9. Apr 27, 2005 #8
    That's a good idea. I didn't think of that.
    Can't u use any point since the whole rod is at rest at that instant?
     
  10. Apr 27, 2005 #9
    Thanks everyone for your help! Ive been sketching various stages in its falling and am carrying out several calculations. But then I had to sleep. Im true in also thinking that mechanical energy is conserved yes?
     
  11. Apr 27, 2005 #10

    OlderDan

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    Yes indeed, and I just worked it out that way.

    [tex]U = mgh + \frac{1}{2}mv^2 + \frac{1}{2}I\omega ^2 = mgh + \frac{1}{2}mv^2 + \frac{1}{2}I\left( {\frac{v}{{L/2}}} \right)^2[/tex]

    [tex]U = mgh + \frac{1}{2}mv^2 + \frac{{2I}}{{L^2 }}v^2 = mgh + \frac{1}{2}mv^2 + \frac{1}{6}mv^2 = mgh + \frac{2}{3}mv^2[/tex]

    [tex]\frac{{dU}}{{dt}} = mg\frac{{dh}}{{dt}} + \frac{4}{3}mv\frac{{dv}}{{dt}} = - mgv + \frac{4}{3}mv\frac{{dv}}{{dt}} = \left[ {\frac{4}{3}\frac{{dv}}{{dt}} - g} \right]mv = 0[/tex]

    [tex]\frac{4}{3}\frac{{dv}}{{dt}} = g[/tex]

    [tex] \frac{{dv}}{{dt}} = a = \frac{3}{4}g[/tex]

    [tex]F = ma = mg - N = \frac{3}{4}mg[/tex]

    [tex]N = mg - \frac{3}{4}mg = \frac{1}{4}mg[/tex]
     
    Last edited: Apr 27, 2005
  12. Apr 27, 2005 #11

    siddharth

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    Dan, I dont think you can use v=(l/2)w in this case. Look at the motion of the point in contact with the ground. The velocity of that point is always parallel to the ground. The center of mass falls in a straight line with its velocity perpendicular to the ground. So the the center of rotation is at a point from which both these velocites are perpendicular.

    So we have
    [tex] mg - N = ma [/tex]
    as the only forces are in vertical direction. In fact this is why the COM falls in a straight line

    [tex] mg\frac{l}{2}\cos\theta = I\alpha [/tex]

    The Torque due to the Normal is 0. Because the perpendicular distance from the Normal force to the center of roation is 0.


    [tex] I=\frac{1}{12}ml^2 + \frac{1}{4} ml^2(\cos\theta)^2 [/tex]

    From parallel axis theorem
    and also

    [tex] a=\frac{l}{2}(\cos\theta)\alpha [/tex]

    From this N can be found.

    EDIT: The angle theta is between the horizontal and the rod.
     
    Last edited: Apr 27, 2005
  13. Apr 27, 2005 #12

    OlderDan

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    The kinetic energy of a translating rotating object can be written as the sum of the translational kinetic energy of the center of mass plus the rotational kinetic energy about the center of mass. As you said, the center of mass in this problem moves in a straight line, and it is the point whose acceleration is along the line of the net force. So that is the center of rotation. The point of contact is not a stationary point. If it were constrained to not move, you would have pure rotation around that point and would use the parallel axis theorem to find the moment of inertia.

    If the point of contact was the center of rotation, the moment of inertia about that point would be constant. It cannot depend on orientation, i.e., it cannot be a function of theta.
     
  14. Apr 27, 2005 #13

    siddharth

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    I am not talking about the point of contact with the ground. I am talking about the instantaneous center of rotation.

    To find it, first Draw the diagram of the rod at an angle theta(with horizontal).
    Draw arrows indicating the direction of velocites of the end of the rod in contact with the ground (parallel to the horizontal) and the velocity of the center of mass. Now draw two perpendiculars to the velocity arrows you have drawn. The point where they meet is the instantaneous center of rotation. About this point the body purely rotates for that instant.

    In your case, you assumed that the COM is falling with a velocity v and angular velocity is w about the Center of Mass. Calculating the velocity of the point in contact with the ground we get, v(P)=-vcm(j) + l/2wsin(theta)(i) + l/2wcos(theta)(j).

    But the velocity of that point is directed along the x-axis only. Therefore the y-component of the velocity is 0.

    From that you get v=l/2wcos(theta) and NOT v=(l/2)w
     
  15. Apr 27, 2005 #14

    OlderDan

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    Then I think you have a contradiction

    The torque in your equation [itex] mg\frac{l}{2}\cos\theta = I\alpha [/itex] is the torque about the contact point. The statement that the torque due to the normal is zero is only true if the contact point is the point of rotation. If there is some other point of rotation, then the normal force produces a torque.

    Your assertion that the velocity of the point of contact is horizontal is clearly correct. The floor constrains the motion. Look at what that velocity is in relation to the CM.

    [tex] v_{cm} = -v_{cm}j[/tex]

    [tex]v_P = -v_{cm}j + \frac{L}{2}\omega sin(\theta)i + \frac{L}{2}\omega cos(\theta)j[/tex]


    and their DIFFERENCE describes the relative motion between them

    [tex]v_{rel} = v_P - v_{cm} = \frac{L}{2}\omega sin(\theta)i + \frac{L}{2}\omega cos(\theta)j[/tex]


    The relative velocity is perpendicular to the rod, and its magnitude is l/2w, as it must be. If you were in the center of mass frame of reference, you would see the end of the rod in circular motion around you at a distance L/2, a fact that you used to figure out the horizontal velocity of the contact point.
     
  16. Apr 27, 2005 #15
    OlderDan,

    But your solution is independent A, the initial angle. How is that possible? If A = 0 then N = mg, not mg/4.
     
  17. Apr 27, 2005 #16

    siddharth

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    First of all i think the both of us are using different theta's.
    Then,

    No, The torque in the above statement is NOT about the contact point but about the Instantaneos center of Rotation. I am unable to explain this clearly without a diagram. Anyway, let me try again

    Draw a straight vertical line along the normal force from the end of the rod. Draw a straight horizontal line from the center of the rod perpendicular to the direction of mg. The point where they meet is the instantaneous center of rotation (not to be confused with the contact point).

    If you find the torque due to normal about the instantaneous center of rotation, you will find it is zero


    From the Ground frame, the velocity of the contact point is only along the x-axis. (As you said, it is constrained to move on the x-axis)

    Also From the ground frame the velocity of the contact point is

    [tex]v_P = -v_{cm}j + \frac{L}{2}\omega sin(\theta)i + \frac{L}{2}\omega cos(\theta)j[/tex]

    But the component of the velocity in the ground frame in the y-direction is zero. So from that you get vcm=l/2wcos(theta).


    I used the fact that v<vector>=r<vector>w
    But there v is not equal to l/2w.
     
  18. Apr 27, 2005 #17

    OlderDan

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    Good point!! There is indeed an error in my analysis. The original premise of conservation of energy is incorrect, not the geometry. I will see if I can make the appropriate correction and still approach the problem from a work energy prespective.
     
  19. Apr 27, 2005 #18

    Doc Al

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    Why wouldn't mechanical energy be conserved?
     
  20. Apr 27, 2005 #19

    OlderDan

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    Because work is being done on the rod by the normal force. Just because the point of contact of the force with the rod is not moving does not mean the object as a whole is not moving. We know the center of mass will move. I need to add a term that accounts for the F(h).dh
     
  21. Apr 27, 2005 #20

    Doc Al

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    Maybe I'm missing something (besides a few marbles), but why not treat this as a straightforward dynamics problem?
    From Newton's 2nd law:
    [tex]mg - N = ma[/tex]

    [tex]N (L/2) \sin A = I \alpha[/tex]

    The rotational inertia about the cm is:
    [tex]I = 1/12 m L^2[/tex]

    The constraint of the contact point slipping along the floor gives:
    [tex]a = \alpha (L/2)\sin A [/tex]

    Solving the above, I get:
    [tex]N = mg/(1 + 3\sin^2 A)[/tex]
     
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