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Torques, forces and COM

  1. Sep 13, 2013 #1
    It's not a homework problem as I am way too old already but it's something that had made me given problem getting to sleep. I'm probably missing smthg.

    let's say you have a horizontal uniform rod lying on ice such as there is no friction and the normal force equal the weight of the rod. Let's say the left end of the rod is at x=0 and the right end is at x=L
    now 10N force is applied in the +Y direction at X=L and a force of 10N is applied in the -Y direction at X=0,9L.
    No other forces acts on the rod therefore : the acceleration of the center of mass should be 0 and if we assume the rod is not moving before the forces are applied then the center of mass at X=L/2 should be fixed. This is somewhat hard for me to vizualize , try this with a pen lying on a table and apply two approx equal force in the opposite direction at two points near one end of the pen you can clearly see a shift of the pen toward one direction the COM is clearly moving.

    Moreover in the example above if you calculate the rotation about x=0 due to the torques you have something like Torque=10N*1L-10N*0,9L... whatever result there is an angular acceleration about x=O making the rod rotate about it . I understand that the point on the rod at x=O is not fixed , is that the origin of my problem ? there is actually a rotation around x=0 but there is also a translational motion of x=0 such that the rotation about it + the T motion make the actual COM fixed when you sum the effects ??? and so the example with the pen on the table is not a good example for visualizing the problem ??
     
    Last edited by a moderator: Sep 13, 2013
  2. jcsd
  3. Sep 13, 2013 #2

    haruspex

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    Your analysis is correct. The problem with pushing the pen with your fingers is that it is just about impossible to arrange that the forces are equal. You will naturally tend to advance the fingers at equal and opposite velocities, the forces being whatever is required to achieve that. You would have a better chance attaching a string at two points on the rod, with the length of the string passing off one side of the table, down to and through a pulley, and back up on the other side of the table. pulling the pulley gently downwards should exert equal forces (subject to friction between rod and table).
     
  4. Sep 13, 2013 #3
    Thank you very much, I understand that actually by trying to use the same force I m actually advancing fingers at equal and opposite velocities. are we not in a situation where I am actually giving opposite but equal linear momentum m.v ? so p1+p2 = MVcom which should be 0 as well then ???
     
  5. Sep 13, 2013 #4

    AlephZero

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    The "pen on the table" is a poor analogy to the ideal situation, because of the friction forces between the pen and the table.

    If you apply "approximately equal" forces at each end, you will start the pen rotating about some point other than its CG. When the pen starts rotating about an off-center point, the friction forces on the parts of the pen moving in opposite directions also affect what happens next.

    If you "push with equal and opposite force near one end",. it's more likely that you are really applying equal and opposite displacements, which is a completely different situation - and the CG does then move, because the forces you apply are not equal and opposite.

    If you really want to do the experiment, you would need to apply the forces using something like two equal weights hanging from strings running over pulleys. And you would also need to take some action to eliminate friction - use an air table so the rod is "floating" on air like a hovercraft, do the experiment on a block of melting ice (provided you can keep the surface horizontal somehow), or at least use something like teflon tape to reduce the friction.
     
    Last edited: Sep 13, 2013
  6. Sep 13, 2013 #5

    haruspex

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    In order that your fingers advance with equal and opposite velocities, the finger closer to the mass centre will exert the greater force. (It is not like striking the rod with two equally massive objects with equal and opposite velocities.)
     
  7. Sep 13, 2013 #6

    stevendaryl

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    If you push on one end of a long object, your push will do two things:
    1. get the object as a whole moving in the direction you are pushing, and
    2. start the object spinning

    The idea of torque is to disentangle these two effects.

    Here's a way to think about it, in the special case of two masses connected by a light, thin rigid rod (like a dumbbell). Imagine the two masses lined up left-to-right, lying on a flat surface. If you push on the left mass, you'll set the dumbbell rotating clockwise. If you push on the right mass, you'll set the dumbbell rotating counterclockwise. There is one spot along the rod so that if you push on that spot, the dumbbell will start moving forward without rotating. That's the center of mass. The amount of rotation imparted by the force clearly depends on the magnitude of the force and how far from the center of mass the force is applied. That's what torque measures: the combination of these two factors.
     
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