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Some Mechanics Problems (Could someone verify my work?)

  1. Nov 18, 2005 #1
    Could someone look over the problems I have before and tell me if my reasoning / answers look alright? I'd really appreciate it.

    For this problem I am going to say that that only iii is true. To prove that the others are wrong (or to give an example), I think of a block on a table. If a force pushes down perpendicular to the table, the block won't move (no work done), but there is an impulse because there was a force exerted for a certain amount of time.

    I set this up as follows:

    [tex]\frac{5m}{8}=\frac{3m}{8}+1\implies m=4\text{kg}[/tex]

    So the answer would be i.

    Using conservation of momentum, I come up with the daughter's final KE being 9 times the original. So answer iv is correct. The first choice is false (it is three times the initial), and the second and third choices are obviously false as well. The last seems false because only the momentum is transferred. I am going to say only choice iv is correct.

    Choice 2 seems true, because the angular velocity does not change with the radius. Three is also true by looking at the parametric equations of a cycloid (also, the direction). 4 is true because it is momentarily at rest at the bottom (as seen in a cycloid). I am going to say answer choices ii and iv are true.

    Thank you very much for your help :smile:.
     
  2. jcsd
  3. Nov 19, 2005 #2

    Galileo

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    Are you sure there is a force acting on the block? If it did it should accelerate according to Newton's 2nd law, but it doesn't. Ofcourse, what you should consider is the NET force. The net force is zero because of the normal force. Recheck the definition of impulse.

    I can't see where you got that equation from.
    If the rod is balanced, the net torque about the pivot point is zero. Equate the torque from the center of mass and the 1kg mass. I got 3 kg.

    Totally correct. :smile:
    The reason kinetic energy is not conserved is because this is not a closed system in a sense. When they push against each other, work is done! This work is precisely equal to the gain in kinetic energy of the system.
     
  4. Nov 19, 2005 #3
    What do you mean by that last sentence? I know that torque is [itex]\tau=\|\vec{r}\times\vec{F}\|[/itex], so in this case it is [itex]\tau=rF[/itex]. What do you mean by equate the torque from the center of mass and the 1kg mass?

    Right, the net force exerted over a period of time. In this case then, the answer would be i only, right?

    Thanks for your help :smile:
     
  5. Nov 19, 2005 #4

    Galileo

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    The total torque about the pivot point is a sum of two contributions. One is ofcourse from the gravitational force acting on the mass at a distance of 15 cm. The other is the torque from gravity acting in the center of mass of the bar (5 cm from the pivot point), this torque has the opposite direction of the other one. The system balances, so the total torque should be zero.
    [tex]\N = \vec r_1 \times \vec F_1+\vec r_2 \times \vec F_2 =0[/tex]
    which, in this case, becomes simply (watch the signs):
    [tex]gm_1r_1-gm_2r_2=0 \iff m_1r_1=m_2r_2[/tex]

    What about iii and iv? Any thoughts about those?
    The impulse equals the change in momentum and the work done equals the change in kinetic energy. Remember that momentum is a vector, while energy is a number. Can you think of a way to change one without the other? (i.e. impuls without work or vice versa?) this may also alter your view on answer i
     
    Last edited: Nov 19, 2005
  6. Nov 19, 2005 #5
    OK that makes sense about the torque; I see how you came up with 3:

    [tex]5m=15\implies m=3\text{kg}[/tex]

    Now, about the impulse / work question:

    I know that momentum is a vector and KE is a scalar, but it still seems that iii and iv are false. The impulse is the net force exerted over a period of time and the work is the net force exerted over a distance, right? If a force is applied for a certain time, the object must move right? Because of this, if an impulse is delivered, work is being done. Now if work is done, the force must be exerted for a period of time, thus an impulse is being delivered as well. If I changed the time a force was exerted, it would also change the distance. If I changed the distance, it would change the time (assuming the force doesn't change). Where am I going wrong here?

    Thanks a lot for the help.
     
  7. Nov 19, 2005 #6

    Galileo

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    Let's look at this a little more closely. If a force is delivered and the object is moving, is work always being done? Let's consider a displacement over a small distance [itex]\vec d[/itex]. The work done when the object moves this small distance is:
    [tex]W=\vec F \cdot \vec d[/tex]
    Is this always nonzero when F and d are nonzero?
    Now can you think of a motion that allows momentum to change while no work is being done?
     
  8. Nov 19, 2005 #7
    [itex]W=\vec{F}\cdot\vec{d}[/itex] is zero when F and d are perpendicular. So are you saying that if a force is applied perpendicular to the direction of motion that there is an impulse? I can't seem to see how that works because 1. the object would either move it the direction of the force or 2. a normal force would cancel out the force. Sorry for being difficult here. I appreciate your help.

    Thanks again.
     
  9. Nov 20, 2005 #8

    Galileo

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    What if the force is always perpendicular to the direction of the motion?
     
  10. Nov 20, 2005 #9
    Would the centripetal force of an object in circular motion fall into this category?
     
  11. Nov 20, 2005 #10

    Galileo

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    Sure does.
     
  12. Nov 20, 2005 #11
    OK that clears things up. Going back to the original problem, this means that answer i is false. This leaves me with iii and iv as true answers. Is this correct?

    Thank you.
     
  13. Nov 20, 2005 #12

    Galileo

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    I agree about answer iii. In circular motion, no work is done on the object (F is perpendicular to v everywhere), so no change in energy, however, since the direction of motion changes continuously its momentum does too.
    However, this does not show that work can be done on an object without changing its momentum. So iv is still open.

    Hint: Can [itex]T=\frac{1}{2}mv^2=\frac{p^2}{2m}[/itex] change while [itex]\vec p=m\vec v[/itex] stays constant?
     
  14. Nov 20, 2005 #13
    No, it cannot (because the mass is constant). So right now, after looking at 4 again, it seems that 1 must be true as well as 3. 2 and 4 I'm not totally sure about, but I'm leaning towards "false" looking at the equation you posted.
     
  15. Nov 21, 2005 #14

    Galileo

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    Doh! Why do you change your mind on statement i?:grumpy:
    You just said last post that you refuted it, because in circular motion an object it receives impulse but no kinetic energy. And that's correct.

    I agree that kinetic energy cannot change without changing the momentum.
    But you can change momentum while keeping the kinetic energy constant. This is because momentum has magnitude AND direction. Circulair motion is exactly when you keep the magintude of p constant whilst changing the direction. That is why I reminded you of the fact that p is a vector and KE is a scalar.
     
    Last edited: Nov 21, 2005
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