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Going Loopy (vertical circular motion)

  1. Jun 15, 2011 #1
    Dear physicists

    I am a physics teacher in London. Whilst I can understand exam style questions on circular motion (you would hope so !) I do have something which is bugging me.

    It is regarding an object completing a "loop-the-loop" inside a track. (like a roller coaster, but lets keep this simple and imagine a block on a smooth surface).

    My problem is this:

    http://www.mpklein.co.uk/loop_the_loop.png

    - At the beginning (travelling horizontal) there is no resultant force on the block as the weight and normal reaction cancel. Newton's 1st Law says the block continues at it's steady speed (I am fine with this)

    - As the incline begins, the normal reaction tilts to the left. The weight is still vertically downwards. If we think about the RESULTANT of these two forces, they no longer cancel. In fact, if we put both forces tip-to-tail, the resultant is down the slope (so the speed of the block should decrease as it follows the path)

    So, how does the block perform circular motion? To perform circular motion we need a RESULTANT force towards the centre of the circle. Which there isn't. I appreciate that the normal reaction points towards the centre of the circle, but this is not a resultant. I cannot even understand why the block moves upwards. I appreciate it has to follow the track which climbs upwards, but surely we should be able to explain this upward motion in terms of forces? The magnitude of the normal reaction is less than that of the weight - there is definitely no resultant force upwards.

    Please could somebody enlighten me in simple steps?

    Thank you

    Matt
     
  2. jcsd
  3. Jun 15, 2011 #2
    Newton's first law is the key. An object keeps its momentum as long as there are no forces acting on it. The "normal reaction" makes sure that the vehicle moves along the loop and the gravitational force pulls it down. There is however no reason for the vehicle to stop as long as those forces are not strong enough. How far it gets depends on the initial velocity and therefore kinetic energy. If it is high enough, the forces will be not strong enough to "drain" the vehicle's momentum in time, so that it can pass the loop.
     
  4. Jun 15, 2011 #3
    Thank you for your reply Polyrhythmic

    It seems that you are suggesting we treat the effect of each type of force separately i.e. one to do the "steering", and one to keep the block on the track.

    In similar examples of circular motion, it is just the resultant that matters. For example, a charged particle moving in a magnetic field.

    Surely if the resultant on the block is "diagonally down to the left" it should not steer upwards?

    It seems as if ONLY the normal reaction is effecting its direction?
     
  5. Jun 15, 2011 #4
    The point is that an object can move in one direction even if all forces acting on it point in the other, as long as it has some initial momentum. If the forces are sufficiently small, they won't slow down the object "instantly". The resultant of your forces could point anywhere, as long as there is some initial momentum which keeps the vehicle going.
     
  6. Jun 15, 2011 #5

    Doc Al

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    Staff: Mentor

    This is incorrect. For uniform circular motion, the resultant force will be centripetal (since there's no tangential acceleration). But this is an example of non-uniform circular motion.
     
  7. Jun 15, 2011 #6
    Hi Matt,
    there are cases where the resultant force may be towards the centre of the circle for example an object being whirled at STEADY speed on a string in a horizontal circle.With these examples the object is accelerating due to a change of direction only.With your example the acceleration is more complex and due to both a change of speed as well as a change of direction.The direction of the resultant force includes the fact that there is a vertical acceleration due to gravity.
     
  8. Jun 15, 2011 #7
    Thank you

    Yes, it does seem I overlooked the fact that this is non-uniform.

    What if we just consider the components of forces perpendicular to the track? Initially this is zero, but on the incline we would find a resultant towards the centre of the circle. Perhaps this is the explanation?

    It does of course make sense that the resultant is down the track (so a block attempting to cruise up will slow down, and possibly slide back the way it came).

    Despite all your patient answers, I just still cannot explain how the block is able to move upwards, other than by saying it follows the track !
     
  9. Jun 15, 2011 #8

    Doc Al

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    Staff: Mentor

    The resultant force will have a component towards the center of the circle. That's what makes the object change direction to follow the track.

    Sure.

    It has an initial velocity. Due to the forces on it supplied by gravity and the track, the velocity changes as it moves along, both in magnitude and direction.
     
  10. Jun 15, 2011 #9
    Thank you Doc Al

    This does make sense when you say that only a component is required towards the centre of the circle ...

    ... BUT ...

    I am fairly sure the resultant is parallel/tangential to the slope. So surely this would never have a component towards the centre?
     
  11. Jun 15, 2011 #10

    Doc Al

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    On what basis do you conclude this?
    If it were true, yes. And then it couldn't move in a circle.
     
  12. Jun 15, 2011 #11
    You may have just helped me by asking that.

    I was thinking that, if you resolve forces for a stationary block on a slope, weight mg, you find that: the component of the weight parallel (down) the slope is mgsin(theta), and perpendicular to the slope it is mgcos(theta).

    So it is probably wrong to assume that the reaction equals the component of weight perpendicular to the slope, mgcos(theta). This would only apply in equilibrium.

    I am guessing then, to enter the circular motion, this normal reaction must be greater than the perpendicular component of weight.

    But how would this be? Where would this larger force come from? What factors might affect the size of the reaction from the slope?
     
  13. Jun 15, 2011 #12

    Doc Al

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    Correct. You can only conclude that because you know that there is no acceleration perpendicular to the slope. (Not so with our curved track problem.)

    Yes.

    You may be thinking that the normal reaction is fixed (as it would be in the inclined plane example). But it's not fixed, it depends on how fast the object is moving at any point. It works like this: The normal force will be whatever it needs to be to prevent the object from pushing through the track. (Of course, if the object moves too fast the track may not be able to provide the needed force--the object breaks through the track.)

    You deduce the required normal force from the resulting kinematics and Newton's 2nd law.
     
  14. Jun 15, 2011 #13
    This makes sense, thank you.

    And this deducing you refer to. Would that be something like this:

    If F is the centripetal force (m*v*v / r), and N is the normal:

    At 6 o'clock : F = N - mg
    At 12 o'clock: F = N + mg
    At 3 o'clock : F = N

    So the track produces the right amount of normal reaction to satisfy N in these equations?
     
  15. Jun 15, 2011 #14

    Doc Al

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    Looks good to me.
     
  16. Jun 15, 2011 #15
    Thank you Doc Al, and everybody else that has contributed. You really have been very helpful.

    The equations in the last post are obviously the cases you are usually asked about in your school/college exams.

    Interesting how I have turned "full circle" going from those equations in class, to being confused, and ending up back at the equations.

    I think it shows how thorough and and helpful you have been. I can't see a button to commend your replies, but otherwise I would.

    Thanks again

    Matt
     
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