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High school circular motion problem

  1. Nov 19, 2014 #1
    1. The problem statement, all variables and given/known data
    The problem is:
    An amusement park is replacing it's stand-up roller coaster with a G-Force ride. Riders will enter the ride at the apex and lean against the wall of the ride. The G-force ride will speed up until riders reach height h above the apex. Determine the minimum speed v needed to keep riders at a constant height h (in terms of h, β, μs, and constant g). Given the wall of the ride make an angle β with the vertical, and the coefficient of static friction between the bodies and the surface of the walls is μs.

    2. Relevant equations
    a= (4*π^2*R)/T^2
    a∆t = v2 - v1
    d = 1/2(v1 + v2)∆t
    d = v1∆t + 1/2a∆t²
    d = v2∆t - 1/2a∆t²
    v2² = v1² + 2ad
    (sinA)/a=(sinB)/b=(sinC)/c
    c^2=a^2 + b^2 - 2abcosC
    Fnet = ma

    3. The attempt at a solution
    I have tried using the equations above in attempt to see if any can be used to solve this problem, but I am completely stuck. I do not know where to start and I am very confused as to how this problem can be solved.

    I have attached a picture to give a better understanding of the problem.
     

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  2. jcsd
  3. Nov 19, 2014 #2

    Bandersnatch

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    Hi Kaliford,

    It's never a good idea to try and mechanically fit a random equation to the problem. Just like with assembling a jigsaw, it's all about seeing the underlying picture.

    Usually the first thing to do when you see a problem involving forces and motion, is to try and draw a free body diagram. This will help you identify the condition that makes the rider stay at height h (remember Newton's 1st law). Have a go at that. Start by identifying all the forces you think act on the rider. Imagine yourself in the ride and which way you'd be pulled/pressed against.
     
  4. Nov 19, 2014 #3
    I doubt this is correct, but this is the free body diagram I came up with.
     

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  5. Nov 19, 2014 #4
    It's not correct. What is Fa?
     
    Last edited: Nov 19, 2014
  6. Nov 19, 2014 #5

    Bandersnatch

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    Use the picture given in the problem description as a template. That slope needs to be there somewhere.
    Imagine you're standing (or leaning against) the slope of the ride. Name each force you're drawing and describe why you think it should be there. Think of what's the difference in tems of forces present when the ride is rotating vs when it's standing still.


    Don't just drop a picture - describe your thought process. Surely, you must have some ideas, however silly they might seem. We can't help you if we don't know what you're thinking.
     
  7. Nov 19, 2014 #6
    And you have to be familiar with the equation of circular motion and the concept of centripetal acceleration in order to solve this problem...or at least that's what I recommend. The kinematics equations you posted won't be that useful.
     
    Last edited: Nov 19, 2014
  8. Nov 19, 2014 #7
    I assume there would be a normal force acting on the person leaning against the wall, and I think that would be perpendicular to the slope. There is also the force of gravity pulling the person downwards. There is would also be friction. As the ride spins, the person is supposed to move upward, but stay against the wall, so my guess would be that there would be static friction going horizontally to keep the person from sliding in that direction, and kinetic friction going along the slope as the person moves up. There must also be a force causing the person to move upward, but I am not sure what that would be. I do not know if any of what I am saying is right. I find the concept of circular motion confusing.
     
    Last edited: Nov 19, 2014
  9. Nov 20, 2014 #8

    Bandersnatch

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    Alright! That's actually not bad thinking.

    You don't need kinetic friction - the question is a bit sloppily worded, and may suggest that, but you can see they only gave you the coefficient for static friction. Think of it as if the person was already stationary at some height, and you'd like to find the balance of forces that will prevent her from slipping up or down.
    Also, all friction needs to be along the slope - it's always in the opposite direction to motion.

    Since this is rotational motion, you'll need either centripetal or centrifugal force. It'll be easier to use the centrifugal force. This is the force that pushes you towards one side when making a turn in a car. It's what you use when looking at rotation as if you were rotating together with it, so the person in question is stationary with respect to the ride (while the outside observer sees her going in circles = definitely not stationary).
    Centrifugal force always acts in the direction perpendicular to the axis of rotation, and outwards (again, imagine being in a car making a sharp turn).

    A good way to do this, would be to first imagine there is no rotation, hence the only real* force acting on the rider is gravity. Draw a free body diagram by splitting the force of gravity into its components perpendicular and parallel to the slope. Then add reaction forces (the normal force and friction).

    Figure out under what conditions(angle, coefficient of friction) the net force is zero.

    Then imagine the same situation without gravity, but with rotation (so, with centrifugal force). Again, draw a free body diagram following the same steps. Again, figure out when the net force on the rider is zero.

    The last step would be to combine the two diagrams.


    *by "real force" I mean the force that is neither a reaction force nor a component of some other force.



    Let us know should you get stuck, and at which step. Again, remember to describe your thought process, as well as drawing the diagram.
     
  10. Nov 20, 2014 #9
    This was very helpful! I understand what you are saying. However, one thing is that my teacher has told the class to not use the term "centripetal" or "centrifugal" forces, saying that they are fictitious forces, and to instead say the force that is causing this kind of motion (for example, when spinning something attached to a string around in circles, the tension is the centripetal force). I do not know whether or not this is the correct way of doing this sort of thing. Perhaps I just do not understand what he was talking about. Do you happen to have any idea what he may have meant by this? I think I am beginning to understand how to draw the free body diagrams and hopefully solve this problem, but I am afraid of using the term "centrifugal" or "centripetal." Is there another or more specific kind of force to use instead?
     
  11. Nov 21, 2014 #10

    Bandersnatch

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    Alright, we can address that.

    You will always have one of the two appear when describing rotational motion. You can't really get rid of them both. They are not "real forces" in different senses, though.

    Take a look at this ride:
    images?q=tbn:ANd9GcSy16wNUIuQsjFHkt6GLffMJ7oilCv-HlYoEySr4lalGhI6GPIrfA.jpg

    It's not too dissimilar to the one in your question. There are only two real forces acting on the riders: gravity and normal force from the seats (supplied by, i.e., equal to, the tension of the chains).

    Try drawing a FBD for one rider, as seen from the ground (observer is stationary, or more precisely, not accelerating).
    You'll have gravity pointing down, and the other force (let's just call it tension, even though no rider is stricly speaking dangling by their neck from the chains) pointing along the chain. If you split the tension into its component forces along the vertical and horizontal axes, you'll end up with gravity being negated by the vertical component (it has to be, otherwise the person would not stay at the same height) and the unbalanced horizontal component pointing towards the axis of rotation.

    Now, this unbalanced force is acting as the centripetal force. That is to say, it's not like there appears some new force that makes the riders go in circles. It's the horizontal component of the tension in the chains that supplies enough force to make them go in circles.

    In this sense, the centripetal force is not "real". There's always some other force that acts as such. However, you can be sure that as long as something is going in circles, there will exist some such force (otherwise according to Newton's 1st law, the object would just travel in a straight line at a constant velocity, or remain stationary).

    To reiterate, you will never add the centripetal force to the FBD and say that it makes something move. You will always point to some other force and say that this one is acting as the centripetal force.
    Since the centripetal force required to keep something in motion can be easily calculated, you can then say something about the magnitude of the supplying force, and how does it need to change as you change the speed of rotation and the distance from its axis.

    In your problem, there are no chains, but there is the equivalent of a chair(the slope against which riders are leaning), whose normal force will supply the unbalanced component pointing towards the axis of rotation.

    Now, as for the centrifugal force. This one is the one that is technically called a fictitious force. These are the forces that appear out of the blue when you try and describe motion from the point of view that is itself accelerating.

    Here's what it means. For the stationary observer, the situations is fully described just with gravity and tension. He sees an unbalanced force, and he sees that the rider is going in circles due to that force. All is well.

    However, imagine you're describing the situation while being a rider yourself. You look towards the centre of the carousel, and at your fellow riders, and see that you are not moving with respect to them.
    Since you are stationary, you must conclude that all the forces acting on you are in perfect balance. There can be no talk of centripetal force, as then you'd be moving rather than staying still.
    So you add a force to exactly compensate for the unbalanced component of the chain tension that we talked about earlier. This force will have exactly the same magnitude, but the opposite direction. It's the force that the rider will think is pushing him outwards.
    You will then see no unbalanced forces, and you will see yourself not moving (again, with respect to the ride). All is well again, even though the landscape is somewhat disconcertingly moving all around you at high speeds for some reason.

    In this case, when the reference frame is non-inertial (which means that it is accelerating in some way - here it is rotating, whcih involves acceleration), a new force appears that wasn't there when the frame of reference was inertial.
    All forces that dissapear if you choose a correct frame of reference are called fictitious forces. These include, but are not limited to, the centrifugal force, the Coriolis force (the thing that makes hurricanes spin in opposite directions on northern and southern hemispheres)* and the Euler force (again, as you ride on a rotating ride and it's only speeding up, the ridder will feel a force pushing him backwars, just as he will feel a force pushing him forwards when the ride slows down to a stop).

    One of the most mind-bendingly interesting things to learn is that gravity can be also modelled as a fictitious force (that's what General Relativity does).


    Anyhow, your teacher probably didn't want you to use centrifugal forces, since he or she mentioned fictitious forces. The centripetal force will be there, but again - supplied by a component of some other force.


    *I have to mention it, as this myth is my pet peeve, it is definitely NOT making water in your toilet spin one way or another as it drains - the effect is too tiny on such scales.


    Alright, I'm done. I hope that was of some help :)
     
    Last edited: Nov 21, 2014
  12. Nov 26, 2014 #11
    Alright, thanks! Sorry it took so long for me to reply. This is the free body diagram I came up with. Am I on the right track?
     

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  13. Nov 26, 2014 #12

    haruspex

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    That's not correct. Centrifugal force is 'fictitious', yes, but centripetal force is real. However, it is not an applied force, such as gravity or mechanical or electrostatic. Instead, it is a resultant force required to achieve a curvilinear motion. It's perfectly ok to refer to centripetal force as long as you treat it that way.
    It's also ok to use centrifugal force in non-inertial frames, but my own preference is to avoid that.
     
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