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Teaching Discrepancy on Centripetal Force?

  1. Nov 25, 2009 #1
    I'm having trouble with circular motion, for the simple reason that my instructor taught it differently than my book explains and how I see it done online. He explains that centripetal force moves towards the outside of the circle, and is an independent force. Meaning, he explains it as a force that maintains and is caused by circular motion.

    For instance, if I were swinging a ball on a rope horizontally, he would say that F=T-Fc, that the sum of the forces are centripetal force and tension acting in opposite directions (tension toward the center, centripetal force toward the outside). Now, I'm seeing it online that Fc=T because tension is the net force that maintains circular motion, and centripetal force and tension are one and the same. Algebraically, that makes sense, and both still produce the correct solution, but I'm wondering, why would my teacher teach that concept incorrectly? Is that a valid method or is he just an idiot? I need some help because I'm pretty confused.

    Also, as a side note, I was wondering what the centripetal force was on a rotating object, such as a merry-go-round or a turntable, because I really have no idea.
     
  2. jcsd
  3. Nov 25, 2009 #2

    D H

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    I feel sorry for you. This is just wrong.

    Fc *is* T in this case. So if Fnet=T-Fc, were correct, there would be no net force acting on the ball. By Newton's first law, that means the ball had either be stationary or be following a straight-line trajectory at a constant velocity.

    There is a viewpoint where this is true.

    250px-Hammerthrow_wire.jpg

    From the perspective of the hammer thrower's eyes, the metal ball is not moving (up 'til the moment he releases it). Your teacher is mixing up centrifugal and centripetal force, and maybe mixing up Newton's third law as well. IMO, teachers should not even start discussing centrifugal force until their students (and the teachers themselves!) have the basics of motion in an inertial frame down pat. There is no centrifugal force in an inertial frame.

    There is a force that is equal but opposite to the force by the rope on the ball. The ball exerts a force on the rope. Newton's third law. What a lot of people don't grok about Newton's third law is in a third law reaction, (a) the action and reaction forces are always the same kind of force, and (b) the action and reaction forces act on different bodies.

    That the ball exerts a reaction force on the rope is a bit irrelevant to understanding the motion of the ball. That reaction force is not acting on the ball, so it does not explain the ball's motion one iota.
     
  4. Nov 25, 2009 #3
    Thank you very much. At least now I understand that I need to ignore that thought process and double check for his future lessons.

    One thing he's never taught us period is this "frame of reference" that I keep hearing about. He explains points of reference (i.e.: torque analysis and rotational statics, picking an arbitrary axis of rotation) but up until I looked around online, I've never heard of a frame of reference. So, you can understand my predicament with my teacher, and I'm one of the students who catch on quick. Imagine the right-brained students in his class...

    Yeah. Not pretty. Anyway, I basically have two questions that need addressed now.

    1. What is a frame of reference?
    2. How would you approach a problem like this in a correct manner:

    Rollercoaster.jpg

    A roller-coaster car speeds down a hill past point A and then rolls up a hill past point B, as shown in the above figure.
    What is the maximum speed the car can have at point B for the gravitational force to hold it on the track?

    Here the force of gravity is the centripetal force...but...would you include normal force? And how would you go about solving it?

    My instructor showed it this (incorrect) way:

    F=ma. Since the car is not moving up and down at point B, the acceleration along the vertical must be zero.
    F=0. And the forces that make up this net force are gravity and the centripetal force (which he says its to the outside).
    0=Fg-Fc
    Fg=Fc
    mg=Vt^2m/r
    Vt=SQRT(gr)

    So it works, because that's what the book says to do but...the logic that got it there is wrong. Help?
     
  5. Nov 25, 2009 #4

    Doc Al

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    This is shockingly bad. As D H said above, your instructor is mixing up centripetal and centrifugal force. No excuse.

    (1) The car is most certainly accelerating at point B--it's moving along the arc of a circle and thus is centripetally accelerated (towards the center of the circle).

    (2) The forces acting on the car are gravity and the normal force. To find the point where the car is just about to lose contact with the road, set the normal force to zero.

    (3) Apply Newton's 2nd law:
    ΣF = ma
    mg = mac = mv2/r

    I'm curious--what book are you using?
     
  6. Nov 25, 2009 #5
    I think I didn't explain very well. The book explains that centripetal force pushes toward the center and the correct way of doing these problems. What I meant was that the resulting [tex]\sqrt{gr}[/tex] supplied the same answer as the back of the book.

    But, to answer the question, Holt Physics. By Holt, Rinehart, and Winston. Doesn't give an edition number.

    So, why is the normal force zero, exactly? And what is a frame of reference?
     
  7. Nov 29, 2009 #6

    Doc Al

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    You are trying to find the point where the car just begins to lose contact with the road.
    Think of it as a set of coordinates used for describing things.
     
  8. Nov 29, 2009 #7

    D H

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    Wrong! The velocity is zero, not the acceleration. Just because velocity is zero at some point in time does not mean the acceleration (and hence the net force) is zero. Your instructor happened to get the right answer here. He or she got lucky.

    So what's the right way to solve this problem?

    Looking at this problem kinematically, the car is moving along a circular path. For both uniform and non-uniform circular motion, the radial component of velocity is identically zero but the radial component of acceleration is [tex]v(t)^2/r[/tex], directed inward. (The tangential component of velocity is non-zero in non-uniform circular motion, but that is irrelevant here.) At the top of the arc, the car must be accelerating downward with an acceleration of [tex]v_t^2/r[/tex] if it is to follow the arc.

    Looking at the problem dynamically, the forces acting on the car are gravity, the normal force, and frictional forces. At the top of the arc, gravity is downward, normal to the arc, and the normal force is upward. Those frictional forces are irrelevant to this problem. All we care about is the radial component of acceleration.

    Equating the kinematic and dynamic points of view lets us solve for the normal force [tex]\vec F_N[/tex]. Denoting "up" as being positive,

    [tex]\aligned
    \vec F_{net,vertical} &= \vec F_g + \vec F_n = -mg + \vec F_N & \text{(dynamics)}\\
    \vec F_{net,vertical} &= -m\frac {v_t^2}{r} & \text{(kinematics)} \\
    \vec F_N &= m\left(g-\frac {v_t^2}{r}\right) & \text{solving for\ } \vec F_N
    \endaligned[/tex]

    The normal force must be non-negative. A negative normal force would mean the car's wheels are in some sealed track that keeps the car from flying off. That is not the case here. Thus the condition for the car to stay on the track is

    [tex]v_t \le \sqrt{gr}[/tex]

    The car will leave the track when the tangential velocity exceeds this limit.
     
  9. Nov 29, 2009 #8

    D H

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    Think of it this way. Forget the car for a bit. Suppose you are standing on a bathroom scale on the surface of the Earth. Above you is a pull-up bar. At first you aren't holding on to this bar. You are just standing on the scale. There are two forces acting on you: Gravity, directed downward and the normal force exerted by the scale, directed upward. The scale will register your weight.

    Now you grab the pull-up bar and start lifting yourself up. The normal force is a constraint force. It supplies whatever force is needed to keep you from sinking into the scale. As you start pulling yourself up, the force exerted by the scale will decrease. The reading on the scale decreases as you exert more and more force on the pull-up bar. When the force on the pull-up bar exceeds your weight you will start moving up. The scale no longer exerts a force on you. It reads zero.

    Back to the roller coaster problem. Suppose that to help yourself understand the problem, you extend x and y axes from point B in the diagram in post #3, with the x axis oriented horizontally and the y axis oriented vertically such that positive x points in the direction of the car's travel positive y points upward.

    You have just established a frame of reference. A nice, simple way to look at frames of reference, or reference frames, is that they comprise an origin and a set of axes with which you can mathematically describe the underlying physics.
     
  10. Nov 29, 2009 #9
    Thank you both a lot. Now this problem makes much more sense. The scale thing really made it make sense.
     
  11. Nov 29, 2009 #10
    In possible defense of the instructor they may have been thrown into this class. Physics is not the easiest class to teach well. In many smaller towns in the United States they just dont have enough teachers. One teacher might teach all the Science classes.

    I am kind of curious to know where the poster lives... rural or urban if from the United States. It should also be noted that many Physics teachers in large school districts get tossed into Physics also from Chemistry.

    From the textbook mentioned, this has to be a high school level class.
     
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