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Homework Help: Centripetal force problem

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A .2kg ball travels in a circle of r=1m, one revolution every second, what is the acceleration.
    What would happen to the force and acceleration if you double the speed?

    2. Relevant equations
    a= (v^2)/r

    3. The attempt at a solution
    2∏1m x ≈ 6.3m

    Speed of the ball = 6.3m/s

    a = (6.3^2)/1 = 40m/s

    Fnet = .2kg 40m/s = 8N

    The solutions manual says that the Fnet and and a would QUADRUPLE if the speed is doubled. Why the heck would it quadruple?

    I'm having a hard time understanding this chapter. Centripetal force makes no sense. If Newton's 3rd law says that for every force there is an equal but opposite force, then why is there not an equal but opposite force OUTWARD when it comes to centripetal force?
  2. jcsd
  3. Sep 19, 2012 #2
    Re: Centripital force problem

    Hi feodalherren.
    the force would quadruple because, as you can see, the acceleration is proportional to the square of the speed.
    so if you multiply the speed by 2, -> 2² so you multiply the acceleration (thus the force) by 4
    the third law does apply.
    suppose the ball that is going in a circle like this is doing so because it is attached to a string. the force you calculated is the one exerted on the ball by this string, but the string tension augments accordingly, that is, the same force is applied outward on the string
  4. Sep 19, 2012 #3
    Re: Centripital force problem

    I still don't understand. I also don't understand why objects stay in orbit if gravity constantly accelerates them at 9.8m/s^2.
    Even if they manage to miss the curvature of the earth because of their speed, should gravity eventually get the upper hand because it increases in force by 10m/s EVERY SECOND. Why, all of a sudden does gravity become constant!?
  5. Sep 20, 2012 #4
    Re: Centripital force problem

    Well, gravity isn't typically constant, it depends on the distance between two objects attracting each other.
    In the case of an object much much bigger than another, like the earth and some satellite, and for a circular orbit, then yes, the center of the earth is approximatively the center of the circle, and the force is constant in magnitude (but not in direction)
    Anyway, gravity is a bit harder to grasp than just the concept of uniform circular motion, so you should make sure you understand that first.
    What did you still not understand about your original question ?
  6. Sep 20, 2012 #5


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    Re: Centripital force problem

    The unit of Force is the Newton.
    The unit of speed is m/s.

    So the force, even if it was increasing, would not be increasing by a certain number of m/s??

    Acceleration means a change in velocity - which can mean a change in magnitude or a change in direction or perhaps both.

    In the case of an orbiting satellite, it is only a change in direction. The satellite does not change speed - it only changes direction so that at all times it is traveling parallel to the curved surface of the Earth below.

    Same thing when you drive a car around a bend on the highway.
    The car is not changing speed, but is changing direction [and thus changing velocity] and never gets any closer to, nor further from, the edge of the road.

    I try to explain an orbiting satellite as follows.
    Suppose you push a pen of the edge of a table - it falls to the ground a few cm from the edge of the table.
    Push it harder instead [ie make it travel faster, but still traveling horizontally] - it still lands, but now perhaps a metre from the table.
    Faster - it may land at the back of the room.
    Now imagine the earth is smooth and uncluttered like a giant billiard ball, except for your table.
    fling the pen off faster - it may land 100m away
    faster!! - it lands 1km away
    faster!!! - it lands 200km away. Note that at this speed, the curvature of the Earth means the pen doesn't land as soon as it would if the earth was modelled as flat.

    FASTER!!! - it may land 1000km away [the curvature of the Earth being even more advantageous.

    FASTER !!!! - now it travels half way round the Earth before hitting the surface.

    STILL FASTER !! - now the pen at all times falls towards the Earth, but the curvature of the Earth means the surface effectively falls away at the same rate, so the pen never gets any closer to the Earth, and approx 90 minutes later it skims across the table top and heads off on its second lap of the globe.
  7. Sep 20, 2012 #6


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    Re: Centripital force problem

    Ok so take your equation..

    a = (v2)/r

    Question asks what happens when the velocity doubles so write two equations for ai and af where i=initial velocity...f=final velocity

    ai = (vi2)/r ..............(1)
    af = (vf2)/r ..............(2)

    and then substitute vf = 2 * vi in eq(2)

    af = ((2vi)2)/r


    af = 4(vi)2/r

    Take the ratio af/ai...

    af/ai = 4 ((vi2)/r) / ((vi2)/r)

    everything cancels except the 4 so..

    af/ai = 4


    af = 4 * ai

    final acceleration is four times the initial.
  8. Sep 20, 2012 #7


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    Speed of the ball =Distance/Time= (2∏*1)m/1s=4∏m/s

    Acceleration = [itex]v^{2}/r[/itex]
    Plug in the value of v and you can find acceleration.

    When you double the speed,
    Now divide the two equations and you will arrive at the answer.
    Do the same for the force

    First you must understand that Centripetal force is not a new force that arises majestically just because a body is moving in a circle. It may be the same forces such as Tension, Friction, Spring etc. When these forces make a body move in a circle we just replace the name by Centripetal force. For your question it says that the body moves in a circle but does not specifies how. It can be because it is attached to a string or because of friction.
    So we just replace these forces by a new name. As far as Newton's 3rd law is concerned, the body does exert a force on the agent which is providing it the "Centripetal Force". So Newton's 3rd law is not violated. For example if you hold a string in your hand the other end of which is tied to a stone and start rotating it the stone follows a circular path. Now if you analyse the forces on the stone it will be the tension force. But wait you said that according to Newton's 3rd law an equal and opposite force must be there. It is there but not on the stone. The stone exerts an equal and opposite force on the string. Thats all.
  9. Sep 20, 2012 #8
    hi utkarshakash
    you misspelled "4" :smile:
  10. Sep 20, 2012 #9
    I think I got the original question now, it makes sense. Thank you.

    I still don't understand the circular orbit though.

    Coming back to the example of the pen. If I dropped this pen off the table, after 1sec it would be traveling at 10m/s, after two seconds, 20m/s etc. Correct?

    Ignoring air resistance, if I throw this pen so that it has an X amount of horizontal force, that doesn't change the fact that it's going to fall faster and faster until it hits the ground. i.e it's accelerating in speed in its vertical direction.
    In other words, thinking about this in a Cartesian coordinate system, the Δ x has no effect on the Δ y. Correct?

    Now, if I throw this pen fast enough for it to "miss" the ground because of the curvature, shouldn't the pen keep falling faster and faster toward the center of gravity until it finally hits it?

    So lets say that it needs to go 5km/s to miss the curvature. The first second it goes 5km/s and the falls 10m because of gravity.
    One second later, it still goes 5km/s but this time it falls 20m/s
    You get the idea.

    Until it finally falls too fast. The question is WHY doesn't the pen's speed accelerate if gravity is 10m/s^2 pulling DOWN?
  11. Sep 20, 2012 #10
    Hi Feodalherren,
    I'm sorry, but you didn't understand yet the circular orbit stuff, otherwise, the question you ask for the gravitation case would also be answered.
    Your mistake is to consider that the acceleration caused by the force has an impact on the velocity magnitude.
    It can have, but this is not necessarily true.
    In the case of the circular orbit, the acceleration is orthogonal to the velocity, so it changes its direction, but it doesn't change its magnitude.
    if you pen was going at 10ms-1, and it is accelerated orthogonally, then after some time, it still goes at 10ms-1, but not in the same direction
  12. Sep 20, 2012 #11
    But isn't gravity ALWAYS pulling everything down? I haven't seen gravity pulling anything to the sides. Further, doesn't gravity increase by 10m/S^2 so eventually the force pulling down should be vastly greater than whatever momentum the object has in any horizontal direction?
  13. Sep 20, 2012 #12
    Ok, I can see where your confusion lies (I think)
    And it is a deep one.
    You are confusing velocity with position.
    When you think about the force 'pulling down' you translate that as 'well the thing now is lower than it was before'
    That would be true if the force had an influence over the position, but it only affect the velocity, that is, the thing was 'going this way', now, because its velocity was pulled down, it is going in a different direction.
    (and, later, the fact that it goes in a different direction will have an effect on its position too)
  14. Sep 20, 2012 #13
    Gravity does not "increase by 10m/S^2". Gravity is constant at 10m/S^2 (at the surface of the earth). The magnitude of the gravitational force acting on an object in circular orbit remains constant.

    If you throw a ball horizontally and it then also accelerates in the y direction, it will hit the earth. But, if you throw it harder horizontally, it will hit the earth further away. In addition, the earth is curving downward in the y direction as the distance x increases (just draw a picture of the earth, and a ball initially moving horizontally). If you throw the ball hard enough (i.e., at high enough velocity), the downward curvature of the earth will just cancel the downward acceleration, and the ball will never hit the earth. It will be in orbit close to the surface of the earth. You can actually calculate what velocity it takes to make this happen, if you neglect air resistance. Or throw the ball horizontally on the moon, where there is no air resistance.
  15. Sep 20, 2012 #14
    I totally understand now. My confusion was that I was confusing acceleration with speed. Thanks guys :).
  16. Sep 20, 2012 #15


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    Oh yes that was a silly mistake. Actually I meant this [itex]a_{final}=(2v_{initial})^{2}/R[/itex]
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