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Centripetal force

  1. Mar 6, 2009 #1
    1. The problem statement, all variables and given/known data

    an inclined plane with angle 30 degrees has a mass on it. The static friction coefficient is 0.2 and the mass is 2 kg. The system then rotated to form a circular path with radius 3 m. Find the angular frequency so the mass will stay stationary.

    2. Relevant equations



    3. The attempt at a solution

    i have drawn the free body diagram and i got that before the system rotated, the mass is moving downwards.

    i got the forces that contributed for centripetal force when it is rotated, which is mg sin 30 and friction, but i think i can't use the static friction because it's moving. i should use kinetic friction instead, but there is no information on the question.

    i try to use the static friction and got the answer ( 1.03 ), but there is no such answer on the choices.....
    thx a lot
     
  2. jcsd
  3. Mar 6, 2009 #2

    LowlyPion

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    Homework Helper

    The static friction is what you use, because the object is to stay stationary on the incline - hence static.

    It is my guess from your statement, and absent a picture, that rotating is in the horizontal plane and the incline rotates about a center with the incline facing toward the center 3m away.

    If the incline is rotating and the object is 3m away on the inclined rotating surface then that would be much different.

    Draw a diagram of your object and sum the forces if the object is to remain stable. You have gravity and static friction resisting the upward component of the centripetal acceleration. Remember that vertical gravity and horizontal centripetal acceleration will both contribute components to the normal force for figuring friction.
     
  4. Jun 29, 2009 #3
    i'm sorry

    here's the picture

    when it's rotated, the direction of the friction is upwards or downwards the inclined plane?

    thx
     

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  5. Jun 29, 2009 #4
    The direction of friction is a good question, actually, since given a high enough frequency of rotation, the mass will 'want' to move up the incline.

    Use a work assumption that the friction points down the incline. You'll end up with a 'sign-ambiguous' expression for it, if you choose to calculate what the friction is, exactly. That means that its magnitude could be either positive, or negative. If it turns out positive, then you assumed correctly. If it turns out negative, it just means that it points the other way, and you need to revise your diagram.

    If you're familiar with fictitious forces, this problem becomes much more simple, in the accelerated frame, the mass is at rest, and is also under the effect of a fictitious force -mA, where A is the centripetal acceleration (As a vector). Find out what the value of this fictitious force is, and where it points (Notice the - sign), and from there on, you have a completely static problem.
     
  6. Jul 2, 2009 #5
    sorry, i'm not familiar with fictitious force

    but i tried to read on wiki about fictitious force and did this:
    i assume that there is fictitious force directed upwards the inclined plane. And in accelerated frame, this fictitious force equals friction and w sin 30o

    so, let fictitious force = F
    F = 0.5 w + fs
    = 0.5 x 2 x 9.8 + 0.2 x w cos 30o
    = 13.195 N

    then F = mA ; A is centripetal acceleration
    A = F/m
    [tex]\omega[/tex]2 r = F/m ; r = 6 m
    [tex]\omega[/tex] = 1.049 rad/s

    do i get it right?

    thx
     
  7. Jul 2, 2009 #6
    No, no, I'm afraid you've misunderstood the article. And just from a short glance it looks like your math is wrong as a result (Friction has two components, one along the x axis, and one along the y axis).
    When you move from an inertial frame (One that has no net force acting on it, as in, that is not accelerated) to an accelerated frame, you notice an extra force.

    Just like how when you accelerate in your car, you feel that you're pushed into your seat, right?

    [tex]\vec F_D = -m\vec A[/tex]
    Where [tex]\vec A[/tex] is the acceleration of the system.

    For simple systems, where the acceleration of the object in question, is the same as that of the system (For instance, when you move your frame of reference onto the object!), you end up with a static problem. As an observer from the object, you see the object at rest, but since you've moved to an accelerated system, you must also take into consideration the fictitious force.
    I think the link 'fictitious force' explains it fairly well.

    It's a mathematical ploy, to a great extent, starting with Newton's second law:
    [tex]\Sigma \vec F = m\vec a[/tex]
    [tex]\Sigma \vec F -m\vec a = 0[/tex]
    Where the net force is the sum of all the actual forces acting on the object.
    In the second line, if we accept [tex]-m\vec a[/tex] to be a force like any other, then we've simplified our problem greatly, by changing it from a dynamic one to a static one.

    Try making a free body diagram from an inertial perspective first. You'll see that the mass has three forces acting on it: Its weight, the normal force and friction.
    Since you know that it is moving in a circular motion, that means that it is being accelerated towards the axis of rotation at an acceleration: [tex]\vec a_r =\omega ^{2} r[/tex]
    Mathematically, that would mean:
    [tex]\Sigma \vec F = m\omega ^{2} \vec r[/tex]
    Rearranging:
    [tex]\Sigma \vec F -m\omega ^{2} \vec r = 0[/tex]

    What does that expression remind you of?
    It's the [tex]\Sigma \vec F[/tex] of the mass, if it were at rest and with a fictitious force acting on it.

    From there, it's a simple static problem, where all you need to do is isolate [tex]\omega[/tex]

    http://img135.imageshack.us/img135/7098/p1000032.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  8. Jul 4, 2009 #7
    on the picture, the friction is directed upwards the inclined plane

    should it be downwards because the mass will likely moves up?

    thx
     
  9. Jul 5, 2009 #8

    LowlyPion

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    Homework Helper

    The friction will oppose movement either up or down.

    Consider the force components acting on the block then.

    You have 3 forces to balance.
    Centripetal acceleration = v2/r = ω2*r
    Gravity down the incline = m*g*sinθ
    Friction = μ*m*g*cosθ

    What ω then is required to keep the block in place?
     
  10. Jul 9, 2009 #9
    i'm still a bit confused about the direction of friction.

    i think the direction of friction will also determine the value of [tex]\omega[/tex]

    based on royalcat's picture, the direction of friction is upwards. In case i take the direction of friction is downwards, i think it will give a different value of [tex]\omega[/tex]

    thx
     
  11. Jul 9, 2009 #10
    That's just a work assumption.

    Developing the equations derived from the FBD, and assuming the direction of friction is unknown, we arrive at the following:
    [tex]\frac{g}{r} \frac {\sin{\theta}\mp \mu \cos{\theta}}{\cos{\theta}\pm \mu \sin{\theta}}=\omega ^2[/tex]
    The right side has to be positive, because it is a squared expression. So the left side has to be positive as well. That leads to either of two conditions.

    (The plus/minus, minus/plus, signs are determined in the following way:
    If you choose to assume that the friction is up the incline, use the top sign, if you choose the opposite, then use the bottom sign)

    If we assume that the friction is up the incline, then we arrive at the following conclusion: [tex]\mu _s < \tan {\theta}[/tex] which means that the mass would be slipping down if it were not for the rotation.

    If we assume that the friction is down the incline, then we arrive at the following conclusion:
    [tex]\mu _s < \cot {\theta}[/tex]

    Based on [tex]\mu _s[/tex], [tex]\tan {\theta}[/tex] and [tex]\cot {\theta}[/tex]

    We'll make or decision on where to point our friction.

    But even if we were to choose incorrectly at first, all that would get us would be a negative value for [tex]\omega ^2[/tex]
    That would let us know that something is wrong with our premises, and since the only place where we have any lee-way is the choice of the sign of friction (Representing its direction) all we would need to do to rectify our poor work assumption would be to substitute [tex]\mu _s[/tex] for [tex]-\mu _s[/tex] and all would be right with the world once more. :3

    Don't mention it. :) It's an interesting problem and I was happy to help with it.
     
    Last edited: Jul 9, 2009
  12. Jul 9, 2009 #11
    oh now i get it.
    sorry if i cause both of you much trouble

    thx a lot royalcat and lowlypion
     
  13. Feb 15, 2010 #12
    Sorry for bringing the old thread up again. I review my lesson for the exam and have trouble with this one (again...)

    I am able to solve this using the accelerated frame as Royalcat taught me (post #6 and #10). Now I'm trying to solve this question using the inertial frame and not getting the same equation for ω2

    Using inertial frame, there are 3 forces acting on the object which are weight, normal, and friction (the picture can be seen on attachment at post #6). The component of forces that contribute to centripetal are N sin θ and fs cos θ. Setting the equation :

    N sin θ - fs cos θ = mω2r , where N = mg cos θ

    mg cos θ sin θ - µs mg cos θ cos θ = mω2r

    ω2 = (g cos θ sin θ - µs g cos2 θ) / r

    Where did I go wrong ?

    Thanks
     
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