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Two Potentially Easy Kinematics Problems (sliding/rotation)

  1. Aug 20, 2015 #1
    Question 1:
    1. The problem statement, all variables and given/known data

    A bug of mass m crawls radially outwards with a constant speed v’ on a disc that
    rotates with a constant angular velocity ω about a vertical axis. The speed v’ is relative to the
    center of the disc. Assume a coefficient of static friction μ, find out where on the disc the bug
    starts to slip.

    2. Relevant equations
    F = ma = ΣF_i

    3. The attempt at a solution
    The question asks when the force of friction is finally overcome, so I think:
    ma = 0 = F_friction - F_cent.
    Or
    F_friction = F_cent.
    μmg = (mv2/r)
    v = wr

    Solving for r:
    r = (1/ω) √(2μg)

    Seems too easy... Is this right?


    Question 2:
    1. The problem statement, all variables and given/known data

    A particle is placed on top of a smooth (frictionless) sphere of radius R. If the particle is slightly
    disturbed, at what point will it leave the sphere?

    2. Relevant equations
    Same as first question, just
    F = ma = ΣF_i

    3. The attempt at a solution
    Similarly, we want to know when the normal force of the sphere on the particle is overcome:
    F_norm = F_cent
    mg CosΘ = (mv2/r)
    CosΘ = y/R (where y is the height above the center of the sphere)

    So:
    y = v2/g

    Finding v2:
    Using conservation of energy, PE_initial = PE_final + KE_final
    mgR = mgy + mv2/2

    Solving for v2
    v2 = 2g(R-y)

    Placing into equation for y:
    y = 2g(R-y)/g = 2(R-y)

    Solving for y:
    y = (2/3) R

    Correct? Or am I making a horrible mistake?
     
  2. jcsd
  3. Aug 20, 2015 #2

    TSny

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    What frame of reference did you choose for setting up Newton's second law? ma ≠ 0 in the earth frame. The bug has a centripetal acceleration that is not zero in the earth frame.

    You can check that the expression on the right does not have the correct dimensions for r.

    Are you using the notation F_norm to denote the magnitude of the normal force that the surface of the sphere exerts on the particle?
    If so, then it is not true that F_norm = F_cent.
    Also, it is not true that F_norm = mg⋅CosΘ (except at Θ = 0). Despite this, your final answer could be right.
     
  4. Aug 20, 2015 #3

    TSny

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    For problem 1, I overlooked the important fact that the bug will have a nonzero tangential acceleration as well as a nonzero centripetal acceleration. The tangential component will involve v'.

    You will need to consider the net acceleration due to combining the centripetal and tangential components.
     
  5. Aug 20, 2015 #4
    No. The velocity vector of the bug is given by:
    $$\vec{v}=v'\vec{i}_r+ωr\vec{i}_θ$$
    where ##\vec{i}_r## and ##\vec{i}_θ## are functions of θ. If the acceleration is equal to the time derivative of the velocity, what is the acceleration vector given by? (Don't forget to include the time derivatives of ##\vec{i}_r## and ##\vec{i}_θ##).

    Chet
     
  6. Aug 20, 2015 #5

    jbriggs444

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    If your coursework has discussed rotating frames of reference and centrifugal and Coriolis forces then an approach using those could be fruitful.
     
  7. Aug 21, 2015 #6
  8. Aug 21, 2015 #7
    You're right, I somehow mixed up that equation in my head. Using my previous method, the position at which the bug slips would be:
    r = μg/ω2

    I agree that the bug undergoes non-zero acceleration, and I use that in my attempt at a solution. But, the point was to find where the bug is unable to crawl at the constant velocity vi and simply flies off the disc. So, at this point, the centripetal acceleration would overcome the force of friction. Immediately before that happens, the two forces should be equal, right? To me, it seemed similar to problems where you find the height at which a steady electric field balances the force of gravity for a charged particle (just set qE=mg, and solve for r).


    You're right, there is indeed some small velocity in the r-hat direction, which I write:
    v=wr φ-hat + vi r-hat
    (I really ought to learn more latex... please bear with my notation)

    If I take the time derivative of ##\vec{v}##, including the time derivatives of the unit vectors, I get:
    a=-ω2vi r-hat + ωvi phi-hat

    But I don't see how this helps me... I am trying to find the position where the bug starts to slip. I could set the acceleration to zero, but I don't see how this approach helps, even for that method.
     
  9. Aug 21, 2015 #8

    TSny

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    OK
    This is not quite correct. When taking the derivative of the first term of the velocity expression, note that r is changing with time.

    You want to find the point where the bug is on the verge of slipping. At that point, the bug has acceleration. You just need to set up Newton's second law at that point.
     
  10. Aug 21, 2015 #9
    Right, I took the time derivative of r as vi (and the bug is moving out at constant velocity vi), and also the time derivatives of the unit vectors, then rearranged, gathered terms, etc. I just didn't show the intermediate steps. If you think the correct derivative is different, please let me know how/where?

    Are you saying I should set the acceleration above in:
    ma = mv^2/r + μmg

    (and presumably only take into account the r-hat direction of a?)

    I just don't know that I agree with that reasoning, or why? That formula for acceleration is only valid while the bug is not slipping.
     
  11. Aug 21, 2015 #10

    jbriggs444

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    That formula is correct, as far as it goes. But it is missing the Coriolis term. The fact that the direction of travel is changing taken together with the fact that there is a constant radial speed means that there is a constant tangential acceleration. That is the Coriolis term.

    Edit: The above is garbled. I jumped from a rotating frame explanation to an inertial frame explanation.

    In the rotating frame, your equation is missing the Coriolis term.

    In the inertial frame, ma already includes mv^2/r and it is incorrect to include it again on the right hand side. The difficulty is computing ma -- which will include the equivalent of both the centrifugal and Coriolis terms.
     
    Last edited: Aug 21, 2015
  12. Aug 21, 2015 #11

    TSny

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    When gathering terms, you should have found two terms that were the same. So, you should have a factor of 2 somewhere in your expression for the acceleration.

    I don't understand this expression, especially the mv2/r on the right side. Newton's second law is : ma = ΣF .
    mv2/r is not a force acting on the particle (assuming that you are working in the inertial frame of the earth). So, it should not appear on the right side of ma = ΣF .
     
  13. Aug 21, 2015 #12
    Ok let's consider another approach:

    Consider a mass m on a plane with static coefficient of friction μ. What is the minimum force necessary to start sliding the mass?
    Fmin = μmg

    Right?

    Ok, now consider another system:
    A mass m is being spun around in a circle on a massless string. The string can bear a tension up to T newtons. The mass is rotated with angular velocity ω. The string is let out slowly to extend its length (slowly enough that the radial velocity is irrelevant). Where does the string break?
    T = mv2/r
    or,
    r = mv2/T

    Now imagine the mass isn't on a string, but on a spinning disc.... T = μmg
    r = μg/ω2

    This is how I look at it. What's wrong with this approach?
     
  14. Aug 21, 2015 #13

    TSny

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    This is correct. However, I can't be sure that you have arrived at this result using correct reasoning since you did not explain how you got this (starting from ΣF = ma). The reason I am being picky here is that I don't believe you are applying ΣF = ma correctly in the bug problem.

    Correct. But again, I don't know for sure how you arrived at the equation T = mv2/r. Can you explain how this comes from a logical application of Newton's second law?

    This is the correct answer for r in this case, but not for the bug problem. Try to state explicitly the steps that lead from ΣF = ma to the answer for r.
     
  15. Aug 21, 2015 #14
    In the first case, the mass will not be moving prior to having the force of friction overcome, so ma = 0 = Sum of F = ugm - applied force, so applied force = ugm.

    In the second case, the mass on string has no radial acceleration, so ma = 0 = Tension - centripetal force, so centripetal force = Tension.

    For the bug, why do these principles no longer apply? What is different about the bug's normal force or centripetal acceleration?
     
  16. Aug 21, 2015 #15

    TSny

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    OK.

    (1) Why do you say ma = 0? A mass moving at constant speed in a circle has nonzero acceleration: a = v2/r in a direction pointing toward the center of the circle.

    (2) The term "centripetal force" gets a lot of students in trouble. "Centripetal" means "toward the center". For the mass on a string, there is one force acting toward the center of the circular motion; namely, the tension force. So, the tension force could be called the "centripetal force" in this problem. You should not think of there being two forces: a tension force and a centripetal force. There is only one force, the tension force.

    You have made two mistakes that conspire to get you the correct answer.

    The acceleration a of the bug has both a centripetal component as well as a tangential component that you have essentially gotten in post #7 (except for the factor of 2 error mentioned in post #11). This acceleration is what goes into the second law.
     
  17. Aug 21, 2015 #16
    Ok, I see about the 2nd case, ma = Tension until the string breaks, after which ma = 0.

    Also, yes I see my error with the factor of 2 in acceleration.

    So, let's consider this: The only force acting on the bug would be the normal force holding it in place, correct? This is the force that "accelerates" it toward the center of the disc (i.e. keeps it from slipping). If that's the case:
    ma = Normal force

    or

    ma = μmg
    a = μg

    And if we use the formula found earlier for a, I can solve for r. But here's the catch: acceleration has both r-hat and phi-hat components! Do I take only the radial part? If so, it's the same as my previous answer. Or do I take the magnitude of a, and use that in this formula? My gut feeling is that the normal force only acts in the radial direction, but that probably isn't true since we just calculated the formula for acceleration which obviously has 2 components...

    So... casting a into cartesian notation by substituting out r-hat and phi-hat for their appropriate cartesian equivalents, I can find the magnitude of a2 as:
    a2 = ω2 (4vi2 + r2ω2)

    Using this in a = μg, I get:
    r = (1/ω) √({μ2g22}- 4vi2)

    Which is a lot more complicated than my previous attempt... Any better?
     
  18. Aug 21, 2015 #17
    I just noticed something: Interestingly, if vi is almost 0, then:
    r ≅ μg/ω2

    Which is my original answer... so I guess the radial velocity might play a role, depending on its value, but if you can take it to be negligible, then the original answer stands. So, this realization makes me feel that my more detailed solution (previous post) is right, since it gives the same answer when vi = 0.
     
  19. Aug 21, 2015 #18

    TSny

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    The standard use of the term "normal force" is the component of force that a surface exerts on an object in a direction perpendicular to the surface. The word "normal" is being used to denote perpendicular direction. So, the normal force on the bug is the upward vertical force that the surface of the disk exerts on the bug.

    Can you identity all of the forces acting on the bug? If you add together these forces (as vectors) you get ΣF, or Fnet. It's always Fnet that goes into the second law: Fnet = ma.
    In the second law Fnet = ma, a is the total acceleration of the particle. So, in terms of magnitudes: Fnet = ma where "a" is the magnitude of the acceleration (including all components).

    There's an error in one of your powers in the last term on the right. [EDIT: My mistake. You have the correct expression for the magnitude of a. Sorry]
     
    Last edited: Aug 21, 2015
  20. Aug 21, 2015 #19

    TSny

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    OK. Looks good now. I hope that you are convinced that the net force is just the friction force even though there are 3 separate forces acting on the bug.
     
  21. Aug 21, 2015 #20
    Yes I see that now. I appreciate all the help, this has been a good review of some concepts I really needed to remember!
     
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