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Physics C Mechanics: torque on a yoyo

  1. Apr 14, 2014 #1
    If you want, the problem is here: q.s 23-24 in http://teachers.sduhsd.net/jdanssae... C review material/AP C MC test mech 2009.pdf

    1. The problem statement, all variables and given/known data
    A solid cylinder of mass m and radius R has a string wound around it (basically a yoyo). A person holding the string pulls it vertically upward, as shown above, such that the cylinder is suspended in midair for a brief time interval Δt and its center of mass doesn't move. The tension in the string is T and the rotational inertia of the cylinder about its axis is 1/2 * mR2

    23) The net force on the cylinder during the time interval Δt is:

    A) T B) mg C) T - mgR D) mgR - T E) zero

    24) The linear acceleration of the person's hand during the time interval Δt is:

    A) (T - mg) / m B) 2g C) g/2 D) T/m E) zero

    2. Relevant equations

    torque = F x R = Iα = (1/2 * mR2) α

    3. The attempt at a solution

    I drew a free-body diagram. The yoyo has mg pulling downward at it's axis (which is its center of mass). The yoyo also has a force F pulling up at its side, creating a torque FR.

    I suppose if it's stationary, the net force is 0. But how on earth can it be stationary? This brings me to the crux of the problem. I can't understand how this phenomenon works.

    I've tried with a yoyo. It's true, if I pull up with a certain acceleration, the yoyo will be stationary for a second. However, this doesn't make sense. I provide a torque on its side, but I do nothing to cancel out the force gravity has on it's center. How does it stay stationary?

    Also, I did not attempt 24 because I did not understand 23.

    Thanks for any help.
     
  2. jcsd
  3. Apr 14, 2014 #2
    It's only stationary because the problem defined it to be so. I suppose you could look at the two extremes:

    1) if you just let the yoyo fall while keeping your then it would accelerate down at a<g (-ve)
    2) if you moved your hand up with a very large acceleration (>>+g) then the yoyo would move up with a very similar acceleration (+ve)

    So somewhere in between the two is an acceleration of the hand upward where the yoyo does not move v=a=0.
     
  4. May 15, 2014 #3
    Could someone please answer the following question to help me understand free-body diagrams better?

    If an object (say a yoyo) has a force mg acting downward at its CM, AND a force T = mg from the string that I'm pulling acting upward at it's edge, what motion will it exhibit? Technically, the forces cancel out so there will be no linear acceleration (i.e. it will stay stationary with respect to the y axis). There is also a net torque so it will spin. Is my analysis correct? Or will it move linearly because the forces do not act on the same point?
     
  5. May 15, 2014 #4
    Yes, it will rotate but the catch is about what point will it rotate?
     
  6. May 15, 2014 #5

    BiGyElLoWhAt

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    One think i want to point out.
    ##T≠ F \times R## ;
    ##T = R \times F = -F \times R##

    Another thing, to give input to your FBD question, Forces don't care about torques or where they're acting about. A force on an object is a force on an object. Period. On the other hand, though, torques are dependant on forces. A standard FBD includes forces only, and can determine your linear acceleration. an extended FBD has forces with radii and includes random torques as well ("a torque of ##6 N \cdot m## acts on the object"), and this is used to determine your angular acceleration.
     
  7. May 15, 2014 #6
    Yes the analysis is correct. It stays stationary (or moving at constant speed) because the net force is zero AND it spins up because the net torque is not zero. It is simple when you decide to trust the equations to mean what they mean.
     
  8. May 15, 2014 #7
    The difference between net Force and net Torque is something you should internalize because doing so also helps you understand things like why we have separate equations for angular momentum vs. linear momentum, or translational KE vs. rotational KE.
     
  9. May 18, 2014 #8
    Got it guys, thanks. I have another question that will help me a lot. I think it's related, so I won't start a new thread.

    If an object, say a book, is on a frictionless surface and experiences a force AT ANY point, not necessarily the CM, it will accelerate with no rotation, correct?

    However, if the book is on a rough surface and is pushed at a non-CM point, it will both accelerate linearly and rotate, correct? Am I right in thinking that there is an external force at a non-CM point AND a frictional force at the CM-- and the combination of these forces creates the torque?

    If you calculate the frictional force at the CM and know the location, magnitude, and direction of the push, is that all you need to know all about the book's motion?
     
  10. May 18, 2014 #9

    haruspex

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    No, There's an unbalanced torque about the mass centre, so it will rotate.
     
  11. May 18, 2014 #10
    No, any force not acting through the CM will cause a rotation, regardless.

    You would also need to know the mass of the book and its mass moment of inertia.
     
  12. May 18, 2014 #11

    haruspex

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    If the book is rotating, you cannot take the friction as acting at the mass centre. Different parts of the book in contact with the surface will be moving in different directions, so the frictional forces will also be in different directions. In pure rotation, the friction acts as a torque. If translating and rotating it gets very messy.
     
  13. May 19, 2014 #12
    Yeah- I thought it would get messy. I saw a thread on here that asked "if a book is pushed across a surface, why does it stop rotating and translating at the same time" and I was mind-boggled.

    It seems that calculating the frictional force at any time would take a messy integral... and calculating the frictional force as a function of time would take an even messier integral.
     
  14. May 19, 2014 #13
    Okay, but what happened to " a force on an object is a force on an object" (paraphrased from somewhere above)?

    I understand why it would rotate. I picture the inertia of the object to be a force that points opposite in direction to the external force.... But given a free body diagram, how can you prove that there is a torque? Again, this example is frictionless. Where is the axis of rotation?

    Thanks.
     
  15. May 19, 2014 #14

    SammyS

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    That (further paraphased) "A force is a force, of course, of course ..." statement was from the BigYellowHat in post #5.
    The sum of the external forces is equal to the mass of the object multiplied by the acceleration of its center of mass:
    [itex]\displaystyle \sum \vec{F}_\text{ext}=M\, a_{\text{cm}}[/itex]​

    Regarding your question about rotation:
    The motion of a rigid body can be described as the motion of its center of mass along with its rotation about the center of mass.

    Furthermore, the rotation of the object about any point is the same as the rotation about any other point, so you can calculate the torque with respect to any convenient point, depending on what forces are acting and where they are acting. -- whatever makes the solution easiest.
     
  16. May 19, 2014 #15

    haruspex

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    Torque, moment, angular momentum etc. are defined with respect to a reference point. Every force exerts a torque about any point that is not in the line of action of that force. It is not meaningful to ask whether a force exerts a torque unless you specify your reference point.
    Usually, the point of interest is the centre of mass of the body acted on. If the sum of the torques about that point is nonzero then the body will undergo angular acceleration.
    An interesting case arises when you have two equal and opposite forces, but along parallel lines of action. It turns out that the reference point no longer matters - whatever point you pick the sum of the torques is the same. Such a system is known as a "couple".
     
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