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Can someone help me with this rotational dynamics problem.

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data
    I've been trying to attempt this problem for the past day but I don't understand how to answer the questions. It's frustrating. If someone could point me in the right direction that would be nice. My TA did the problem in class but I still don't quite understand it. This is my first time taking a physics course.

    A plank with a mass M = 5.60 kg rides on top of two identical, solid, cylindrical rollers that have R = 4.40 cm and m = 2.00 kg. The plank is pulled by a constant horizontal force of magnitude 6.40 N applied to the end of the plank and perpendicular to the axes of the cylinders (which are parallel). The cylinders roll without slipping on a flat surface. There is also no slipping between the cylinders and the plank.

    (a) Find the initial acceleration of the plank at the moment the rollers are equidistant from the ends of the plank.
    magnitude: m/s2
    direction:

    (b) Find the acceleration of the rollers at this moment.
    magnitude: m/s2
    direction:

    (c) What friction forces are acting at this moment? (Let fp be the frictional force exerted by each roller on the plank, and let fg be the rolling friction exerted by the ground on each roller.)
    fp = N
    direction:

    fg = N
    direction:


    2. Relevant equations

    Ʃτ=Fd

    v=ωr
    a=[itex]\alpha[/itex]r



    3. The attempt at a solution

    Not sure where to start. Any help would be well appreciated.
     
  2. jcsd
  3. Nov 8, 2012 #2

    tiny-tim

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    welcome to pf!

    hi ben! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)
    try (a) first …

    start by giving things names: call the acceleration of the plank "a", the angular acceleration of the cylinders "α", and the friction force between the plank and each cylinder "C"

    write out the F = ma equation for the plank, and the τ = Iα equation for each cylinder, and also the "rolling constraint" equation that relates a to α …

    what do you get? :smile:
     
  4. Nov 8, 2012 #3
    Ok so I got, F=ma, thus 6.4N=5.6a, a=6.4/5.6≈1.14

    and a=αr

    1.14=α(.044)

    α=25.97

    The for torque I got the following equation T=Iα where I=1/2mr^2

    plugging in the respective values the equation looks like T=1/2(mr^2)(α)=1/2(2(.044^2))(25.97)=.0502Nm
     
  5. Nov 8, 2012 #4

    tiny-tim

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    nooo :redface:

    F - C = ma​

    (ie you have to use the net force, F - C;

    you'll also have to use C to find the angular acceleration)
     
  6. Nov 8, 2012 #5
    Alright so, I got: 6.4N-C=5.6a

    a=6.4-C/5.6

    and

    α=(6.4-C/5.6)/.044

    And Torque: T=Iα= 1/2(2(.044^2))((6.4-C/5.6)/.044)=.00194((6.4-C/5.6)/.044)

    My problem is, I don't know where to go from here. I'm trying to find the angular acceleration. So would C=Normal force x mu? Then I can plug that into α? But mu isnt given? Frustrating!
     
  7. Nov 8, 2012 #6

    tiny-tim

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    but what is T?

    and about what point are you measuring T, and the moment of inertia?
     
  8. Nov 8, 2012 #7
    Torque is Radius x Force so would it be Radius x 6.4-C? Then set that equal to my Torque function and solve for C? Then I can plug that into my a) equation to find the acceleration?
     
    Last edited: Nov 8, 2012
  9. Nov 8, 2012 #8
    Also, thank you for helping me.
     
  10. Nov 8, 2012 #9

    tiny-tim

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    yes :smile:, but
    i] the 6.4N is not a force on the cylinders, so it does not contribute to the torque
    ii] the friction with the ground does contribute to the torque
    iii] you don't want to have to find the friction, so about which point should you measure the torque?

    (and iv] don't forget there are two cylinders)
     
  11. Nov 8, 2012 #10
    But the 6.4N is the cause for the rotation of the wheels which is parallel to the cylinders. If there were no 6.4 N they wouldn't rotate. The only other force I can think of that is applied to the cylinder is the downward force of mg on from the plank. Is friction the only force that contributes to the torque? Excuse my ignorance, I'm trying every opportunity I see.
     
  12. Nov 8, 2012 #11

    SammyS

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    Draw a free body diagram for the plank and for a cylinder. (The free body diagram for the second cylinder is identical to that for the first cylinder.)
     
  13. Nov 8, 2012 #12
    We have, mg from the plank pointing down, Normal force for the plank pointing up, Force of friction pointing to the left, and 6.4N pointing to the right.
     
  14. Nov 8, 2012 #13

    SammyS

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    That force of friction on the plank if 2fp, or as called it, 2C.

    Now, for a free body diagram for either cylinder...
     
  15. Nov 8, 2012 #14
    For either cylinder, we have a normal force pointing up, the mg of the plank down onto the cylinder plus the mg of the cylinder itself. The side ways force of the plank's force pointing to the right.
     
  16. Nov 8, 2012 #15

    SammyS

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    How about fg, "rolling friction exerted by the ground on each roller".

    If there were no friction between the ground & the cylinder, the cylinders would rotate at a different rate than they do with that friction.
     
  17. Nov 9, 2012 #16
    F_p is to the right on the top of the cylinder and F_g is pointing to the right on the ground. I got that but how do I go about finding the initial acceleration when the rollers are equidistant to the ends of the plank. I'm lost in the abyss .
     
  18. Nov 9, 2012 #17

    SammyS

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    fp acting on the cylinder is to the left if the cylinder is being pulled to the right.

    The cylinder will exert a force of fp on the ground which is to the right.
     
    Last edited: Nov 9, 2012
  19. Nov 9, 2012 #18

    tiny-tim

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    hi ben! :smile:

    (just got up :zzz:)
    yes that's right :smile:

    do you now see how important it is to draw a separate free body diagram for each differently-moving part?

    if you don't, you'll get confused as to which force is acting on which body … only directly acting forces count! :wink:
    the only significance of the "equidistant" thing is that it means that the normal forces are the same, and so the friction forces are also the same

    (it just makes the arithmetic easier! :biggrin:)

    ok, you now have a free body diagram which tells you there are only two forces that can give a torque to each cylinder: the plank friction force (Fp), and the ground friction force

    you have a completely free choice as to which point you take torques about …

    if you only want Fp in the equation, which point should you choose? :smile:
     
  20. Nov 9, 2012 #19
    Would I choose the top of the cylinder because it's the area in perpendicular contact with the friction force to the plank?
     
  21. Nov 9, 2012 #20

    SammyS

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    The plank exerts zero torque about the top of the cylinder.

    The ground exerts zero torque about the bottom of the cylinder.

    Now, back to tim's question... Which point should you choose for calculating torque, if you don't want to mess with fg, the frictional force exerted by the ground?
     
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