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Homework Help: Tangential acceleration, Inertia, and Torque

  1. Dec 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A solid rod of mass M = 1.35 kg and length L = 83 cm is suspended by two strings, each with a length d = 71 cm (see Figure), one at each end of the rod. The string on side B is cut. What is the magnitude of the initial tangential acceleration of end B?
    http://gauss.vaniercollege.qc.ca/webwork2_files/CAPA_Graphics/Gtype20/prob13a.gif


    The string on side B is retied and now has only half the length of the string on side A. What now is the magnitude of the initial tangential acceleration of end B if it's cut again?
    http://gauss.vaniercollege.qc.ca/webwork2_files/CAPA_Graphics/Gtype20/prob13b.gif

    2. Relevant equations

    a(tangential)= rα
    α=Torque/I
    I=1/3MR^2


    3. The attempt at a solution
    Given: m=1.35kg
    L=.83m
    d=.70m

    I= (1/3)(1.35)(.83)^2=.31 kg*m^2
    Torque= Frsinθ
    F=ma
    F=(1.35*9.8)=13.23N
    Torque=13.23(0.83)=10.98N*m
    α=10.98/.31= 35.42kg*m^2
    a(tangential)= 0.83(35.42)=29.4 m/s^2

    I haven't tried b) yet because a) isn't the right answer
     
  2. jcsd
  3. Dec 15, 2012 #2

    TSny

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    Did you use the correct value for the "lever arm" for the force of gravity?
     
  4. Dec 15, 2012 #3
    83 cm converted to .83 m... also tries using half the mass because each rope held 1/2 the weight of the rod, it didn't work either
     
  5. Dec 15, 2012 #4
    Ohhh, do I have to use the length of the string?
     
  6. Dec 15, 2012 #5

    TSny

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    The rod is an extended object. Each particle of the rod has some weight. So, the weight doesn't actually act at just one point of the rod. Nevertheless, every object has a special point where you can imagine all the weight of the object to be concentrated for the purpose of calculating torque due to weight.

    Do you know what that special point is called? Do you know where that point is located on the rod?
     
  7. Dec 15, 2012 #6
    Center mass?
     
  8. Dec 15, 2012 #7

    TSny

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    Yes, center of mass (or center of gravity). If you support an object at the center of gravity, it will balance. Suppose you wanted to balance a rod on the tip of your finger. At what point of the rod would you place your finger in order to have the rod balance? That point is the center of gravity.
     
  9. Dec 15, 2012 #8
    Okay, but how do I apply this to the problem^ Do I use the length 0.83/2?
     
  10. Dec 15, 2012 #9

    TSny

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    Yes. You need to use the distance from the axis of rotation to the center of mass.
     
  11. Dec 15, 2012 #10
    Torque=13.23(.83/2)=5.49N*m
    alpha= 5.49/.31=17.71rad/s^2
    a(tangental)= 17.71*.83/2=7.35m/s^2

    Where did I go wrong again?
     
  12. Dec 15, 2012 #11

    TSny

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    I think you now have the correct answer for alpha. But to get the tangential accelertion of end B, you need to use the correct distance from the axis to end B.
     
  13. Dec 15, 2012 #12
    It worked! Thank-you so much!, I'm going to try b) now..
     
  14. Dec 15, 2012 #13
    So, for b), moment of Inertia would be the same? Then for the torque=Frsin(theta), do I find the vertical component of gravity?
     
  15. Dec 15, 2012 #14

    TSny

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    Gravity is already vertical. In ##\tau = Frsin\theta##, ##\theta## is the angle between ##r## and the force of gravity. ##r## is directed from the axis to the center of mass.
     
  16. Dec 16, 2012 #15
    I'm still not getting the right answer, is the center of mass located at the center or is it more to the left now?
     
  17. Dec 16, 2012 #16

    TSny

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    The center of mass of the rod is always at the center of the rod. So in [itex]\tau[/itex] = Frsinθ, r will still be the distance from the axis to the center of the rod. However, you need to figure out the value of θ, which is the angle between the r direction and the force direction.
     
  18. Dec 16, 2012 #17
    sin90/0.83=sin(theta)/0.355
    sin-1= 23.8deg
    then, I followed the same steps as I did for part a).
    Torque=1.35*9.8*.355*sin23.8
    Inertia is the same
    alpha=T/I
     
  19. Dec 16, 2012 #18

    TSny

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    You need to find the angle labeled θ in the attached figure. It looks like you calculated [itex]\phi[/itex]. (You might want to recheck the calculation - I think you plugged .335 into your calculator instead of .355).

    Anyway, once you have [itex]\phi[/itex], it should be easy to see what θ is.
     

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  20. Dec 16, 2012 #19
    I'm so sorry!!
    I got the angle theta to be 115.322deg
    1.35(9.8)(.355)sin115.322=4.2453
    alpha=4.2453/.31=13.6948
    a=13.6948*.83=11.3666
    I know it's wrong because tangential acceleration should be greater than the one found in part a)
     
  21. Dec 16, 2012 #20

    TSny

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    Are you sure the answer should be greater? Note that for the tilted rod, the "lever arm" distance of the weight force is smaller than for the horizontal rod. So, less torque for the tilted rod.
     
  22. Dec 16, 2012 #21
    Ok, but the answer is still wrong and I don't know where my mistake is...
     
  23. Dec 16, 2012 #22

    TSny

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    OK. After reconsidering the problem I see that it's more complicated when the rod is released from a tilted position. Point A of the rod is not a fixed axis of rotation.

    Point A will accelerate to the left when the string at B is cut such that the center of mass will initially have a linear acceleration vertically downward. This is because the direction of the acceleration of the center of mass is determined by the direction of the net force acting on the rod. Both the tension in the string at A and the force of gravity are vertical forces at the instant the string at B is cut. So, the initial acceleration of the center of mass must be vertical.

    I can help you work through the problem. But first, it would help me if you could tell me what level course this is and what dynamics principles you are familiar with that you think are relevant.

    [EDIT: Also, due to the fact that point A is not a fixed axis of rotation, it's not clear to me what is meant by "tangential acceleration" of point B. Tangent to what? But it is possible to find the linear acceleration of point B just after the string is cut. The acceleration of B will not be perpendicular to the rod.]
     
    Last edited: Dec 16, 2012
  24. Dec 16, 2012 #23
    It's mechanics, I'm guessing around grade 12, but I'm from Quebec, so here it's CEGEP. It's a crap system, but anyways, here is what I know from dynamics

    -the 3 basic equations for rotational kinematics
    -Torque=I(alpha)
    I=mr^2 for a point on a mass
    I know how to calculate I for a continuous rod or disk
    I know how to relate linear and angular variables

    Maybe I missed a few things, but that's in general.
     
  25. Dec 16, 2012 #24

    TSny

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    (Students in the US would probably not encounter a problem like this until their first engineering dynamics course, which they usually take after a year of calculus-based college physics.)

    When the string at B is cut, you have essentially 4 unkowns: the x and y components of the acceleration of the center of mass, the angular acceleration of the rod, and the tension in the string at A.

    You have 3 dynamical equations that relate these 4 unknowns: F = ma (x and y equations) and [itex]\tau=I\alpha[/itex] (about the center of mass).

    A 4th relation can be obtained through a kinematical constraint: the acceleration of point A can only be horizontal immediately after the string at B is cut due to the constraint of the string.

    The acceleration of point A, ##\vec{a}_A##, is related to the acceleration of the center of mass, ##\vec{a}_C##, through the kinematics equation

    ##\vec{a}_A = \vec{a}_C + \vec{a}_{A/C}##

    where ##\vec{a}_{A/C}## is the acceleration of point A relative to the center of mass C.
     
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