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Homework Help: Angular and Tangential Variables

  1. Nov 24, 2007 #1
    [SOLVED] Angular and Tangential Variables

    1. The problem statement, all variables and given/known data
    A thin rod (length = 1.5m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge. The mass of the rod may be ignored, compared to the mass of the object fixed to the top if the rod. The rod, starting from rest, tips over and rotates downward. (a) What is the angular speed of the rod just before it strikes the floor? (Hint: consider using the principle of conservation of mechanical energy.) (b) What is the magnitude of the angular acceleration of the rod just before it strikes the floor?

    2. Relevant equations
    mgh = 1/2 mv2
    v=st ?

    3. The attempt at a solution
    I'm really having trouble even setting this one up. I drew a picture of the setup, I know the bar is initially at rest. I'm completely lost on how to even go about it. I've tried the PE=KE equation and canceled out the mass variable which gives me v = 5.4, then I plugged that velocity into the w=vt using a time I got from using 1/2gt2 but that shouldn't really work as its not a free fall problem I think.
    Any ideas on how to setup the problem? I think I can figure it out if I can just wrap my brain around the proper setup.

    Thanks in advance!
    Last edited: Nov 24, 2007
  2. jcsd
  3. Nov 24, 2007 #2

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    Your energy eqn is correct.

    Torque = I*Angular accn.

    Take torque of the forces acting on the rod about the fixed pt.
  4. Nov 24, 2007 #3
    Ok. But how would I find torque without the mass of the rod?
    T = LF sin theta...right?
    T = (1.5m)(mg) sin (theta) ?
  5. Nov 24, 2007 #4

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    How did you get the ang velo? Same way.
    (BTW, I presume you meant the ang speed when you said v=5.4.)
  6. Nov 24, 2007 #5
    I found angular velocity using w=vt. However its not correct. The t (time) I got from using [tex]\frac{1}{2}[/tex]gt[tex]^{2}[/tex] but that isn't correct as its not a free fall problem, right? So my angular velocity of (5.4)(1.5) = 8.1 rad/s is incorrect. The book gives an answer of 3.61 rad/s.

    Am I thinking completely wrong on this one?
  7. Nov 24, 2007 #6

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    Let's start again.

    The rot KE just before it strikes the floor is the initial grav PE of the rod. If I is the MI of the rod about the fixed pt, and w is the ang speed, then can you write down any eqn?
  8. Nov 24, 2007 #7
    Lets see...
    mgh=1/2 mv^2
    So the KE = [tex]\frac{1}{2}[/tex] I[tex]\omega^{2}[/tex]

    Does the PE = Igh ?

    so Igh = [tex]\frac{1}{2}[/tex] I[tex]\omega^{2}[/tex]

    Wouldn't I cancel leaving gh = [tex]\frac{1}{2}[/tex][tex]\omega^{2}[/tex]

    Then I could just solve for [tex]\omega[/tex] ?
  9. Nov 24, 2007 #8

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    The grav PE of a body wrt a plane is mgh, where h is the height of the CG of the body above the plane. This would be equal to Iw^2/2. Can you find w now?
  10. Nov 24, 2007 #9
    I'm not sure I understand why Iw^2 is divided by 2 in your last post.

    For a rod setup this way, I = 1/2 ML^2 ?
    However, I'm missing the mass. So I dont think I could find w just yet, right?
    I'm sorry, I'm really not trying to be purposefully dull.
  11. Nov 24, 2007 #10

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    The expression for rotational KE about a point is Iw^2/2, when the MI of the body about that point is I. It's a formula, like mv^2/2 for translational KE.

    For a rod, the MI about one of its ends is mL^2/3, where L is the length of the rod.

    Now can you find w, as I said in my last post?
  12. Nov 24, 2007 #11
    Ok, my teacher used the equation 1/2 Iw^2 but thats the same as Iw^2/2. Sorry about that. So, mgh = Iw^2/2 ?

    Then, I = mL^2/3....
    So we get the equation as...

    mgh = [tex]\frac{(mL^2/3)w^2}{2}[/tex]

    I can if I had the mass of the rod to insert for m in the equation. However the problem said to ignore the mass...
    So the full equation should be...

    gh = [tex]\frac{(L^2/3)w^2}{2}[/tex]

    If so, then yes, I can find w.
    w = 6.26

    However the answers in the book have the angular velocity as 3.61 rad/s
    and the angular acceleration as 6.53 rad/s^2
  13. Nov 24, 2007 #12

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    What is the height h of the CG of a rod?
  14. Nov 24, 2007 #13
    Hmm...didn't think of that.
    It would be W[tex]_{1}[/tex]X[tex]_{1}[/tex] / W[tex]_{1}[/tex] ?

    But the rod is oriented vertically so there is no distance and weight is not given, right?
  15. Nov 24, 2007 #14

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    The CG of a uniform rod is always at its mid point. So, in the above formula, put L/2 in place of h, and then find w again.
  16. Nov 24, 2007 #15
    Ah....man I feel stupid. Thank you so much for your help, I think I really understand what was going on now.

    Lets see...

    g(L/2) = [tex]\frac{(L^2/3)\omega^2}{2}[/tex]

    so that gives [tex]\omega[/tex]=4.43 which is still different from the book answer of 3.61. However sometimes the book gives an answer a bit different from the right one. This is a bit larger than the normal difference though. Does that look right to you?
  17. Nov 24, 2007 #16

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    I read your first post again. The rod is massless. There's a mass at the top end of the rod. Do the calc again. It's much easier.
  18. Nov 24, 2007 #17
    Heh, ok so just remove the mass component (I) from the equation?
    I've done it as gh=w^2/2 thats still not right.

    i'm sorry I'm a bit confused I think.
  19. Nov 24, 2007 #18

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    Grav PE = mgL. Rot KE just before hitting the ground = (1/2)Iw^2. I=mL^2. So,

    mgL = (1/2)mL^2*w^2 => w= 3.61.
  20. Nov 24, 2007 #19
    wow, I made this so much harder than it was supposed to be.

    I really appreciate your time and help.

    Thanks alot!
  21. Nov 24, 2007 #20

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    I'm sorry that I made a mistake while reading the problem. On the brighter side, you would be able to do the other kind of problem from now on. Find the angular accn now.
  22. Nov 24, 2007 #21
    No problem, yeah that was free info for me!

    Ok, so...

    [tex]\alpha[/tex] = 3.61 / .55s

    [tex]\alpha[/tex] = 6.56

    So does that mean that the equation I used for finding the time is correct? If so, how? doesn't the rigid stature of the rod change the time it takes to fall through the distance?
  23. Nov 24, 2007 #22

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    No, this is incorrect.

    For finding ang accn, look at what I'd said before -- torque etc.
  24. Nov 24, 2007 #23
    Ok, so Torque = I * Angular Acceleration

    We dont know Torque or I because I involves mass. We can use the same rationale as before and say I = mL^2 but we still are missing mass.

    Couldn't we use w[tex]^{2}[/tex] = w[tex]_{o}[/tex][tex]^{2}[/tex] + 2 [tex]\alpha[/tex] ([tex]\theta[/tex])

    And use 1/4 2[tex]\pi[/tex] as [tex]\theta[/tex] ?
  25. Nov 24, 2007 #24

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    That formula you've given is valid for uniform ang accn.

    The torque of the weight of the mass at the point before striking the floor is force*dist from pt of rotation = mg*L. I remains the same as above, i.e., mL^2.

    Can you complete the eqn now and find the value with proper units?
  26. Nov 24, 2007 #25
    Well since the rod is massless we cannot find mg

    the problem did say to ignore the mass so do you simply remove m from the equation?
    Giving just gL or (9.8)(1.5m) = 14.7

    So 14.7 = 2.25 a
    a = 6.53 rad/s^2

    Hah, ok so why can we just drop mass out of an equation and still be ok?
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