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Rotational Dynamics about an axis parellel to the axis at the center of mass

  1. Aug 16, 2012 #1
    1. The problem statement, all variables and given/known data
    A uniform sign is nailed into a wall at its top two corners. The nail on the upper-right corner breaks, and the sign begins to rotate about the remaining nail in the upper-left corner. The mass of the sign is 69 kg, L (length) = 1.6 m, H (height) = 2.2 m, and the moment of inertia of the sign about the upper-left corner is 170 kg · m2.

    What is the moment of inertia about an axis that is parallel to the one through the upper-left nail but passes through the center of mass of the sign?


    What is the angular acceleration of the sign just after the nail breaks?


    2. Relevant equations
    Icm (rectangular sign) = [m*(H^2] + L^2)]/12

    I(p) = I(cm) + m*d^2
    Where d is the perpendicular distance to the axis of rotation

    Ʃτ = I*[itex]\alpha[/itex]

    ?????

    3. The attempt at a solution
    So I am just confused on the second part. Using the first equation that I provided I was able to find the moment of inertia about a parallel axis at the center of mass. For the next part though, I tried using the equation for a mass rotating about a "point" at a certain distance from the center of mass to find this distance. From there I stopped. I don't really know what to do now? I thought of maybe using torque but wasn't sure how.
    Any help is appreciated! Thanks :)
     
  2. jcsd
  3. Aug 16, 2012 #2

    cepheid

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    The gravitational force can be thought of as acting at the position of the centre of mass. Since the sign is uniform (i.e. its density is constant), the centre of mass corresponds to the geometric centre of the sign. So, you have a force equal to the weight of the sign acting at its centre. What torque does this force produce around the pivot point?
     
  4. Aug 16, 2012 #3
    So torque is equal to Force crossed with the distance.

    τ = F x d

    And in this case we would want the perpendicular distance from the center of the sign to the pivot point, which is 1/2 the length. The force, as you said, will be the weight, which is perpendicular to this distance. So we can simplify the cross product.

    τ = F*d = (m*g)*(1/2*L)

    Then can I set this torque equation equal to this one?

    τ = I*[itex]\alpha[/itex]
     
  5. Aug 16, 2012 #4

    cepheid

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    Yeah, of course. If one expression 1 is equal to the torque, and expression 2 is also equal to the torque, then expression 1 must be equal to expression 2.
     
  6. Aug 16, 2012 #5
    Sweet! Thanks so much for your help. I was getting all mixed up.

    Now I have to find the tangential acceleration. This is fairly easy. However, I just wanted to double check one point. In the equation where a (tan) = [itex]\alpha[/itex] * r, is r the distance from the center to the point of rotation? Or is it the perpendicular distance as I used in my last part of the problem? Just a clarification would be helpful.
     
  7. Aug 16, 2012 #6

    cepheid

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    The problem as stated in your original post doesn't ask for the linear (tangential) acceleration at any point, it only asks for the angular acceleration, ##\alpha##. The angular acceleration is the same for every point in the rigid body, whereas the tangential acceleration will be different at different radii from the axis of rotation. Since the problem as you've stated it doesn't ask for a tangential acceleration, I can't really advise you on the radius at which you should compute it.
     
  8. Aug 16, 2012 #7
    Oops! Sorry, I didn't copy the full problem. Here is the part referring to the tangential acceleration. Feel free to contribute or not.

    What is the magnitude of the tangential acceleration of the center of mass of the sign just after the nail breaks?

    So I am assuming that I am using the distance from the point of the center of mass to the point of rotation...
     
  9. Aug 16, 2012 #8

    cepheid

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    Yeah. It asks for the acceleration of the centre of mass, so you have to use the 'r' of the centre of mass. The linear acceleration is different for different points in the sign, so you can't use some other 'r' corresponding to some other point. You have to use the radial distance of the point whose acceleration you're interested in.
     
  10. Aug 16, 2012 #9
    Hmmm that was a bit confusing to me. So basically am I going to have to manipulate the equation pertaining to the parallel axis theorem to solve for the radial distance?

    (i.e. I (p) = I (cm) + m*r^2)
     
  11. Aug 16, 2012 #10

    cepheid

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    Then let me give a less confusing answer:


    YES.

    Of course you have to use the radius of the point whose acceleration you are trying to compute. That's all I was trying to say in my previous post.

    What the heck does the parallel-axis theorem have to do with this part of the problem?
     
  12. Aug 16, 2012 #11
    I really appreciate your help. I am sorry if I have aggravated you. I will stop posting.
     
  13. Aug 16, 2012 #12

    cepheid

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    No, you haven't aggravated me, just puzzled me quite a bit. If you still are having trouble solving the problem and don't know what to do, then keep posting.
     
  14. Aug 16, 2012 #13
    Thank you. I think I can handle it from here. I really do appreciate your patience, and sorry for puzzling you. Sometimes it takes a while to wrap my head around even the simplest physics concepts. It's very frustrating. I completely understand what you were trying to say now. I don't know where the heck the parallel axis theorem came from. My bad.

    Thanks!
     
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