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Homework Help: Relationship between the angle and time of a tilted falling object

  1. May 28, 2012 #1
    1. The problem statement, all variables and given/known data

    find relationship between the angle and time of a tilted falling object

    2. Relevant equations

    any that is applicable

    3. The attempt at a solution

    As seen in picture below, and that's what I have done so far. It's a tilted object, only gavity and the support force(floor) are acting on it.

    • Is it valid? (using conservation of energy)
    • or do I need to do a force analysis?
    • if it's in-valid, what should I google for?

    any input is appreciated, it's a real word problem, thanks :smile:
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. May 28, 2012 #2
    What kind of movement is the body going through? You used velocity, but what kind of velocity exactly?
  4. May 28, 2012 #3
    the object is simply fall on the floor, from almost up right, due to gravity. there is no other force.

    image you try to put a stick up right on the floor, it simply can't stay up right from the moment you take your hand off it.

    please ask if i am not making sense, as I am not sure if i describe it correctly
  5. May 28, 2012 #4
    You're making sense, I understand the question.

    My question for you is how does the stick fall? What kind of velocity does it gain?
    For example, a car has translational velocity, while a pinwheel has rotational velocity.
  6. May 28, 2012 #5
    that's part of the question, the answer is from analyzing the situation, that's the part that I am not sure if my equation is valid.

  7. May 28, 2012 #6
    You wrote in your work that mgh = 1/2mv2. This is assuming that the falling stick is transferring all of its potential energy into translational kinetic energy.
    But this isn't true, its potential energy is gradually being transferred into rotational kinetic energy. So your energy statement should have three parts: the total energy (that it starts with), the potential energy at some height h (or some angle theta), and the rotational kinetic energy.
  8. May 29, 2012 #7
    that makes sense, so how do I fix the formula? from wikipedia, I found rotational kinetic energy is
    Code (Text):
    , so the new formula is something like:

    Code (Text):
    and it can be simplied as:

    Code (Text):
    Is this one valid now?

    if it is valid, can you help me help out with I in the formula, as I have no idea what that is.

  9. May 29, 2012 #8
    I don't know why you added an m into the term for rotational kinetic energy (in that middle code block).

    And I is the moment of inertia. It depends on the body that is being rotated, and for a uniform rod it is 1/3mL2. Since we already defined h as being the distance from the end of the rod to the center of mass, what can you say L is? And what does the equation simplify into now?
    (And what is ω? It may be helpful to turn it into an equivalent expression.)
  10. May 29, 2012 #9
    sorry about the added m into the term for rotational kinetic energy, it's a typo.

    as of the moment of inertia I=1/3mL2, assuming m is the mass, L is the length of the object, so the new equation is:

    Code (Text):
    mgh=mgh(1-sinθ)+0.5Iω[SUP]2[/SUP], I=1/3mL[SUP]2[/SUP]
    mgh=mgh(1-sinθ)+(1/2)(1/3)m(2h)[SUP]2[/SUP]ω[SUP]2[/SUP], ω=θ/t
    g=g(1-sinθ)+(2/3)hω[SUP]2[/SUP], ω=θ/t
    Is it valid now?

    Last edited: May 29, 2012
  11. May 29, 2012 #10
    All of the work is correct with the exception of ω=θ/t, as ω (the rotational velocity) is the change in θ over the change in time, so it should read ω=dθ/dt.
  12. May 29, 2012 #11
    Thanks Villyer, for all your help :smile:

    this should be the last equation now?

    mgh=mgh(1-sinθ)+0.5Iω2, I=1/3mL2
    mgh=mgh(1-sinθ)+(1/2)(1/3)m(2h)2ω2, ω=dθ/dt
    g=g(1-sinθ)+(2/3)hω2, ω=dθ/dt

  13. May 29, 2012 #12
    That all looks right to me :cool:

    I attempted to solve this differential equation, but I got stuck trying to evaluate ∫√{csc(θ)}dθ, which wolfram alpha gives as involving an 'elliptic integral of the first kind' (http://www.wolframalpha.com/input/?i=integral+√csc(x)+dx) which I don't understand.

    Maybe if the domain is limited to (0,π/2] the integral is easier.
  14. May 29, 2012 #13
    As in my attached pic in OP, θ is between 0 to Pi/2, but I think solving this need more advance mathematical method, which I don't have a clue :confused:

    Hope someone with the skills and knowledge jump in and give me a hand

  15. May 29, 2012 #14


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    Hello bug13. Welcome to PF !


    The motion of the stick is more complicated than it might seem initially.

    You can solve this with energy principles, as you are doing, or with force & torque.

    If the contact of the stick with the floor is frictionless, then the C.M. (center of mass) of the stick falls exactly vertically downward.

    If you assume that the bottom of the stick maintains contact with the floor (yes, I realize it's sliding) then you will find that the solution resulting from this will have that the angular velocity, ω, of the stick achieves some maximum, then decreases. This can't happen unless the floor pulls down on the end of the stick. Once you reach the point (*) at which ω is a maximum, ω will remain constant and the center of mass will fall with acceleration of g.

    Added in Edit:

    * At this point it's also true that the acceleration of the C.M. would become greater than g, downward.

    Also, the top end of the stick will contact the floor first, before any other part does.
    Last edited: May 29, 2012
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