physic/machanic: relationship between the angle and time of a tilted falling object

Discussion in 'Homework Help' started by bug13, May 28, 2012.

  1. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    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 :)


    [​IMG]
     
    Last edited: May 28, 2012
  2. WBahn

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    Mar 31, 2012
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    Nice try, but there are a two important things you are overlooking:

    1) Gravity is not the only force acting on the stick. There are also the support forces at the ground. The stick moves in response to the net influences of ALL the forces acting on it.

    2) Energy is not only going into motion of the center of mass, but also into the rotational motion of the stick about its center of mass.

    Start with a free-body diagram (FBD) to identify all of the forces on the stick. This may require being a bit more specific about the support arrangement; for instance, is the assumption that the bottom end is either pinned or that the floor produces enough friction to keep the end of the stick from sliding sideways? Then analyze the kinematics to identify the constraints on how the stick can move. For instance, the rate at which the center of mass falls is directly coupled to the rotational speed of the stick about its center of mass. Here you will need to modify the FBD to account for forces that aren't there when the object is stationary and just tilted. Then apply the equations of motion, both for translational and rotational. At this point, you should have enough to solve for the angle as a function of time.
     
  3. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    Thanks for you input WBahn, your third para. is a bit too much for me, so here is what I come up with (one step at a time):

    E(InitPotential)=E(Petential)+E(RotationalKineticEnergy)

    Code ( (Unknown Language)):
    1. mgh=mgh(1-sinθ)+0.5Iω^2, I=1/3mL^2
    2. mgh=mgh(1-sinθ)+(1/2)(1/3)m(2h)^2ω^2, ω=θ/t
    3. g=g(1-sinθ)+(2/3)hω^2, ω=θ/t
    4. g=g(1-sinθ)+(2/3)h(θ/t)^2
    what to do next?

     
  4. WBahn

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    The final paragraph was intended primarily to give you some terms to google for.

    You have two kinds of kinetic energy. Translational (movement of the center of mass) and rotational (movement about the center of mass).

    If we assume that the end on the ground is pinned so that the stick has to rotate about that point, then you can take care of both of the kinetic energies by using the moment of inertia of a long thin stick about one end.

    At any given angle of rotation, you can calculate how much gravitational potential energy has been turned into kinetic energy. You can then use the above to determine the rotational velocity at that angle.
     
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  5. bug13

    Thread Starter Well-Known Member

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    I try to stay away from using FBD as in this situation, is troublesome, my idea is using conservation of energy, this is a more simple solution for me.

    so my above equation is valid?

     
  6. WBahn

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    You are saying ω=θ/t. This is analogous to saying v = x/t, which would only be true if the object were travelling at a constant velocity or, in this case, if it were rotating at a constant rate. But we know that this is not the case. So you need the general case expression, which is ω=dθ/dt.
     
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  7. t_n_k

    AAC Fanatic!

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    Why isn't the angular velocity ω at any angle θ simply proportional to the loss in the rod's potential energy at that angle?

    So is it not a matter of equating

    Initial PE - current PE = Increase in KE

    Rod Length = L
    Rod Mass = M

    Initial PE = (MgL)/2

    Increase in KE = (Iω^2)/2

    For the rod I=(ML^2)/3
     
  8. WBahn

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    Because rotational kinetic energy is not proportional to angular velocity, but rather to the square of it.

    But finding the angular velocity as a function of angle is only part of the story. He appears to be looking for the angle as a function of time.
     
  9. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    that make sense, so anything else need to be changed? sine I produce this equation in the assumption that the object is rotating at a constant rate.

    Code ( (Unknown Language)):
    1. mgh=mgh(1-sinθ)+0.5Iω^2, I=1/3mL^2
    2. mgh=mgh(1-sinθ)+(1/2)(1/3)m(2h)^2ω^2, ω=dθ/dt
    3. g=g(1-sinθ)+(2/3)hω^2, ω=dθ/dt
    4.  
    5. g=g(1-sinθ)+(2/3)h(dθ/dt)^2


     
  10. t_n_k

    AAC Fanatic!

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    Another approach is to use torque relationships

    At angle \theta the torque on the rod is

    \tau=Mg\frac{Lcos(\theta)}{2}=I\alpha=-I\ddot{\theta}=-\frac{ML^2}{3}\ddot{\theta}

    Integrating the resulting equation in \ddot{\theta} w.r.t \theta & having regard to the boundary conditions would give the required \dot{\theta}
     
  11. t_n_k

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    EDIT: I can now see why doing this by numerical methods is a good option.
     
  12. WBahn

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    Code ( (Unknown Language)):
    1. mgh=mgh(1-sinθ)+0.5Iω^2, I=1/3mL^2
    This is saying

    Total Energy, mgh, is equal to PE energy converted, mgh(1-sinθ), plus rotational KE, 0.5Iω^2.

    Does that make sense?

    Conservation of energy says that total energy is constant, which would be stated more along the lines of:

    Total Energy, mgh, is equal to PE energy remaining, mgh(sinθ), plus rotational KE, 0.5Iω^2.

    Or, equivalently:

    PE energy converted, mgh(1-sinθ) equals rotational KE, 0.5Iω^2.

    To other things to note:

    h = L/2

    and, since you have chosen θ to be the angle CW from the negative horizontal axis, recognize that ω corresponds to the rate of rotation in the CW direction. So you expect it to be negative.

    Solving this analytically may be impossible because you have a nonlinear differential equation for two reasons: (1) it involves ω^2, and (2) it involves trig functions.

    NOTE: I'm ignoring t_n_k's posts right now, but you shouldn't. His last post is the approach I originally tried to steer you toward. Now I am merely trying to work with you on the approach you have chosen. At the end of the day, I recommend you solve the problem using both approaches and be satisfied how both work and how both yield the same result. But I think you will be well served by slugging through the fine points of your approach. Also, note that his approach is leading to the same problem, namely the same kind of nonlinear differential equation; this is actually reassuring because it means both approaches are (potentially) leading to the same place. However, his only has the trig issue and I think it may be possible to work through that using the chain rule.
     
  13. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    another point taken :)

    so here is the rework equation:

    Code ( (Unknown Language)):
    1. mgh=(1-sinθ)=(1/2)Iω^2, I=(1/3)mL^2, ω=(dθ/dt), L=2h
    put them together, and I get:

    Code ( (Unknown Language)):
    1. g(1-sinθ)=(2/3)h(dθ/dt)
    I can only take it one step at a time for now, I will look into your step 2 and 3 after I get my formula right, I will get confused if I try to do too much at a time :)

     
  14. bug13

    Thread Starter Well-Known Member

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    Hi t_n_k,

    can you lay down some basic for me please, your approach is all new to me, like something I can google for :)

     
  15. WBahn

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    Check your units! Always, always, always check your units!

    Code ( (Unknown Language)):
    1. mgh=(1-sinθ)=...
    Left side has units of energy. Right side is dimensionless. => You KNOW it is wrong.

    Code ( (Unknown Language)):
    1. g(1-sinθ)=(2/3)h(dθ/dt)
    Left side has units of m/s^2. Right side has units of m/s. So what do you know?
     
  16. bug13

    Thread Starter Well-Known Member

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    Opooos! typo :p

    but it's a very good point, I will add "check units" to exam/test review routine.

    it should be:

    Code ( (Unknown Language)):
    1. g(1-sinθ)=(2/3)h(dθ/dt)^2


     
  17. WBahn

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    All he is doing is applying the equivalent of F=ma in the rotational world. The equivalent of Force (the thing that makes something tend to move in a straight line) is torque (the thing that makes something tend to rotate). The equivalent of mass (the inertia that tends to resist changes in motion in a straight line) is moment of inertia (the inertia that tends to resist changes in rotational motion). The equivalent of acceleration (the measure of the change in motion from a constant speed in a straight line) is angular acceleration (the measure of change in rotational motion.

    The torque on the stick due to gravity is the component of the gravitational force that is acting tangential to the stick's rotation multiplied by the distance to the center of mass (the point where the gravitational force is effectively concentrated). Think about using a wrench to loosen a bolt. The further away from the bolt you apply the force, the greater the effect for the same amount of force. But, you also intuitively know that you get the best effect by applying the force perpendicular to the wrench (i.e., tangent to the circle you want the bolt to rotate in). You wouldn't apply the force partway toward the bolt or away from the bolt unless you had to, because you know that part of your force is being wasted. The same is true here. With the stick tilted, part of the gravitational force is acting along the axis of the stick toward the pivot point and is accomplishing nothing. So you have to resolve the gravitational force into two components - one acting parallel to the stick and one acting perpendicular to the stick. Only the latter contributes to the torque. This turns out to be equivalent to using the full weight but only using the perpendicular distance between that line of force and the pivot point. Either results in the same torque, namely mgLcos(θ)/2.
     
  18. WBahn

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    You're actually still missing the point. Don't add "check units" to your "exam/test review" routine. Add "check units" to your "how I work every problem I ever work" routine.

    It doesn't matter if you are working a problem on an exam, designing something for your boss, or figuring out how much concrete to order for that patio you are planning; track the units from the first step to the last step of your work. Always. Without exception. Would you like to have to pay for nine times as much concrete as you really needed because you divided by 3 to convert feet to yards when what you needed to do was divide by 27 to convert from cubic feet to cubic yards? Tracking your units will not only catch that a mistake was made, but provide a really big hint as to where the error was made.

    Do just two things and you will eliminate a huge majority of the mistakes you will ever make working problems:

    1) Tracking units throughout your work and make sure they work out.
    2) Always ask if the answer makes sense.
     
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  19. bug13

    Thread Starter Well-Known Member

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    thanks a lot, that's actually so true, I just realized that is one of the easiest way to eliminate most of the mistakes I have ever make.

    Thanks for pointing it out, I think these two points will serve me well in the future when I am working, really appreciate it! :)
     
  20. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    now I don't seem to know how to make θ as the subject.:mad:
     
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