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Dynamics problem

  1. Jul 5, 2006 #1
    Hi, I'm from Chile, so i had to translate the problem from spanish and my english ain't that good so if you don't understand something, please tell me.

    A particle of mass m is initially at rest at the edge of a horizontal semicilinder of radius R (theta (t=0)=0). The particle is tied to an ideal string with the other end of the string (P) loose on the other side of the semicilinder. P is pulled vertically downwards, starting at rest and with a constant acceleration (a0). Depending on the value of a0 it is possible that in some point along the trayectory the particle detaches from the surfaceof the semicilinder or that the string looses its tension.

    a)assuming the particle detaches from the surface before the string looses its tension, find an equation for the angle theta of detachment

    b)assuming the string looses its tension before the particle detaches from the surface, find the angle theta in which this happens.

    This is what i've got:

    [tex]m(-\rho \dot{\theta^2} \hat{\rho} + \rho \ddot{\theta} \hat{\theta}) = N \hat{\rho} + T \hat{\theta} - mg\sin(\theta)\hat{\rho} - mg\cos(\theta)\hat{\theta}[/tex]
     
  2. jcsd
  3. Jul 5, 2006 #2
    sorry, here goes an image of the system

    [​IMG]

    jajajaja, it doesn't look too good, but you'll get the picture
     
  4. Jul 5, 2006 #3

    andrevdh

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    a) The force that keeps the particle on the cylinder is the radial component of its weight (it is forced to travel along an semicurcular path) - the centripetal acceleration is provided by it. This puts an upper limit on the speed of the particle if it is to stay on the cylinder. It will fly off the cylinder if a0 increases its speed beyond this value.
     
  5. Jul 5, 2006 #4
    Are you sure? because a0 will generate an acceleration in [tex]\hat{\theta}[/tex] and the radial component of the wight generates the acceleration in [tex]\hat{\rho}[/tex]

    i know the condition for the first question is that the normal (N) becomes 0and for the second one that the tension becomes 0, but in both cases, working either in [tex]\hat{\theta}[/tex] or [tex]\hat{\rho}[/tex] i get a nonlinear first order edo that i can't solve, at least not [tex]\theta[/tex] dependently.
     
  6. Jul 6, 2006 #5

    andrevdh

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    The motion is governed by the two force components. One that controls the curvature and one that wants to keep it going in a straight line. At the point of departure the tangential one outweights the radial one - this is the condition as stated by the problem itself!
     
  7. Jul 6, 2006 #6
    I'm sorry, you're absolutely right. I wasn't thinking on the meaning of the ecuations, just trying to solve them and overlooked the fact that [tex]\rho \dot{\theta^2}[/tex] is [tex]\frac{v_t}{r}[/tex]. [tex]v_t[/tex] being tangential velocity.

    So, the ecuation for [tex]\hat{\rho}[/tex] would be:

    [tex]m(-\rho \dot{\theta^2}) = N \hat{\rho} - mg\sin(\theta)\hat{\rho}[/tex]

    [tex]\Longleftrightarrow m(-\rho \dot{\theta^2} + sin(\theta)) = N[/tex]

    so, N = 0 [tex]\Longleftrightarrow \rho \dot{\theta^2_l} = sin(\theta_l)[/tex], l for liftoff

    the only problem with this is that i can't integrate this since i've already imposed my restriction. Maybe i made a mistake somplace else?
     
  8. Jul 6, 2006 #7

    andrevdh

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    The tangential acceleration will be a0 and since there is no radial motion for a particle moving in a circle

    [tex]a_0 = r\ddot{\theta}[/tex]

    which tell you that

    [tex]\ddot{\theta} = \frac{a_0}{r} = constant[/tex]

    also for circular motion we have that (again no radial motion in a circle)

    [tex]v = r\dot{\theta}[/tex]

    Note that

    [tex]r\ddot{\theta}^2 = \frac{v^2}{r} = g\sin(\theta)[/tex]

    as we all know. This gives you the value of the speed of the particle at lift-off. Which is what originally made me thought that the problem can be solved.

    At this point I am stuck (I have'nt worked with this type of theory for a long time). Maybe someone else can help you further (try and keep the thread alive. Someone else might decide to chip in.)?
     
    Last edited: Jul 7, 2006
  9. Jul 8, 2006 #8

    andrevdh

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    I was hoping that our little Paquete will by now have shown us that he could solve the problem by himself, since I have given him all the neccessary info. But alas...

    We came to the conclusion that the tangential acceleration of the particle will be a0. That is it will start out with a tangential velocity of 0 and accelerate until its tangential velocity becomes

    [tex]v_1 = \sqrt{Rg\sin(\theta_1)} [/tex]

    where the 1 subscript indicates the values at lift-off. The problem therefore boils down to determining the angle at which this speed is reached by the particle. The time at which this occurs can be found from the fact that the tangential acceleration of the particle is given by

    [tex]a_0 = R\ddot{\theta}[/tex]

    which upon integration leads to

    [tex]\dot{\theta} = \frac{a_0}{R}t + C[/tex]

    since [tex]\dot{\theta} = 0 \ at \ t=0[/tex] comes to

    [tex]\dot{\theta} = \frac{a_0}{R}t[/tex]

    which comes as no surprise since this boils down to [tex]v=at[/tex]

    The time at which we get lift off will therefore be

    [tex]t_1 = \frac{v_1}{a_0}[/tex]

    which gives us

    [tex]t_1 = \frac{\sqrt{Rg\sin(\theta_1)}}{a_0} [/tex]

    so the next step is to determine the angle at this stage in the motion.
     
    Last edited: Jul 8, 2006
  10. Jul 8, 2006 #9
    Sorry, i had a final exam yesterday, so i checked your answer but i planned on responding today.

    i have a different approach

    [tex]a_0 = R\ddot{\theta} \Longleftrightarrow a_0 = R\frac{d\dot{\theta}}{d\theta} \frac{d\theta}{dt}[/tex]

    [tex]\Longleftrightarrow a_0d\theta = R\dot{\theta}d\dot{\theta}[/tex]

    if we now integrate the left side between [tex]\theta = 0[/tex] and [tex]\theta = \theta_1[/tex] and the right side between [tex]\dot{\theta} = \dot{\theta} (\theta = 0)[/tex] and [tex]\dot{\theta} = \dot{\theta} (\theta = \theta_1)[/tex]

    we get then [tex]a_0\theta_1 = R\frac{\dot{\theta^2_1}}{2}[/tex]

    Replacing [tex]\dot{\theta_1}[/tex] with [tex]\frac{v_1}{R}[/tex] would give the equation that they ask in the problem

    But i think there's a mistake, because you assume that the tangential acceleration of the particle is a0 but there's also the tangential component of the weight. Is this correct?
     
    Last edited: Jul 8, 2006
  11. Jul 8, 2006 #10
    Sorry, i made a mistake

    [tex]a_0 = R\ddot{\theta} \Longleftrightarrow a_0 = R\frac{d\dot{\theta}}{d\theta} \frac{d\theta}{dt}[/tex]

    [tex]\Longleftrightarrow a_0d\theta = R\dot{\theta}d\dot{\theta}[/tex]

    if we now integrate the left side between [tex]\theta = 0[/tex] and [tex]\theta = \theta_1[/tex] and the right side between [tex]\dot{\theta} = \dot{\theta} (\theta = 0)[/tex] and [tex]\dot{\theta} = \dot{\theta} (\theta = \theta_1)[/tex]

    we get then [tex]a_0\theta_1 = R\frac{\dot{\theta^2_1}}{2}[/tex]

    Replacing [tex]\dot{\theta_1}[/tex] with [tex]\frac{v_1}{R}[/tex] would give the equation that they ask in the problem

    But i think there's a mistake, because you assume that the tangential acceleration of the particle is a0 but there's also the tangential component of the weight. Is this correct?
     
  12. Jul 9, 2006 #11

    andrevdh

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    I get the same result as you do for the angle.

    The statement of the problem says that the string is pulled with an acceleration of a0. This means that P is adjusted during the motion so that the tangential acceleration of the particle will be a0, irrespective of other forces acting on the particle (which simplifies the problem immensely).

    Upon further investigation it seems that separation occurs immediately if

    [tex]a_0 > 0.5g[/tex]

    and at 90 degrees if

    [tex]a_0 = 0.318g[/tex]

    so for the range

    [tex]0.318g \leq a_0 \leq 0.5g[/tex]

    we do get a solution to problem (a). What is interesting to note is that the angle at which separation occurs is not dependent on the radius of the cylinder.
     
    Last edited: Jul 9, 2006
  13. Jul 9, 2006 #12
    Ok, i see what you mean. But is there any way to do the problem with this ecuation?

    [tex]m(-\rho \dot{\theta^2} \hat{\rho} + \rho \ddot{\theta} \hat{\theta}) = N \hat{\rho} + T \hat{\theta} - mg\sin(\theta)\hat{\rho} - mg\cos(\theta)\hat{\theta}[/tex]


    i mean, imposing the conditions in it.

    Thanks for all the help by the way.
     
  14. Jul 10, 2006 #13

    andrevdh

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    Yes, but it boils down to the same maths. It is easier to separate the problem into its components and treat them separately. The equation looks very impressive, but makes it hard to solve the problem.

    We need to incorporate our understanding of the physical situation into the maths in order to solve the problem. Solving mechanical problems requires one to have a keen insight into the involved physics. This is a prerequisite to solve the problem successfully. First make sure that you really understand the problem. Then use the maths to solve it. The maths can only give you back what you have put in, it is not a magic wand that you can wave at the problem, but a tool, that needs to be applied skillfully. If you do not impose the right conditions on the maths it will not give you the required results.

    Firstly the normal force dissappears at lift-off as you rightly remarked. Then the tension and cos component of the weight (theta components) can be replaced by +ma0 - As long as the string is wrapped around the cylinder, and the string is under tension (not slack), there will be a one to one motion of the end of the string (acceleration a0) and the particle. That is if the end of the string moves a distance delx in a time interval dt the particle will motion through a similar dels distance in the same time interval. The tangential acceleration of the particle are therefore a0 as implied by the statement of the problem. If you did not realize that you would not have been able to solve the problem correctly, because it is one of the conditions given by the problem. So you have to incorporate it into the maths.
     
    Last edited: Jul 10, 2006
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