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Weight of the support of a pendulum with initial speed

  1. Mar 3, 2012 #1

    I'm interresting about the weight of the support of the pendulum with an initial speed. It is the http://en.wikipedia.org/wiki/Pendulum but with an initial speed. The mass come at top with straight initial speed and after turn around the pendulum. I attempt to have the weight of the support but I have more than the real weight. Example, if I put 10 N turn around with an initial speed of 10 m/s and a radius of 1 m I have 11.36N for the weight of the support. I don't find my error, if you can help me ?

    I have put my equations, and if you can find the problem I would be happy.

    Attached Files:

  2. jcsd
  3. Mar 3, 2012 #2
    Fpy=mgcosθ ? y-component of weight is mg.
  4. Mar 3, 2012 #3
    Ok, mg is y-component but it's the weight on the stem I need. You have a stem, mg is not full in the stem all the time. For example at 90° (horizontal position) 0 weight is added to the stem, the mg only accelerate the mass m. And more, if I put mg it's worst. It's the weight on the support I need to find. Imagine with a rope, how the mg could be added to the support when the mass is in horizontal position ?
    Last edited: Mar 3, 2012
  5. Mar 3, 2012 #4
    Thanks for the explanation. I am a just a learner in this topic and I don't understand why you need to integrate to calculate the force, My solution to this problem follows. I am I doing wrong?
    You have velocity [itex]\vec{v(θ)}[/itex], from this you can get acceleration [itex]\vec{a(θ)}[/itex]:


    Two forces acting on the mass [itex]\vec{T}[/itex] along the stem and weight [itex]\vec{W}[/itex] in -y direction. Newton's second law:


    You can find [itex]\vec{T}[/itex] from the above equation and by multiplying it by cosθ, you get ist vertical component. Of course calculation of [itex]\frac{\vec{dv}}{dt}[/itex] is a little bit tricky.
  6. Mar 3, 2012 #5
    I don't know your method, I'm trying.

    For you:

    T=[itex]( m*\frac{v}{r} * \frac{dv}{d\theta} - m*g ) * cos\theta[/itex] , is that ? Without integrate ?

    I need to integrate the forces because I calculate all forces at each point divised by speed, because if the speed is great the force is not here for a long time. After, I divided by the period.

    I have calculated for a pendulum without initial speed at start and from -90° to +90° and I found 10N. I think it's possible like that but I don't find my error in this case.

    Thanks for your help ;)
    Last edited: Mar 3, 2012
  7. Mar 3, 2012 #6
    First of all, there is an errors in your expression of v(θ). Replace m with R. Then

    As for my method, you should treat the terms as vectors:

    The results becomes:


    since [itex]v^{2}=v_{0}^{2}+2Rg(1-cosθ)[/itex]

    We end up with


    I don't know how to validate the formula!
    Last edited: Mar 3, 2012
  8. Mar 3, 2012 #7
    In fact, I would like to find where my solution is false. Because it works without initial speed. Someone has an idea ?
  9. Mar 3, 2012 #8
    Simpler that what we thought!

    Your errors or mistakes:

    1. [itex]v=\sqrt{v_{0}^{2}+2Rg(1-cosθ)}[/itex]

    2. [itex]F_{y}=mgcos^{2}θ[/itex] because mgcosθ is along the stem. You need to multiply it by cosθ once again.

    add it to Fcy and you are done!

    I don't think you need any integration.
  10. Mar 3, 2012 #9
    Right for the speed but with R=1 and m=1 this don't change the result.

    I don't understand what you said it's cos², could you explain ?
  11. Mar 3, 2012 #10
    The first multiplication by cosθ is as you said before. The component of weight force perpendicular to the stem only accelerates the mass tangentially. It exert no force on the support. The component along the stem is mgcosθ. This is transferred to the support with the same direction of the stem. But you need the vertical force on the support, you need to multiply it by cosθ then.
  12. Mar 3, 2012 #11
    That's ok like that ! thanks
    But the integration done 10.04 N not exactly 10 N I know it is numerical integration but I have put 1000000 of steps and 8 digits for numbers. Could you verify the integration and say if it's exactly 10 N ? I work with big precision for the numbers.
  13. Mar 3, 2012 #12
    Now I understand why you divide the force by velocity and integrate. You want to calculate the average force during one rotation (2[itex]\pi[/itex] radian). The result is expected to be equal to the weight of the mass.

    The error is not a numerical error. I got a very good precision even with 1000 divisions. The formula of v(θ) has another mistake. The correct one is:

  14. Mar 3, 2012 #13
    yes, all is fine thanks !
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