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Tension in a Pendulum Problem - 30kg, 4m cord, 4m/s @ bottom

  1. Jun 17, 2006 #1
    The ball has a mass of 30 kg and a speed of [itex]4\,\frac{m}{s}[/itex] at the instant it is at its lowest point, [itex]\theta\,=\,0[/itex]. Determine the tension in the cord and the rate at which the ball's speed is decreasing at the instant [itex]\theta\,=\,20[/itex]. Neglect the size of the ball. NOTE: The cord length is 4 m.

    Here is what I have found so far, the acceleration:

    [tex]\sum\,F_t\,=\,-W\,sin\,\theta\,=\,m\,a_t[/tex]

    [tex]-m\,g\,sin\,\theta\,=\,m\,a_t[/tex]

    [tex]a_t\,=\,-g\,sin\,\theta[/tex]

    [tex]a_t\,=\,-\left(-9.81\,\frac{m}{s^2}\right)\,sin\,(20)[/tex]

    [tex]a_t\,=\,3.36\,\frac{m}{s^2}[/tex]

    I cannot get the right answer for the tension in the cord though. Please help!
     
    Last edited: Jun 18, 2006
  2. jcsd
  3. Jun 17, 2006 #2

    Doc Al

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    What did you get for the tension in the cord? (Show how you solved it.)
     
  4. Jun 17, 2006 #3
    [tex]\sum\,F_{\theta}\,=\,T\,-\,W\,=\,m\,a[/tex]

    [tex]T\,=\,m\,a\,+\,W[/tex]

    [tex]T\,=\,(30\,kg)\,\left(3.36\,\frac{m}{s^2}\right)+\,(30\,kg)\,\left(9.81\,\frac{m}{s^2}\right)[/tex]

    [tex]T\,=\,395\,N[/tex]

    The answer in the back of the book is [itex]T\,=\,361\,N[/itex] though.
     
    Last edited: Jun 18, 2006
  5. Jun 18, 2006 #4

    Hootenanny

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    HINT: Think centripetal force and energy.
     
  6. Jun 18, 2006 #5
    I thought that was what I was doing?

    please help, I have no idea what it is that I am doing wrong.
     
  7. Jun 18, 2006 #6

    Hootenanny

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    What you have there is the tangental acceleration, which is not the same an the centripetal acceleration. The centripetal acceleration is given by;

    [tex]a_{c} = \frac{v^2}{r}[/tex]

    You can calculate the velocity using conservation of energy. You will need to use the velocity at the bottom of the swing and the change in height. Note, that at any point in the swing the resultant force must equal the centripetal force;

    [tex]\sum\vec{F} = m\frac{v^2}{r} = T - mg\cos\theta[/tex]

    Can you go from here?
     
  8. Jun 18, 2006 #7
    Here is how I solved for the velocity at the instant:

    [tex]s\,=\,\theta\,r[/tex]

    [tex]ds\,=\,d\theta\,r[/tex]

    [tex]-g\,sin\,\theta\,=\,a_t\,=\,v\,\frac{dv}{ds}\,=\,v\,\frac{dv}{d\theta\.r}[/tex]

    [tex]-g\,r\,sin\,\theta\,d\theta\,=\,v\,dv[/tex]

    [tex]-g\,r\,\int_{0}^{\theta}\,sin\,\theta\,d\theta\,=\,\int_0^v\,v\,dv[/tex]

    [tex]v^2\,=\,2\,g\,r\,\left(cos\,\theta\,-\,1\right)[/tex]

    [tex]v\,=\,\sqrt{2\,\left(-9.81\,\frac{m}{s^2}\right)\,(4\,m)\,\left[cos\,(20)\,-\,1\right]}\,=\,2.18\,\frac{m}{s}[/tex]

    Then I used the above result for this equation below:

    [tex]T\,=\,(30\,kg)\frac{\left(2.18\,\frac{m}{s}\right)^2}{4\,m}\,+\,(30\,kg)\,\left(9.81\,\frac{m}{s^2}\right)\,\cos\,(20)[/tex]

    [tex]t\,=\,35.64\,+\,276.6\,=\,312.3\,N[/tex]

    This answer is still not correct though!
     
    Last edited: Jun 18, 2006
  9. Jun 18, 2006 #8

    Hootenanny

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    Perhaps conservation of energy, would be a easier approach than calculus? I'll start;

    [tex]\frac{1}{2}mv_{i}^{2} = \frac{1}{2}mv_{f}^{2} + mgh[/tex]

    [tex]v_{i}^{2} = v_{f}^{2} + 2gh[/tex]

    Now, using trig we find that h is given by;

    [tex]h = 4 - 4\cos\theta[/tex]

    If you can't see this, try drawing a diagram where the pendulum is displaced from the equilibrium position. Thus, we obtain;

    [tex]v_{i}^{2} = v_{f}^{2} + 2g( 4 - 4\cos\theta)[/tex]

    You need to solve for vf. Can you go from here?
     
    Last edited: Jun 18, 2006
  10. Jun 18, 2006 #9
    What is the definition of h?

    I used your [itex]v_f[/itex] in the above T equation and it is still an incorrect answer. Is the T equation right?
     
  11. Jun 18, 2006 #10

    Hootenanny

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    h is change in vertical height. Yes, your tension equation is correct. I think you have made an arithmatic error somewhere, as I obtained 361N using the above method which agrees with the answer given. I'll take the next step for you;

    [tex]T = m\frac{v_{i}^{2} - 2g(r - r\cos\theta)}{r} + mg\cos\theta[/tex]

    Just sub your number directly in. Can you go from here?
     
  12. Jun 18, 2006 #11
    I found something about pendulums and the definition of h: http://en.wikipedia.org/wiki/Pendulum

    [tex]T = (30 kg)\frac{\left(4 \frac{m}{s}\right)^{2} - 2\left(9.81 \frac{m}{s^2}\right)\left[(4 m) - (4 m)\cos (20)\right]}{4 m} + (30 kg)\left(9.81 \frac{m}{s^2}\right)\cos (20)[/tex]

    [tex]T = (30 kg) \frac{11.26 \frac{m^2}{s^2}}{4 m} + 276.6 N[/tex]

    [tex]T = 84.50 N + 276.6 N[/tex]

    [tex]T = 361 N[/tex]

    Thats cool! But how do I solve it using the other method that was mentioned?
     
    Last edited: Jun 18, 2006
  13. Jun 18, 2006 #12

    Hootenanny

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    I will say again check your calculations, you are using the correct the formula and are plugging in the correct numbers; you are simply making an arithmetic error. Try doing the calculation is stages.
     
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