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Magnitude of acceleration of a pendulum

  1. Apr 25, 2013 #1
    1. The problem statement, all variables and given/known data

    A 2 kg mass hangs from the end of a 1 meter long string which is Øxed at its other end. The mass is displaced so that the string is at 53° to the vertical and released (remember sin(53°) = cos(37°)=0.8). Find the magnitude of the total acceleration of the mass when the string is at 37°
    to the vertical.

    2. Relevant equations

    T=2π√1/g), ω/2π=1/T, a=-ω2xmaxcos(ωt+phi)

    3. The attempt at a solution

    The period of the pendulum is ~2 seconds, so ω=π can plug that into a=-ω2xmaxcos(ωt+phi) and set phi=0.

    I'm not sure how to find xmax, or even what it represents in this case

    Alternatively I tried making a free body diagram and see the two forces are from gravity and the tension in the string
     
  2. jcsd
  3. Apr 25, 2013 #2

    cepheid

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    I like the second approach. This is a dynamical problem (a problem involving forces and their effects). You've correctly identified the two forces that act. Now, resolve the gravitational force into components parallel to and perpendicular to the string. Which of these components is balanced by something else, and which is not? Okay, so what is the net (unbalanced) force on the object, then? Okay, so what is the acceleration?
     
  4. Apr 25, 2013 #3
    T is parallel to mgcosθ so the net force on the pendulum is mgsinθ=15.6N

    using f=ma, a=7.83m/s2

    Is this correct? The answer key I have says the acceleration is 7.2m/s2
     
  5. Apr 25, 2013 #4

    cepheid

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    No, it's not correct. What you're calculating is the tangential component of the acceleration (tangent to the circle). There is also a radial or centripetal component to the acceleration (perpendicular to the tangential component, and pointing towards the centre of the circle). It is not 0. The reason is that that the tension is not equal to the component of the weight parallel to the string. (I forgot about this). It is *larger* than it. The difference is the centripetal force. You have to compute the centripetal force. To do this, you need to know the speed and the radius. The radius is given (1 m). The speed can easily be found from conservation of energy. You just need to find the difference in height of the mass between the initial position when it was at 53 degrees, and the final position at 37 degrees.

    The total acceleration will be the vector sum of the radial and tangential components.
     
  6. Apr 25, 2013 #5
    Δh = .153m, PE=3J

    using KE, v=1.732m/s

    so Fcp=6N

    so now I have to break this into its horizontal component: Fx=6sin37=3.61

    I JUST REALIZED I MISCLACULATED ABOVE mgsin(37)=11.8

    add this to the tangential force of 11.8N = 15.4N

    and I get 7.7m/s2

    I'm confused. The coordinate system I chose has Tension on the y axis. the force of gravity is in the 4th quadrant. this gives mgsin(37) along the x axis and mgcos(37) along the negative y axis. I need to vector add the Tension and mgsin(37) to get the sum force? Why is this force not tangent to the circle?
     
  7. Apr 25, 2013 #6

    cepheid

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    I get a different Δh. I think yours is wrong. How did you calculate it? Post your work, please.

    No, you don't have to "break it into its horizontal component." Why would you want to do that?

    You can't just "add them together." They are perpendicular to each other. You have to take their vector sum. I said that before, in my previous post.

    I actually have no idea what you are talking about, sorry.
     
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