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Conservation of energy in a pendulum

  1. Oct 13, 2008 #1
    A pendulum consists of a 2.0kg bob attached to a light 3.0m long string. While hanging at rest with the string vertical, the bob is struck a sharp horizontal blow, giving it a horizontal velocity of 4.5m/s. At the instant the spring makes an angle of 30degrees with the vertical.

    What is:
    a) the speed
    b)gravitational potential energy(relative to its value is at the the lowest point)
    c)the tension on the string
    d)what is the angle of the string with the vertical when the bob reaches its greatest height

    The work from the tension of the string is 0 because it is perpendicular to the displacement of the bob.
    Ki = 0 since it starts out at rest

    The conservation of energy is:
    Ui + Ki = Uf + Kf
    mgh + 0 = 0 1/2 m Vf^2
    vf = root of 2gh

    Is this correct approach to answer part A?

    For Part B
    b)gravitational potential energy(relative to its value is at the the lowest point)
    Im not sure what to do. Is this the potential energy of the bob?

    c)the tension on the string
    Isn't the tension just MG?

    d)
    d)what is the angle of the string with the vertical when the bob reaches its greatest height
    Any hints on how to start this one?
     
  2. jcsd
  3. Oct 13, 2008 #2

    Doc Al

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    The initial KE is not zero. The bob starts with a speed of 4.5m/s.

    Yes. Find the height of the bob.

    No. Analyze the forces acting on the bob at the moment in question.

    What's the final KE in this case?
     
  4. Oct 13, 2008 #3
    So it is mgh + 1/2 m*(4.5m/s)^2 = 0 + 1/2 m * Vf^2
    2gh + (4.5m/s)^2 = Vf^2

    The height wouldn't be 3m correct?
    Would the height be 3m - 3m * cos 30?

    Forces acting on the Bob is
    T - mg = m a
    Since it is a pendulum, the acceleration of the bob is centripetal correct?
    a = V^2/R
    where R = the length of the string

    D)
    The final KE would be 0 since it will stop moving? Im not too sure on w hat you mean.
     
  5. Oct 14, 2008 #4

    Doc Al

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    It starts out at the bottom (h = 0) and ends up at the height h, so:
    0 + 1/2 m*(4.5m/s)^2 = mgh + 1/2 m * Vf^2

    Right!

    The forces on the bob are tension (acting radially) and gravity (acting down).
    There are two components to the acceleration: The centripetal (radial) component, which you just described, and a tangential component. To find the tangential component, find the force component in the tangential direction.

    Right.
     
  6. Oct 14, 2008 #5
    What is:
    a) the speed
    0 + 1/2 m*(4.5m/s)^2 = mgh + 1/2 m * Vf^2
    where h = 3m - 3m*cos 30
    I get Vf = 3.5m/s

    b)gravitational potential energy(relative to its value is at the the lowest point)
    MGH = 2kg * 9.81m/s * 3m-3m* cos 30
    = 7.9J



    c)the tension on the string
    So this equation is correct? T - mg = m a

    I have to find the force that made it go horizontal with the speed of 4.5m/s and using that find the horizontal component of it?

    d)what is the angle of the string with the vertical when the bob reaches its greatest height
    How would I find the angle?
    The final KE is = 0
    The greatest height would be final potential energy correct? which is mgh.
     
  7. Oct 14, 2008 #6

    Doc Al

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    Good.


    No, since tension and mg act in different directions. To find the tension, analyze the radial components of the forces on the bob and apply Newton's 2nd law in the radial direction. Hint: What's the radial component of the weight?


    No. You only have to worry about the forces acting on the bob when it's at 30 degrees.

    Good. Find the height, then use a bit of trig to find the angle.
     
  8. Oct 14, 2008 #7
    Im not too familiar with the radial components and tangential components...,
    Im not sure how to calculate them either,
    the weight of the bob is mg...


    For Part D:
    finding the height since the KE = 0 and Uf = mgh and Ui = 0
    Ki = Uf
    1/2 m vi ^2 = mgh
    and solve for h?

    The h would be the length for the opposite side of the triangle
    The hyp would be the length of the string?
    Using inverse Sin to find the angle?
     
  9. Oct 14, 2008 #8

    Doc Al

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    Radial just means "along the radius", which in this case means parallel to the string. What's the component of the weight parallel to the string?

    Yes.

    You know how to find h given the angle, just reverse the process.
     
  10. Oct 14, 2008 #9
    oooh. So it is mg cos 30?
     
  11. Oct 14, 2008 #10

    Doc Al

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    Yes! But what sign should it have?

    I suggest making the tension positive--towards the center of the arc = positive, away = negative. (So what would that make the sign of the centripetal acceleration?)
     
  12. Oct 14, 2008 #11
    The centripetal acc. will be negative?
    Im still little confuse here.
    So is it T - mg cos 30 = m a where a is the centripetal acc.?
     
  13. Oct 14, 2008 #12

    Doc Al

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    Staff: Mentor

    Which way does the centripetal acceleration point?
    Yep.
     
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