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Homework Help: Minimum velocity at bottom of pendulum

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data

    A pendulum is made of a rigid rod (mass m, length l) and a small bob of mass M attached at one end. The rod is pivoted on the other end. What should be the minimum speed of the bob at its lowest point so that the pendulum completes a full circle?

    2. Relevant equations

    anything that'll work

    3. The attempt at a solution

    I thought of finding the energy of the system when the bob is at the highest position. It came as E=(M+m/4)v21/2 where v1 is the velocity of the bob at topmost position. Then to get the energy at the bottom position, and then equating the two, this is what I get:
    (2M+m)gl + (M+m/4)v21/2 = (M+m/4)v22/2 , where v2 is the velocity of the bob at the bottom position

    But then what?
  2. jcsd
  3. Apr 10, 2013 #2


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    At the lowest possible speed of the bob at the bottom of the swing that still allows full circle oscillation, the bob and the rod would have lost all kinetic energy at the top of the swing and therefore be at momentary rest. So there's no need to consider KE when bob is at the highest position - it's zero. All kinetic energy has been converted to gravitational PE at the top of the swing.
  4. Apr 10, 2013 #3
    So then E(top) = 2Mgl+mgl/2 and E(top)=E(bottom)=(M+m/4)v2/2

    Solving v=√4gl and this is one of the options in the answer.

  5. Apr 10, 2013 #4


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    Your PE equation looks wrong. Remember that the differences in height between the top and bottom are 2l for the bob M and l for the rod m (centre of mass of rod m has moved by height ½*l*2 = l). It's like going from the 6 o' clock to the 12 o' clock position.

    Better recheck your working. I couldn't get the mass terms to cancel out.
  6. Apr 10, 2013 #5
    :P right!! But then doing 1/2 * (M+m/4)* v^2 = 2Mgl + mgl gives v=√[8(2M+m)/(4M+m)] that's not an option! am I going wrong somewhere else?
  7. Apr 10, 2013 #6


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    I'm sorry, I think I was on the wrong track.

    The conservation equation should be (bottom) rotational KE of rod + translational KE of bob = (top) gravitational PE of rod + gravitational PE of bob.

    The rod's moment of inertia ##\displaystyle I## about its end (pivot) is given by ##\displaystyle \frac{1}{3}ml^2##. Its rotational KE is then ##\displaystyle \frac{1}{2}I\omega^2##, where ##\displaystyle \omega = \frac{v}{L}##, where ##\displaystyle v## is the speed of the bob.

    You still can't cancel out the masses, but I think this should be the right approach. The rod is not a point mass, and rotational kinetic energy cannot be ignored.

    Also, can you please list the choices you have?
  8. Apr 11, 2013 #7
    ok... will try that, meanwhile, these are my choices:

    1. sqrt [ g*l*(12M+6m)/(3M+m) ]
    2. sqrt [ 4*g*l ]
    3. sqrt [ 5*g*l ]
    4. sqrt [ g*l*(15M+6m)/(3M+m)
  9. Apr 11, 2013 #8


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    Yes, the answer is an option on that list. It's either 1 or 4 because I've told you we can't cancel out the masses, now you just have to determine which it is.
  10. Apr 11, 2013 #9
    doing as you said,
    1/2 v^2(M+m/6) = Mg*2l + mg*l ....... is the equation which gives v= sqrt [ g*l*(12M+6m)/(3M+m/2) ] which is close to option 1, but with (3M + m/2) in the denominator instead of (3M+ m).

    Is there still something wrong?
  11. Apr 11, 2013 #10


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    Shouldn't the rotational kinetic energy of the rod be ##\displaystyle \frac{1}{2}I\omega^2 = (\frac{1}{2})(\frac{1}{3})ml^2 \omega^2 = \frac{1}{6}mv^2##?

    So total KE of bob and rod = ##\displaystyle \frac{1}{2}Mv^2 + \frac{1}{6}mv^2 = \frac{1}{6}(3M + m)v^2##.
  12. Apr 11, 2013 #11
    Yeah, that's what I did, but made a simple calculation error!! Option 1 is coming to be correct. Thanks for your time and help.
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