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Homework Help: Question on rotational motion

  1. Oct 10, 2009 #1
    Imagine a trebuchet with a stiff rod of 3m and neligible mass.Two masses 60kg and 0.12kg are at its end with the bigger mass 0.14m away from the pivot point. Find the maximum speed the smaller mass attain.

    Alright the question is somewhat like this and i start of by finding the moment of inertia and the torque which gives me the angular acceleration but how do i continue from here?
  2. jcsd
  3. Oct 10, 2009 #2


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    Hi semc! :smile:

    They're point masses, so I wouldn't bother with moment of inertia …

    just use (ordinary) PE + KE = constant, to find the speeds when the heavy mass is at the bottom. :wink:
  4. Oct 10, 2009 #3
    Hi tim i don't understand why the two masses can be treated as point mass since the question doesn't specify that the masses have uniform density distribution. So if we are asked to find the velocity of the mass at a given specific time we should calculate the angular acceleration and from there find the velocity or is there a easier method?

    I have another question: I learn that force cross d ,where d is the perpendicular distance from the point the force is acting to the pivot point, gives the torque cause by this force correct? So what does the direction of the torque tell us? Does it merely tell us which way the object is rotating or does it gives us information like the direction of the force? Ain't torque the tendency to rotate an object by a force?
  5. Oct 10, 2009 #4
    The forces can be treated as point masses because we have to :smile:

    It's indeed true that we don't know what the shape or density distribution of the masses is.
    If the bigger mass would be an iron sphere it would have a radius of about 0.12 m so it certainly matters. The smaller mass can be treated as a point mass with very high accuracy.

    Even if you assume that [tex]\frac{1}{m} \int r^2 dm [/tex] of the larger mass is 0.14
    you still wouldn't know what its centre of mass is, which you need for calculating it's gravitational potential energy.

    In any case, it's much easier to only consider potential and kinetic energy and no forces or torques.
  6. Oct 10, 2009 #5


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    Hi semc! Hi willem2! :smile:
    The direction of the torque is the axis of rotation, nothing more.
    As willem2 :smile: says, there's no choice … we're not told the shape or size, so we have to assume they're point masses!
    Almost correct. :smile: it's the perpendicular distance from the line of the force to the pivot point. :wink:

    Since the only force is gravity, and the line of that force is always vertical, that means the perpendicular distance in this case keeps changing, so the torque keeps changing …

    so if you use torque, you'll have a very messy unnecessary integral. :frown:

    As willem2 says, it's much easier in this case to only consider PE and KE and no forces or torques (or angular anything). :rolleyes:

    Try it! :smile:
  7. Oct 10, 2009 #6
    I understand now. Thanks for he help guys!!:biggrin:
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