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Linear vs. Angular Momentum

  1. Oct 25, 2012 #1
    In classical mechanics,

    p = mv
    L = Iω

    These correspond to linear and angular momentum, respectively. They're both called momentum, but...they don't have the same units. Why is that?? How can we call them both momentum when they don't seem to represent the same physical quality? Can we set up equations of momentum conservation when going from a system of linear motion to one of rotational motion, like a sticky clay ball thrown in the air onto the edge of a rotating disk? If so, how do we account for this difference in the definitions of linear and angular momentum?
  2. jcsd
  3. Oct 25, 2012 #2
  4. Oct 25, 2012 #3
    Thanks, Jimmy!
  5. Oct 26, 2012 #4
    p = mv

    L = rxp

    So of course they have different units. Angular momentum is basically linear momentum multiplied by distance

    If you take the time derivative of mv you get ma So the rate of change of momentum is equal to the net force acting. Internal forces cancel out by the equal and opposite rule so you're left with only external forces. If there are no external forces the rate of change of momentum is zero, so it does not change... hence conserved.

    For angular momentum if you take the time derivative of rxp you get rxF which is torque. By the same logic you're left with only external torques.

    So no force means no change in linear momentum. No torque means no change in angular momentum. The idea behind them is similar. One is just the rotational analogy of the other, but they're different.

    Oh and just in case you haven't seen angular momentum as rxp that's the definition. If something's moving in a circle then r and p are perpendicular and v can be expressed as ωr turning rxp into (if we only care about magnitude) rmωr = mr2w

    I = mr2 so Iω
  6. Oct 26, 2012 #5


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    And you have conservation of energy that also has to account for linear motion and rotational motion.

    Instead of your example, set a ball at the top of a ramp, figure out how fast it should accelerate according to the laws of motion (keeping in mind only a portion of the gravitational acceleration is in the direction of motion), and then test it out. The ball takes too long to reach the bottom and is moving too slow when it reaches the bottom. Some of the ball's potential energy was converted to rotational kinetic energy instead of linear kinetic energy.

    Energy doesn't seem quite so strange because at least you're working in the same units for both linear and rotational.

    Go back to just a linear example. What's the rate of change in kinetic energy? (which is 1/2mv^2)

    Now go back to a purely rotational example. What's the rate of change in rotational energy? (which is 1/2 Iw^2)
  7. Oct 27, 2012 #6


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    If a stationary object is in a collision with an object with angular momentum rxp
    And later, at rest , is in a collision with an object with linear momentum mv.
    And given that all 3 objects have same mass. And the linear speed and angular speed
    are equal: Linear speed = 2m/sec. and angular speed = 1 radian/sec , radius = 2m In which collision will the stationary object have greater acceleration ?
    Last edited: Oct 27, 2012
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