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Centrifugal force, rotation

  1. Mar 23, 2008 #1
    I attached a string to the center point of one end of a 2 foot length of steel pipe, and then while holding the loose end of the string overhead I swung the pipe in a circle. When I released the string the pipe flew off in a straight line as I expected, but what I didn't expect was, the pipe instantly began to rotate in the same direction (counter clockwise) that I was spinning it in before I released the string.
    I had expected the pipe to fly off in a straight line with no rotation just as an arrow would fly through the air.
    Why did the pipe immediately begin to rotate when I released the string?
  2. jcsd
  3. Mar 23, 2008 #2


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    The leading/outer edge of the pipe has a higher velocity in that direction than the trailing edge. There is of course a smooth transition in velocity from the leading edge to the trailing edge (it reduces).
  4. Mar 23, 2008 #3
    The pipe already had angular momentum, equal to the rate that it was being spun at. It simply continued to retain this angular momentum once it was released. To stop this angular velocity, a countering torque force would be required.
  5. Mar 24, 2008 #4

    Doc Al

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    Staff: Mentor

    It was rotating when you were swinging it, so it just kept rotating. (Imagine someone viewing the circling pipe from above--they would clearly see the pipe rotating about its center in addition to rotating about you.)

    Note that the center of mass of the pipe will fly off in a straight line (subject to gravity, of course).
  6. Mar 25, 2008 #5

    So that's angular momentum. Well I'm tickled pink, that actually makes sense to me, and I can imagine that smooth transition in velocity as I visualize the pipe rotating through the air.

    Thank you both for your clear explanations :-)
  7. Mar 25, 2008 #6


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    I have a question .

    From mvr, we can see that while you are swinging the pipe, the pipe has an angular momentum of m * v * r, where r is the center of mass of the pipe to your hand. So, While it flies off, what is that r again in this case??
  8. Mar 25, 2008 #7

    Doc Al

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    If the pipe were a point mass (and thus not spinning about its axis) it would have an angular momentum about the person equal to [itex]\vec{r}\times m\vec{v} = mvr\sin\theta[/itex]. R is the position vector describing the location of the "pipe" with respect to the origin. Once the pipe flies off (ignoring gravity) that angular momentum doesn't change.

    The total angular momentum of the spinning pipe is the sum of (1) The angular momentum of its center of mass about the origin, and (2) the angular momentum of the pipe about its center of mass. (Your comment only addressed the first piece.)
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