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Homework Help: What is the drum's angular acceleration?

  1. Aug 1, 2005 #1
    A drum of radius R rolls down a slope without slipping. Its axis has acceleration 'a' parallel to the slope. What is the drum's angular acceleration? Please help me solve in polar coordinates.
  2. jcsd
  3. Aug 1, 2005 #2
    since the drum rolls without slipping:
    Angular Acceleration = α = a/R

    to understand this, what's the relationship between the linear motion of the drum's axis (moving down the ramp) to the rolling outer surface circumference? remember, there's no slipping.
    Last edited: Aug 1, 2005
  4. Aug 2, 2005 #3
    Linear speed [itex]v = \omega R[/itex]
    So if T is the time period, distance in one revolution = circumference
    [itex]vT = 2\pi R[/itex]

    Please note that the motion here is downhill(clockwise) so there will be changes in the sign if we take into account the direction. So how do I make these changes?
  5. Aug 2, 2005 #4
    remember that linear velocity (v), linear acceleration (a), angular velocity (ω), and angular acceleration (α) are all vector quantities. when solving a problem, one of the first jobs is to select convenient (orthogonal) coordinate axes into which these vector quantities can be projected into their coordinate components.

    for your problem, it's typical to consider the ramp with highest level on the left, lowest level on the right, with motion from "left-to-right". then it's also typical to establish a coordinate system consisting of:
    1) "x" axis parallel to ramp surface, (+) direction left-to-right
    2) "z" axis normal (perpendicular) to ramp surface, (+) direction upward from ramp surface
    3) "y" axis into page (normal to both above axes), (+) direction into page

    for your problem, the linear motion is "left-to-right" along the "x" axis and thus its x-component is (+). the angular motion is clockwise, which according to the "right-hand rule", produces a (+) "y" component.
    thus, for this problem, all quantities have (+) values for their components.

    with the above in mind, and working with the aformentioned components, you basically have solved the problem.
    beginning where you left off:

    If T is the period of revolution, distance in 1 revolution = 1 circumference:

    [tex]vT \ = \ 2\pi R[/tex]

    [tex]v \ = \ (2\pi /T) R[/tex]

    [tex]v \ = \ (\omega) R[/tex]

    [tex]\frac{dv}{dt} \ = \ a \ = \ \frac{d(\omega R)}{dt} \ = \ \frac{d\omega}{dt} \cdot R \ = \ \alpha R [/tex]
    Last edited: Aug 2, 2005
  6. Feb 13, 2010 #5
    EDIT: eh crap... made a mistake and your solution right I think
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