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**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.

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since the drum rolls without slipping:Reshma said:downa 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.

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.

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Linear speed [itex]v = \omega R[/itex]geosonel said: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.

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

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remember that linear velocity (Reshma said: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 isdownhill(clockwise) so there will be changes in the sign if we take into account the direction. So how do I make these changes?

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]

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EDIT: eh crap... made a mistake and your solution right I think

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