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a= -w.w.y

where a is the centripetal acceleration , w is the angular velocity and y is the displacement[?]

- Thread starter chense
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- #1

- 5

- 0

a= -w.w.y

where a is the centripetal acceleration , w is the angular velocity and y is the displacement[?]

- #2

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'-' sign is there because here 'a' is component of acceleration in the radial direction which is normally taken

cheers

- #3

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Chense,Originally posted by teddy

'-' sign is there because here 'a' is component of acceleration in the radial direction which is normally takenpositive outwards.As the centripetal acceleration istowardsthe center its component along radial direction is negative,hence the negative sign ('w' and 'y' are positive in the above equation,'w' being themagnitudeof angular velocity and 'y' being themagnitudeof distance from the axis of rotation)

cheers

Can you just clarify what you do mean here - are you talking about circular motion, in which case teddy is right. But normally one would use 'r' for the radial distance - while you used 'y' and called it 'displacement'. Which makes me think you might mean 'amplitude' of an oscillation.

Cheers,

Ron.

- #4

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Not exactly the motion of an oscillator.

- #5

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My point exactly - which is why I'd like chense to clarify these terms.Originally posted by plus

Not exactly the motion of an oscillator.

Cheers,

ron.

- #6

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- #7

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Got you now!Originally posted by chense

Think of the particle moving in a circular track of radius r at constant angular velocity w. Now imagine looking at the track from the side - the particle (or, more correctly, its projection) will oscillate up and down. Let's say that the instantaneous distance of the particle above or below the central axis is y. The

The equation for y is:

y = r sin w.t

Differentiating twice gets you the acceleration:

a = - r.w.w sin w.t = - w.w.y

Mathematically, that's where the minus comes from. As 'plus' mentioned, its physical significance is that as the particle gets further from y = 0, then it DEcelerates (because of the minus) and comes back again.

Cheers,

Ron.

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