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Old Dec 6th 2016, 09:48 AM   #1
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Calculating centripetal acceleration

When a chainsaw is in operation, the chain moves with a linear speed of v=6.0m/s. At the end of the saw, the chain follows a semicircular path with a radius of r=0.049m.

I thought the formula for centripetal acceleration was v^2/r, but apparently that isn't it. Am I unable to use linear speed in the centripetal accel formula?
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Old Dec 6th 2016, 10:56 AM   #2
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You are correct - the magnitude of radial acceleration for a particle moving on a circular course at constant speed is v^2/r. So here the magnitude of acceleration is:

$\displaystyle | \vec a | = \frac {v^2} r =\frac {(6 \ \frac m s )^2}{0.049m}\ =\ 735 \frac m {s^2}$

The direction of the acceleration is inwards toward the center of rotation, so in vector form the acceleration is:

$\displaystyle \vec a = -735 \hat r \ \frac m {s^2}$

where $\displaystyle \hat r$ is the unit vector in the radial direction. Note the minus sign - could that be what you're missing?

Last edited by ChipB; Dec 6th 2016 at 11:04 AM.
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Old Dec 6th 2016, 12:07 PM   #3
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Physics does linear kinematics, then circular. Real design requires vector analysis of motion. Vectors are needed. If you know the derivatives of the sine and csine - these Items might make sense to you.

1.25 Vector Basis: Circular Motion | THERMO Spoken Here!
Altitude of Geostationary Orbits | THERMO Spoken Here!

Just a Thought... JP
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Old Dec 6th 2016, 12:55 PM   #4
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apparently the answer was 3.1*10^5 m/s^2 ... can anyone explain how to get to that answer?
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Old Dec 6th 2016, 02:14 PM   #5
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Originally Posted by ilovephysics View Post
apparently the answer was 3.1*10^5 m/s^2 ... can anyone explain how to get to that answer?
I can't explain it. But clearly that answer is wrong - an acceleration of that magnitude is equivalent to 30,000 g's! No chain saw blade could withstand that!
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