A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a radius of 10 cm

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A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a radius of 10 cm

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Lệ Thu 4 years 2021-08-17T14:44:14+00:00 1 Answers 26 views 0

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  1. giabao
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    2021-08-17T14:45:16+00:00
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    Answer:

    a_{r} = 1006.382g \,\frac{m}{s^{2}}

    Explanation:

    Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:

    a_{r} = \omega^{2}\cdot R

    Where:

    \omega – Angular speed, measured in radians per second.

    R – Radius of rotation, measured in meters.

    The angular speed is first determined:

    \omega = \frac{\pi}{30}\cdot \dot n

    Where \dot n is the angular speed, measured in revolutions per minute.

    If \dot n = 3000\,rpm, the angular speed measured in radians per second is:

    \omega = \frac{\pi}{30}\cdot (3000\,rpm)

    \omega \approx 314.159\,\frac{rad}{s}

    Now, if \omega = 314.159\,\frac{rad}{s} and R = 0.1\,m, the resultant acceleration is then:

    a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)

    a_{r} = 9869.588\,\frac{m}{s^{2}}

    If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:

    a_{r} = 1006.382g \,\frac{m}{s^{2}}

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