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Math Help - Physics- centripetal and tangential acceleration

  1. #1
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    Unhappy Physics- centripetal and tangential acceleration

    You are designing a centrifuge to spin at a rate of 15,650 rev/min.
    (a) Calculate the maximum centripetal acceleration that a test-tube sample held in the centrifuge arm 15.7 cm from the rotation axis must withstand.

    (b) It takes 1 min, 17 s for the centrifuge to spin up to its maximum rate of revolution from rest. Calculate the magnitude of the tangential acceleration of the centrifuge while it is spinning up, assuming that the tangential acceleration is constant.


    So this is what I did...

    (a)
    15650 rev/min * 1min/60 sec = 260.833 rev/sec

    With r = 15.7 cm, c=2(3.14)r = 98.646 cm

    Covering that distance in 1 sec gives a velocity of 0.98646 m/s.

    Then a=(v^2)/r = (0.98646^2)/0.157=6.19811 m/s^2

    However, it said that my answer is wrong.

    (b)
    Total t=77 sec so i thought that youíre going from 0 to 6.19811 over 77 seconds, so itís (6.19811/77 m/s) per second. That gives you 0.080495 m/s^2 of acceleration.

    Unfortunately, this answer is also wrong. Can someone figure out where I am messing up?
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  2. #2
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    Quote Originally Posted by horan6 View Post
    You are designing a centrifuge to spin at a rate of 15,650 rev/min.
    (a) Calculate the maximum centripetal acceleration that a test-tube sample held in the centrifuge arm 15.7 cm from the rotation axis must withstand.

    (b) It takes 1 min, 17 s for the centrifuge to spin up to its maximum rate of revolution from rest. Calculate the magnitude of the tangential acceleration of the centrifuge while it is spinning up, assuming that the tangential acceleration is constant.
    (a) you made the incorrect assumption that the linear speed is a constant.

    a_c = r\omega^2

    a_c = (.157 \, m)\left(15650 \, \frac{rev}{min} \cdot \frac{1 \, min}{60 \, sec} \cdot \frac{2\pi \, rad}{1 \, rev}\right)^2 = 4.22 \times 10^5  \, \frac{m}{sec^2}

    (b)  a_T = r\alpha

    \alpha = \frac{\Delta \omega}{\Delta t} = \frac{\frac{1565\pi}{3} \, \frac{rad}{sec}}{77 \, sec}

    \alpha = \frac{1565\pi}{231} \, \frac{rad}{sec^2}

    a_T = (.157 \, m)\left(\frac{1565\pi}{231} \, \frac{rad}{sec^2}\right) = 3.34 \, \frac{m}{sec^2}
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