At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setting it takes

Question

At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setting it takes t = 9.5 s to uniformly accelerate to its final angular speed ω2 = 750 rad/s.

A) Calculate the angular acceleration of the centrifuge α1 in rad/s2 over the time interval t.

B) Calculate the total angular displacement (in radians) of the centrifuge, Δθ, as it accelerates from the initial to the final speed.

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Ngọc Khuê 5 months 2021-08-31T06:49:34+00:00 1 Answers 12 views 0

Answers ( )

  1. Helga
    0
    2021-08-31T06:51:08+00:00
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    Answer:

    Part(a): The angular acceleration is 5.63~rad~s^{-2}.

    Part(b): The angular displacement is 2629~rad.

    Explanation:

    Part(a):

    If \omega_{1},~\omega_{2}~and~\alpha be the initial angular speed, final angular speed and angular acceleration  of the centrifuge respectively, then from rotational kinematic equation, we can write

    \alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)

    where ‘t‘ is the time taken by the centrifuge to increase its angular speed.

    Given, \omega_{i} = 250~rad~s^{-1}, \omega_{f} = 750~rad~s^{-1} and t = 9.5~s. From equation (I), the angular acceleration is given by

    \alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}

    Part(b):

    Also the angular displacement (\Delta \theta) can be written as

    &&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad

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