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.

Answer 1

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

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