Question

A centrifuge in a medical laboratory rotates at an angular speed of 3600 rev/min. When switched off, it rotates through 52.0 revolutions before coming to rest. Find the constant angular acceleration of the centrifuge.

Answer 1

`-217.508\ \text{rad/s}^2`

Step-by-step explanation:

`\omega_i` = Initial angular velocity =
`3600*(2\pi)/(60)=377\ \text{rad/s}`

`\theta` = Angular displacement =
`52\ \text{rev}=52* 2\pi=326.72\ \text{rad}`

`\omega_f` = Final angular velocity = 0

`\alpha` = Angular acceleration

From the kinematic equations of angular motion we have

`\omega_f^2-\omega_i^2=2\alpha\theta\\\Rightarrow \alpha=(\omega_f^2-\omega_i^2)/(2\theta)\\\Rightarrow \alpha=(0-377^2)/(2* 326.72)\\\Rightarrow \alpha=-217.508\ \text{rad/s}^2`

The constant angular acceleration of the centrifuge is
`-217.508\ \text{rad/s}^2`.