Chapter 8, Problem 60A

To determine

The required angular velocity in revolutions per minute.

Answer to Problem 60A

  $1.11\times {10}^5\text{}\text{rev/min}$

Explanation of Solution

Given:

Centripetal acceleration of a laboratory centrifuge is $a_c=0.35\times {10}^{6}g$ .

Distance of the point of centripetal acceleration from the axis, $r=2.50\text{cm = 0}\text{.025}\text{m}$ .

Formula used:

The centripetal acceleration is:

  $a_c=\frac{v^2}{r}$

Where, v is the linear velocity, and r is the radius of rotation.

Also, for a rotating and rigid object, its linear velocity( v ) can be related to its angular velocity $\left(\omega \right)$ as,

  $v=r\omega$

Where r is the radius of the object.

Calculation:

Now, the centripetal acceleration can be written in terms of angular velocity as,

  $\begin{array}{c}{a}_c=\frac{v^2}{r}\\ =\frac{{\omega }^2{r}^2}{r}\\ =r{\omega }^2\end{array}$

Using given values in above, the angular velocity will be,

  $\begin{array}{c}\omega =\sqrt{\frac{a_c}{r}}\\ =\sqrt{\frac{0.35\times {10}^{6}g}{0.025}}\\ =\sqrt{\frac{0.35\times {10}^6\times 9.8}{0.025}}\\ =11713.24\text{rad/s}\times \frac{1\text{}\text{rev}}{2\pi \text{rad}}\times \frac{60\text{}\text{s}}{1\text{min}}\\ =1.11\times {10}^5\text{}\text{rev/min}\end{array}$

Conclusion:

Thus, the required angular velocity is $1.11\times {10}^5\text{}\text{rev/min}$ .