Chapter 8, Problem 43P

To determine

To find: The energy that is required to being it from rest.

Answer to Problem 43P

The value of the energy is $1.4\times {10}^4\text{J}$ .

Explanation of Solution

Given information:

A centrifugal rotor has a moment of inertia of $3.75\times {10}^{-2}\text{kg}\cdot {\text{m}}^{\text{2}}$ . The energy required to bring it from rest to 8250 rpm.

Formula used:

  $\omega =2\pi f$

Here,

  $\omega$ is the angular velocity,

  $f$ is the frequency.

Calculation:

Moment of inertial of centrifuge $I=3.75\times {10}^{-2}\text{kg}\cdot {\text{m}}^{\text{2}}$

Frequency of rotation $f=8250\text{rpm}$

  $\begin{array}{l}=\frac{\text{8250}\text{rpm}}{\text{60}\text{s/m}}\\ =137.5\text{rev/s}\end{array}$

Angular velocity $\omega =2\pi f$

  $\begin{array}{l}=2\pi \times 137.5\\ =864\text{rad/s}\end{array}$

The Rotational Kinetic Energy is $\frac{1}{2}I{\omega }^2$

  $\begin{array}{l}K.E=\frac{1}{2}I{\omega }^2\\ =\frac{1}{2}\times 3.75\times {10}^{-2}\times {864}^2\\ =1.4\times {10}^4\text{J}\end{array}$

  $\text{Energy}=1.4\times {10}^4\text{J}$

Conclusion:

The value is $\text{Energy}=1.4\times {10}^4\text{J}$ .