Question

A centrifuge rotor has a moment of inertia of 3.20 10-2 kg·m2. How much energy is required to bring it from rest to 500...

A centrifuge rotor has a moment of inertia of 3.20 10-2 kg·m2. How much energy is required to bring it from rest to 5000 rpm?

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Answer #1
Concepts and reason

The concept required to solve the given problem is work-energy theorem in rotational kinematics.

Initially, convert the angular speed of centrifuge from rpm to radians per second by using conversion factor. Then, use work-energy theorem to find the work done on the centrifuge.

Fundamentals

The rotational kinetic energy K of a rotating object is given by following expression.

K=12Iω2K = \frac{1}{2}I{\omega ^2}

Here, I is the moment of inertia, and ω\omega is the angular speed.

According to work-energy theorem in rotational kinematics, the work done W on a rotating object is equal to change in rotational kinetic energy of the object.

W=12Iωf212Iωi2W = \frac{1}{2}I\omega _{\rm{f}}^2 - \frac{1}{2}I\omega _{\rm{i}}^2

Here, I is the moment of inertia, ωf{\omega _{\rm{f}}}is the final angular speed, and ωi{\omega _{\rm{i}}}is the initial angular speed.

The final angular speed of the centrifuge is,

ωf=5000rpm{\omega _{\rm{f}}} = 5000\,{\rm{rpm}}

Convert the units for final angular speed from rpm to radians per second.

ωf=5000rpm(2πrad1rev)(1min60s)=523.59rad/s\begin{array}{c}\\{\omega _{\rm{f}}} = 5000\,{\rm{rpm}}\left( {\frac{{2\pi \,{\rm{rad}}}}{{1\,{\rm{rev}}}}} \right)\left( {\frac{{1\,{\rm{min}}}}{{60\,{\rm{s}}}}} \right)\\\\ = 523.59\,{\rm{rad/s}}\\\end{array}

According to work-energy theorem in rotational kinematics, the work done W on a rotating object is equal to change in rotational kinetic energy of the object.

W=12Iωf212Iωi2W = \frac{1}{2}I\omega _{\rm{f}}^2 - \frac{1}{2}I\omega _{\rm{i}}^2

Substitute 523.59 rad/s for ωf,{\omega _{\rm{f}}},0 rad/s for ωi,{\omega _{\rm{i}}},and 3.20×102kgm23.20 \times {10^{ - 2}}\,{\rm{kg}} \cdot {{\rm{m}}^2}for I in the above equation.

W=12(3.20×102kgm2)(523.59rad/s)212(3.20×102kgm2)(0)2=4386.3J\begin{array}{c}\\W = \frac{1}{2}\left( {3.20 \times {{10}^{ - 2}}\,{\rm{kg}} \cdot {{\rm{m}}^2}} \right){\left( {523.59\,{\rm{rad/s}}} \right)^2} - \frac{1}{2}\left( {3.20 \times {{10}^{ - 2}}\,{\rm{kg}} \cdot {{\rm{m}}^2}} \right){\left( 0 \right)^2}\\\\ = 4386.3\,{\rm{J}}\\\end{array}

Ans:

The work done is 4386.3 J.

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