Question

(I) A centrifuge rotor has a moment of inertia of\({\bf{3}}{\bf{.25}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{2}}}}{\bf{ kg}} \cdot {{\bf{m}}^{\bf{2}}}\).How much energy is required to bring it from rest to 8750 rpm?

Short answer

The energy of the centrifuge rotor is \(13643.53{\bf{ }}{\rm{J}}\).

Step-by-step solution

Identification of given data 

The given data can be listed below as:

• The moment of inertia of the centrifuge is\(3.25 \times {10^{ - 2}}{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).
• The rotation of the rotor is\(N = 8750{\rm{ rpm}}\).
• The initial rotation of the rotor is \({N_0} = 0{\rm{ rpm}}\).

Understanding the energy of a centrifuge rotor

The energy of a centrifuge rotor is stored in the form of the change in the rotational kinetic energy.

The rotational kinetic energy is dependent on the moment of inertia of the centrifuge and the rotation of the rotor.The energy unit of a centrifuge is expressed in terms of Joules.

Determination of the energy of the rotor

The change in the rotational kinetic energy of the rotor can be expressed as:

\(\begin{align}K &= {K_f} - {K_i}\\K &= \frac{1}{2}I{\omega ^2} - \frac{1}{2}I\omega _0^2\\K &= \frac{1}{2}I \times {\left( {\frac{{2\pi N}}{{60}}} \right)^2} - \frac{1}{2}I \times {\left( {\frac{{2\pi {N_0}}}{{60}}} \right)^2}\\K &= \frac{1}{2}I \times {\left( {\frac{{2\pi N}}{{60}}} \right)^2} - \frac{1}{2}I \times {\left( {\frac{{2\pi \times 0{\rm{ rpm}}}}{{60}}} \right)^2}\end{align}\)

Here,\({K_f}\)is the final rotational kinetic energy and\({K_i}\)is the initial rotational kinetic energy.\({\omega _0}\)is the initial angular speed of the rotor,\(\omega \)is the angular speed of the rotor, and\({N_0}\)is the initial rotation of the rotor whose value is 0 rpm.

This can be further solved as:

\(K = \frac{1}{2}I \times {\left( {\frac{{2\pi N}}{{60}}} \right)^2}\)

Here, K is the energy of the rotor.

Substitute the values in the above equation.

\(\begin{align}K &= \frac{1}{2} \times 3.25 \times {10^{ - 2}}{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}} \times {\left( {\frac{{2\pi \times 8750{\rm{ rpm}}}}{{60}}} \right)^2}{\left( {\frac{{1{\rm{ rps}}}}{{1{\rm{ rpm}}}}} \right)^2}\\K &= 13643.53{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ J}}}}{{1{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}}}} \right)\\ &= 13643.53{\bf{ }}{\rm{J}}\end{align}\)

Thus, the energy of the rotor is \(13643.53{\bf{ }}{\rm{J}}\).