/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 A centrifuge has a rotational in... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 10

A centrifuge has a rotational inertia of $6.5 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}$ How much energy must be supplied to bring it from rest to 420 rad/s \((4000 \text { rpm }) ?\)

Short Answer

Expert verified
\) Answer: The energy required to bring the centrifuge from rest to an angular speed of \(420\,\mathrm{rad/s}\) is approximately \(572.94\,\mathrm{J}.\)

Step by step solution

01

Understand the important parameters and energy conservation principle

The exercise provides us with the rotational inertia (I) of the centrifuge, \(6.5\times10^{-3}\,\mathrm{kg}\cdot\mathrm{m}^2,\) and the final angular speed (蠅), \(420\,\mathrm{rad/s}.\) The centrifuge is initially at rest, which means its initial angular speed (\(\omega_0\)) is \(0\,\mathrm{rad/s}.\) Using the conservation of energy, we can calculate the energy needed to bring the centrifuge from rest to the given angular speed.
02

Calculate the initial and final rotational energies

First, we will calculate the initial and final rotational energies of the centrifuge using the formula for rotational kinetic energy: Rotational Kinetic Energy, \(K=\frac{1}{2}I\omega^2.\) Initial Rotational Energy (\(K_0\)) = \(\frac{1}{2}I\omega_0^2 = \frac{1}{2}\times(6.5\times10^{-3}\,\mathrm{kg}\cdot\mathrm{m}^2)\times0^2=0\,\mathrm{J}\) Final Rotational Energy (\(K_f\)) = \(\frac{1}{2}I\omega^2 = \frac{1}{2}\times(6.5\times10^{-3}\,\mathrm{kg}\cdot\mathrm{m}^2)\times(420\,\mathrm{rad/s})^2\)
03

Calculate the energy that must be supplied to the centrifuge

Now we will calculate the required energy to bring the centrifuge from rest to the given angular speed using the conservation of energy principle: Energy supplied (\(E\)) = Final Rotational Energy (\(K_f\)) - Initial Rotational Energy (\(K_0\)) \(E= K_f - K_0\) Substitute the values of \(K_0\) and \(K_f\) into the equation and compute the energy supplied: \(E = \frac{1}{2}\times(6.5\times10^{-3}\,\mathrm{kg}\cdot\mathrm{m}^2)\times(420\,\mathrm{rad/s})^2 - 0\,\mathrm{J}\) \(E \approx 572.94\,\mathrm{J}\) So, the energy that must be supplied to bring the centrifuge from rest to 420 rad/s is approximately \(572.94\,\mathrm{J}.\)

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Most popular questions from this chapter

A diver can change his rotational inertia by drawing his arms and legs close to his body in the tuck position. After he leaves the diving board (with some unknown angular velocity), he pulls himself into a ball as closely as possible and makes 2.00 complete rotations in \(1.33 \mathrm{s}\). If his rotational inertia decreases by a factor of 3.00 when he goes from the straight to the tuck position, what was his angular velocity when he left the diving board?
A child of mass \(40.0 \mathrm{kg}\) is sitting on a horizontal seesaw at a distance of \(2.0 \mathrm{m}\) from the supporting axis. What is the magnitude of the torque about the axis due to the weight of the child?
A skater is initially spinning at a rate of 10.0 rad/s with a rotational inertia of \(2.50 \mathrm{kg} \cdot \mathrm{m}^{2}\) when her arms are extended. What is her angular velocity after she pulls her arms in and reduces her rotational inertia to \(1.60 \mathrm{kg} \cdot \mathrm{m}^{2} ?\)
A spoked wheel with a radius of \(40.0 \mathrm{cm}\) and a mass of $2.00 \mathrm{kg}$ is mounted horizontally on friction less bearings. JiaJun puts his \(0.500-\mathrm{kg}\) guinea pig on the outer edge of the wheel. The guinea pig begins to run along the edge of the wheel with a speed of $20.0 \mathrm{cm} / \mathrm{s}$ with respect to the ground. What is the angular velocity of the wheel? Assume the spokes of the wheel have negligible mass.
A uniform diving board, of length \(5.0 \mathrm{m}\) and mass \(55 \mathrm{kg}\) is supported at two points; one support is located \(3.4 \mathrm{m}\) from the end of the board and the second is at \(4.6 \mathrm{m}\) from the end (see Fig. 8.19 ). What are the forces acting on the board due to the two supports when a diver of mass \(65 \mathrm{kg}\) stands at the end of the board over the water? Assume that these forces are vertical. (W) tutorial: plank) [Hint: In this problem, consider using two different torque equations about different rotation axes. This may help you determine the directions of the two forces.]
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