/*! 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 62 A student sits on a rotating sto... [FREE SOLUTION] | 91Ó°ÊÓ

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A student sits on a rotating stool holding two \(3.0-\mathrm{kg}\) objects. When his arms are extended horizontally, the objects are \(1.0 \mathrm{~m}\) from the axis of rotation and he rotates with an angular speed of \(0.75 \mathrm{rad} / \mathrm{s}\). The moment of inertia of the student plus stool is \(3.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and is assumed to be constant. The student then pulls in the objects horizontally to \(0.30 \mathrm{~m}\) from the rotation axis. (a) Find the new angular speed of the student. (b) Find the kinetic energy of the student before and after the objects are pulled in.

Short Answer

Expert verified
The new angular speed of the student is 1.63 rad/s. The initial and final kinetic energies of the system are 2.51 J and 5.48 J, respectively.

Step by step solution

01

Calculate Initial Angular Momentum

The initial total moment of inertia is the sum of the moment of inertia of the student and stool (3.0 kg·m²) and the two 3.0 kg objects at 1.0 m distance from the axis of rotation. Since the moment of inertia for a point mass is given by \( mr^{2} \), the total initial moment of inertia, \( I_{i} \), is \( I_{i} = 3.0 \, kg \cdot m^{2} + 2 \cdot 3.0 \, kg \cdot (1.0 \, m)^{2} = 9.0 \, kg \cdot m^{2} \). The initial angular momentum \( L_{i} \) is the product of the initial moment of inertia and the initial angular speed, \( L_{i} = I_{i} \omega_{i} \) = 9.0 kg·m² · 0.75 rad/s = 6.75 kg·m²/s.
02

Calculate Final Angular Momentum

The final moment of inertia is the sum of the moment of inertia of the student plus stool (3.0 kg·m²) and the two 3.0 kg objects at 0.30 m distance from the axis of rotation. Plugging into the formula, the total final moment of inertia \( I_{f} \) is \( I_{f} = 3.0 \, kg \cdot m^{2} + 2 \cdot 3.0 \, kg \cdot (0.30 \, m)^{2} = 4.14 \, kg \cdot m^{2} \). By conservation of angular momentum \( L_{i} = L_{f} \), the final angular speed \(\omega_{f}\) is \( L_{i} / I_{f} \) = 6.75 kg·m²/s ÷ 4.14 kg·m² = 1.63 rad/s.
03

Determine Initial and Final Kinetic Energy

The initial kinetic energy \( K_{i} \) is calculated using the formula \( K = \frac{1}{2} I \omega^{2} \) which gives \( K_{i} = \frac{1}{2} \cdot 9.0 \, kg \cdot m^{2} \cdot (0.75 \, rad/s)^{2} = 2.51 \, J \). The same formula is used for the final kinetic energy \( K_{f} \) which results in \( K_{f} = \frac{1}{2} \cdot 4.14 \, kg \cdot m^{2} \cdot (1.63 \, rad/s)^{2} = 5.48 \, J \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
Moment of inertia is a physical quantity that reflects how mass is distributed in relation to an axis of rotation. It's a rotational analogue to mass in linear motion, and it influences the resistance of an object to change in its rotational motion.

To calculate the moment of inertia, you use the formula \( I = mr^2 \) for point masses, where \( m \) is the mass and \( r \) is the distance from the axis of rotation. However, for extended bodies, the calculation involves summing up or integrating all the small mass elements \( dm \) multiplied by their respective distance squared from the axis. In the given problem, when the student holds the weights further away, the moment of inertia is greater because the masses are at a larger radius from the axis, making it harder for the student to spin.
Angular Speed
Angular speed refers to how quickly an object rotates or revolves relative to another point, expressed in radians per second (rad/s). It represents the angle turned, in radians, through a given amount of time.

In our example, the student's initial angular speed was given as \(0.75 \) rad/s. This means for every second that the student spins, they sweep out an angle of \(0.75 \) radians. Angular speed can change if the moment of inertia changes, assuming angular momentum is conserved, as seen when the student pulls the weights closer to the axis.
Kinetic Energy
In rotational motion, kinetic energy quantifies the energy an object possesses due to its rotation and is given by the formula \( K = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular speed.

For the student, the initial and final kinetic energies are calculated before and after moving the objects closer to the axis. The kinetic energy increases with the square of the angular speed, which explains why even with a smaller moment of inertia, the student's kinetic energy can increase, as seen when they pull the objects nearer.
Conservation of Angular Momentum
Conservation of angular momentum is a principle stating that if no external torque acts on a system, the total angular momentum of that system remains constant. Angular momentum is given by \( L = I\omega \), the product of moment of inertia and angular speed.

In the scenario where the student pulls the weights closer, the system's angular momentum is conserved. Initially, with arms extended, the system has a certain angular momentum that is equal to the product of the larger moment of inertia and the slower angular speed. When the student pulls the weights in, decreasing the moment of inertia, the angular speed increases to keep the angular momentum constant.
Rotational Motion
Rotational motion is the circular movement of an object about a central axis. It includes both spinning and revolving motions — like a planet rotating on its axis or revolving around a star. The study of rotational motion involves quantities such as angular displacement, angular velocity, and angular acceleration, analogous to their linear counterparts.

In the given textbook problem, we analyze the student’s spinning motion on a stool, which is a perfect example of rotational motion. The change in how the weights are positioned affects the rotational motion through changes in moment of inertia, directly influencing the angular velocity and kinetic energy of the system.

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Most popular questions from this chapter

QIC S A cylinder with moment of inertia \(I_{1}\) rotates with angular velocity \(\omega_{0}\) about a frictionless vertical axle. A second cylinder, with moment of inertia \(I_{2}\), initially not rotating, drops onto the first cylinder (Fig. P8.65). Because the surfaces are rough, the two cylinders eventually reach the same angular speed \(\omega\). (a) Calculate \(\omega\). (b) Show that kinetic energy is lost in this situation, and calculate the ratio of the final to the initial kinetic energy.

A model airplane with mass \(0.750 \mathrm{~kg}\) is tethered by a wire so that it flies in a circle \(30.0 \mathrm{~m}\) in radius. The airplane engine provides a net thrust of \(0.800 \mathrm{~N}\) perpendicular to the tethering wire. (a) Find the torque the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane when it is in level flight. (c) Find the linear acceleration of the airplane tangent to its flight path.

\(\mathbf{Q} \mid \mathbf{C}\) A rope of negligible mass is wrapped around a \(225-\mathrm{kg}\) solid cylinder of radius \(0.400 \mathrm{~m}\). The cylinder is suspended several meters off the ground with its axis oriented horizontally, and turns on that axis without friction. (a) If a \(75.0-\mathrm{kg}\) man takes hold of the free end of the rope and falls under the force of gravity, what is his acceleration? (b) What is the angular acceleration of the cylinder? (c) If the mass of the rope were not neglected, what would happen to the angular acceleration of the cylinder as the man falls?

A meter stick is found to balance at the \(49.7-\mathrm{cm}\) mark when placed on a fulcrum. When a \(50.0\)-gram mass is attached at the \(10.0-\mathrm{cm}\) mark, the fulcrum must be moved to the \(39.2-\mathrm{cm}\) mark for balance. What is the mass of the meter stick?

A horizontal \(800-\mathrm{N}\) merry-go-round of radius \(1.50 \mathrm{~m}\) is started from rest by a constant horizontal force of \(50.0 \mathrm{~N}\) applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go- round after \(3.00 \mathrm{~s}\). (Assume it is a solid cylinder.)

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