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A professor doing a lecture demonstration stands at the center of a frictionless turntable, holding 5.00 -kg masses in each hand with arms extended so that each mass is \(1.20 \mathrm{~m}\) from his centerline. A (carefully selected!) student spins the professor up to a rotational frequency of 1.00 rpm. If he then pulls his arms in by his sides so that each mass is \(0.300 \mathrm{~m}\) from his centerline, what is his new angular speed? Assume that his rotational inertia without the masses is \(2.80 \mathrm{~kg} \mathrm{~m}^{2}\), and neglect the effect on the rotational inertia of the position of his arms, since their mass is small compared to the mass of the body.

Short Answer

Expert verified
When a professor pulls the two 5.00 kg masses towards him, changing their positions from 1.20 m to 0.300 m, his new angular speed increases to approximately 0.717 rad/s.

Step by step solution

01

Convert the Rotational Frequency to Angular Speed

To convert the rotational frequency (in rpm) to angular speed (in rad/s), we can use the following conversion: \(\omega [\text{rad/s}] = 2\pi (\text{frequency [\text{rpm}]}) \times \frac{1 \mathrm{~min}}{60 \mathrm{~s}}\) Hence, we have \(\omega_1 = 1.00 \times 2\pi \times \frac{1}{60} \approx 0.1047 \mathrm{~rad/s}\).
02

Calculate the Initial Moment of Inertia

We can find the moment of inertia of each mass using the formula for a point mass, \(I = mr^2\), and summing their inertia along with the professor's initial rotational inertia (\(I_0\)): \(I_1 = I_0 + m r_1^2 + m r_1^2 = 2.80 + 5(1.20)^2 + 5(1.20)^2 = 2.80 + 14.40 + 14.40 \approx 31.60 \mathrm{~kg.m^2}\)
03

Calculate the Final Moment of Inertia

Now, we'll calculate the final moment of inertia after the professor pulls his arms in. Using the same formula for a point mass, we have: \(I_2 = I_0 + m r_2^2 + m r_2^2 = 2.80 + 5(0.30)^2 + 5(0.30)^2 = 2.80 + 1.8\mathrm{~kg.m^2}\).
04

Use Angular Momentum Conservation to Find the New Angular Speed

Now, we'll use the conservation of angular momentum equation: \(I_1 \omega_1 = I_2 \omega_2\) Solving for \(\omega_2\) : \(\omega_2 = \frac{I_1}{I_2} \omega_1\) \(\omega_2 = \frac{31.60}{4.6} \times 0.1047 \mathrm{~rad/s} \approx 0.717 \mathrm{~rad/s}\) Hence, the professor's new angular speed is approximately 0.717 rad/s.

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Key Concepts

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

Rotational Motion
Understanding rotational motion is essential when studying moving objects that turn around a fixed point, like a spinning wheel, a turning planet, or even a professor doing a demonstration on a turntable. The key aspects of rotational motion involve angular speed, rotational inertia (moment of inertia), and the conservation of angular momentum. Angular speed measures how quickly an object spins, which is typically expressed in radians per second (rad/s) or rotations per minute (rpm). It's a scalar quantity that represents the rate of rotation about an axis.

When solving problems involving a change in rotational speed, it is important to first establish the initial angular speed and then identify any factors that influence this value. For instance, in the provided example, the professor initially spinning with an angular speed can alter this speed by changing the distribution of his mass relative to the axis of rotation. This change directly affects the moment of inertia and consequently, the angular speed, due to the conservation of angular momentum—a fundamental principle in rotational dynamics.
Moment of Inertia
The moment of inertia, often symbolized as 'I', is akin to mass in linear motion but pertains to rotational motion. This quantity indicates how difficult it is to change the rotational speed of an object. The moment of inertia depends on both the mass of the object and the distribution of this mass relative to the axis about which it rotates. For a single point mass, the moment of inertia formula is simply 'I = mr^2', whereby 'm' is the mass of the object and 'r' is the distance from the rotational axis.

In the example of the professor and the turntable, each hand-held mass contributes to the overall moment of inertia and must be included in the calculation. It is clear when the masses are closer to the centerline, the moment of inertia decreases, which is why 'I' decreases when the professor pulls his arms in. Understanding how altering the radius influences 'I' is crucial for discerning changes in rotational behavior.
Angular Momentum Conservation
Angular momentum conservation is a principle stating that if no external torque acts on a system, its total angular momentum remains constant. This concept is analogous to the conservation of linear momentum in translational motion. The conservation law can be stated as 'L = Iω', where 'L' is the angular momentum, 'I' is the moment of inertia, and 'ω' is the angular speed.

To apply this principle, as in our exercise, you must equate the initial angular momentum with the final angular momentum, providing you with an equation to solve for an unknown, usually a change in angular speed. The professor's demonstration on the turntable perfectly encapsulates this principle: even as he pulls his arms in, changing his moment of inertia, the product of the inertia and his angular speed remains constant, resulting in a higher angular speed when his arms are close to his torso. Recognizing that the product of the moment of inertia and angular speed before and after a change must be equal is integral to solving many rotational dynamics problems.

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Most popular questions from this chapter

The Earth has an angular speed of \(7.272 \cdot 10^{-5} \mathrm{rad} / \mathrm{s}\) in its rotation. Find the new angular speed if an asteroid \(\left(m=1.00 \cdot 10^{22} \mathrm{~kg}\right)\) hits the Earth while traveling at a speed of \(1.40 \cdot 10^{3} \mathrm{~m} / \mathrm{s}\) (assume the asteroid is a point mass compared to the radius of the Earth) in each of the following cases: a) The asteroid hits the Earth dead center. b) The asteroid hits the Earth nearly tangentially in the direction of Earth's rotation. c) The asteroid hits the Earth nearly tangentially in the direction opposite to Earth's rotation.

A solid sphere rolls without slipping down an incline, starting from rest. At the same time, a box starts from rest at the same altitude and slides down the same incline, with negligible friction. Which object arrives at the bottom first? a) The solid sphere arrives first. b) The box arrives first. c) Both arrive at the same time. d) It is impossible to determine.

A solid cylinder, a hollow cylinder, a solid sphere, and a hollow sphere are rolling without slipping. All four objects have the same mass and radius and are traveling with the same linear speed. Which one of the following statements is true? a) The solid cylinder has the highest kinetic energy. b) The hollow cylinder has the highest kinetic energy. c) The solid sphere has the highest kinetic energy. d) The hollow sphere has the highest kinetic energy. e) All four objects have the same kinetic energy.

A wheel with \(c=\frac{4}{9},\) a mass of \(40.0 \mathrm{~kg},\) and a rim radius of \(30.0 \mathrm{~cm}\) is mounted vertically on a horizontal axis. A 2.00 -kg mass is suspended from the wheel by a rope wound around the rim. Find the angular acceleration of the wheel when the mass is released.

In experiments at the Princeton Plasma Physics Laboratory, a plasma of hydrogen atoms is heated to over 500 million degrees Celsius (about 25 times hotter than the center of the Sun) and confined for tens of milliseconds by powerful magnetic fields ( 100,000 times greater than the Earth's magnetic field). For each experimental run, a huge amount of energy is required over a fraction of a second, which translates into a power requirement that would cause a blackout if electricity from the normal grid were to be used to power the experiment. Instead, kinetic energy is stored in a colossal flywheel, which is a spinning solid cylinder with a radius of \(3.00 \mathrm{~m}\) and mass of \(1.18 \cdot 10^{6} \mathrm{~kg}\). Electrical energy from the power grid starts the flywheel spinning, and it takes 10.0 min to reach an angular speed of \(1.95 \mathrm{rad} / \mathrm{s}\). Once the flywheel reaches this angular speed, all of its energy can be drawn off very quickly to power an experimental run. What is the mechanical energy stored in the flywheel when it spins at \(1.95 \mathrm{rad} / \mathrm{s}\) ? What is the average torque required to accelerate the flywheel from rest to \(1.95 \mathrm{rad} / \mathrm{s}\) in \(10.0 \mathrm{~min} ?\)

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