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Chapter 11: Problem 58

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a mass of \(150 \mathrm{~kg}\), a radius of \(2.0 \mathrm{~m}\), and a rotational inertia of \(300 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the axis of rotation. A \(60 \mathrm{~kg}\) student walks slowly from the rim of the platform toward the center. If the angular speed of the system is \(1.5 \mathrm{rad} / \mathrm{s}\) when the student starts at the rim, what is the angular speed when she is \(0.50 \mathrm{~m}\) from the center?

Short Answer

Expert verified
The angular speed increases to \(2.57\, \text{rad/s}\).

Step by step solution

01

Identify the Given Values

We know the mass of the platform is \(150 \text{ kg}\) and its radius is \(2.0 \text{ m}\). The rotational inertia of the platform is \(300 \text{ kg} \cdot \text{m}^2\). The mass of the student is \(60 \text{ kg}\). The initial angular speed is \( \omega_i = 1.5 \text{ rad/s}\). We want to find the angular speed when the student is \(0.50 \text{ m}\) from the center.
02

Recall the Conservation of Angular Momentum

The angular momentum of a system remains constant if no external torques act on it. Thus, the initial angular momentum \(L_i\) is equal to the final angular momentum \(L_f\). Mathematically, this is represented as: \(L_i = L_f\).
03

Calculate Initial Angular Momentum

The initial angular momentum \(L_i\) is the sum of the platform's and the student's angular momentum:\[ L_i = I_\text{platform} \cdot \omega_i + I_\text{student} \cdot \omega_i \]The initial moment of inertia of the student when she's at the rim is \(I_\text{student_initial} = m_s \cdot r^2\), where \(m_s = 60 \text{ kg}\) and \(r = 2.0 \text{ m}\).Calculate the initial moment of inertia:\[ I_\text{student_initial} = 60 \times (2)^2 = 240 \text{ kg} \cdot \text{m}^2 \]Thus, the initial angular momentum is:\[ L_i = (300 + 240) \cdot 1.5 \text{ rad/s} = 810 \text{ kg} \cdot \text{m}^2/\text{s} \]
04

Calculate Final Angular Momentum

The end moment of inertia of the student when she is \(0.50 \text{ m}\) from the center is:\[ I_\text{student_final} = m_s \cdot (0.50)^2 = 60 \times 0.25 = 15 \text{ kg} \cdot \text{m}^2 \]Add this to the platform's rotational inertia to get the total final rotational inertia:\[ I_\text{total_final} = 300 + 15 = 315 \text{ kg} \cdot \text{m}^2 \]
05

Solve for the Final Angular Speed

Use the conservation of angular momentum:\[ L_i = L_f \]\[ I_\text{total_initial} \cdot \omega_i = I_\text{total_final} \cdot \omega_f \]\[ 810 \text{ kg} \cdot \text{m}^2/\text{s} = 315 \cdot \omega_f \]Solving for \(\omega_f\):\[ \omega_f = \frac{810}{315} = 2.57 \text{ rad/s} \]

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Key Concepts

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

Rotational Inertia
Rotational inertia, also known as the moment of inertia, is a property that quantifies how difficult it is to change the rotational motion of an object. Imagine trying to stop a spinning merry-go-round 鈥� the effort required depends largely on its rotational inertia. In general, rotational inertia depends on two main factors:
  • The mass of the object
  • How that mass is distributed relative to the axis of rotation
For a point mass, the rotational inertia is calculated as:\[ I = m imes r^2 \]where "\(m\)" is the mass, and "\(r\)" is the distance from the axis of rotation.
In our example, the student acts as a point mass walking from the rim towards the center. Her contribution to the system鈥檚 rotational inertia changes as she moves, affecting the overall rotational behavior of the platform. This change is why her position impacts the final angular speed.
Angular Speed
Angular speed is a measure of how quickly an object rotates or revolves relative to another point. Unlike linear speed, which is about covering distance, angular speed is about covering angles in a given time.Typically denoted by \(\omega\), angular speed has the unit of radians per second (rad/s). In many real-life scenarios, like our example with the rotating platform and student, angular speed can change due to shifts in the distribution of mass.In our scenario, angular speed is influenced by the conservation of angular momentum. As the student's position changes, it impacts the inertia and, consequently, the rotational speed of the disk. Initially, when the student is at the edge, the system has a certain angular speed. As she moves closer, the improved moment of inertia causes the platform to spin faster to conserve angular momentum, illustrating the inverse relationship between rotational inertia and angular speed.
Moment of Inertia
The moment of inertia is a crucial concept in rotational dynamics, similar to mass in linear dynamics. It gives insight into how mass is spread in relation to an axis of rotation. It determines how much torque is required for a desired rotational acceleration.The formula varies with the shape and mass distribution:
  • For a point mass, \( I = m imes r^2 \)
  • For more complex shapes, like disks, the formulas incorporate their specific geometry.
In our detailed example, the platform and the student contribute to the total moment of inertia. The platform has a constant moment of inertia as it doesn't change, while the student鈥檚 default moment of inertia shifts as she moves towards the center. This variable moment of inertia impacts the system鈥檚 angular momentum, affecting how the rotational speed adjusts to maintain momentum conservation.

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Most popular questions from this chapter

A yo-yo has a rotational inertia of \(950 \mathrm{~g} \cdot \mathrm{cm}^{2}\) and a mass of \(120 \mathrm{~g} .\) Its axle radius is \(3.2 \mathrm{~mm}\), and its string is \(120 \mathrm{~cm}\) long. The yo-yo rolls from rest down to the end of the string. (a) What is the magnitude of its linear acceleration? (b) How long does it take to reach the end of the string? As it reaches the end of the string, what are its (c) linear speed, (d) translational kinetic energy, (e) rotational kinetic energy, and (f) angular speed?

Two particles, each of mass \(2.90 \times 10^{-4} \mathrm{~kg}\) and speed \(5.46 \mathrm{~m} / \mathrm{s}\), travel in opposite directions along parallel lines separated by \(4.20 \mathrm{~cm} .\) (a) What is the magnitude \(L\) of the angular momentum of the two-particle system around a point midway between the two lines? (b) Is the value different for a different location of the point? If the direction of either particle is reversed, what are the answers for (c) part (a) and (d) part (b)?

a wheel is rotating freely at angular speed 800 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with twice the rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the angular speed of the resultant combination of the shaft and two wheels? (b) What fraction of the original rotational kinetic energy is lost?

In a playground, there is a small merry-go-round of radius \(1.20 \mathrm{~m}\) and mass \(180 \mathrm{~kg} .\) Its radius of gyration (see Problem 79 of Chapter 10 ) is \(91.0 \mathrm{~cm}\). A child of mass \(44.0 \mathrm{~kg}\) runs at a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) along a path that is tangent to the rim of the initially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round. Calculate (a) the rotational inertia of the merry-go-round about its axis of rotation, (b) the magnitude of the angular momentum of the running child about the axis of rotation of the merry-go-round, and (c) the angular speed of the merry-go- round and child after the child has jumped onto the merry-go-round.

A Texas cockroach of mass \(0.17 \mathrm{~kg}\) runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has radius \(15 \mathrm{~cm}\), rotational inertia \(5.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and frictionless bearings. The cockroach's speed (relative to the ground) is \(2.0 \mathrm{~m} / \mathrm{s}\), and the lazy Susan turns clockwise with angular speed \(\omega_{0}=\) \(2.8 \mathrm{rad} / \mathrm{s}\). The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?
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