/*! 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 69 A disk-shaped merry-go-round of ... [FREE SOLUTION] | 91Ó°ÊÓ

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A disk-shaped merry-go-round of radius \(2.63 \mathrm{m}\) and mass \(155 \mathrm{kg}\) rotates freely with an angular speed of \(0.641 \mathrm{rev} / \mathrm{s} .\) A 59.4 -kg person running tangential to the rim of the merry- goround at \(3.41 \mathrm{m} / \mathrm{s}\) jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. What is the final angular speed of the merry-go-round?

Short Answer

Expert verified
The final angular speed of the merry-go-round is approximately 0.444 rev/s.

Step by step solution

01

Calculate Initial Angular Momentum

The initial angular momentum consists of the angular momentum of the merry-go-round alone. It is given by \( L_i = I_i \omega_i \). The moment of inertia \( I \) for a disk is \( \frac{1}{2}MR^2 \), where \( M \) is the mass and \( R \) the radius.\[I_i = \frac{1}{2} \times 155 \times (2.63)^2 = 536.39225 \, \text{kg}\cdot\text{m}^2\]The initial angular speed \( \omega_i \) in radians per second is \( \omega_i = 0.641 \times 2\pi = 4.027 \) rad/s.Thus, the initial angular momentum is:\[L_i = 536.39225 \times 4.027 = 2161.83 \, \text{kg}\cdot\text{m}^2/\text{s}\]
02

Calculate Person's Contribution to Angular Momentum

The person is running tangentially to the rim and his contribution to the angular momentum when he jumps on is \( L_p = m v r \), where \( m = 59.4 \text{ kg} \) (mass of the person),\( v = 3.41 \, \text{m/s} \), and \( r = 2.63 \, \text{m} \).\[L_p = 59.4 \times 3.41 \times 2.63 = 533.706 \, \text{kg}\cdot\text{m}^2/\text{s}\]
03

Calculate Total Angular Momentum After

The total initial angular momentum from the merry-go-round and the person is conserved and will be equal to the final angular momentum.\[L_{total} = L_i + L_p = 2161.83 + 533.706 = 2695.536 \, \text{kg}\cdot\text{m}^2/\text{s}\]
04

Calculate Final Moment of Inertia

The final moment of inertia is the sum of the merry-go-round's moment of inertia and the person's moment of inertia treated as a point mass at the rim:\[I_f = I_i + mr^2 = 536.39225 + 59.4 \times (2.63)^2 = 967.46525 \, \text{kg}\cdot\text{m}^2\]
05

Solve for Final Angular Velocity

Using conservation of angular momentum \( L_{total} = I_f \omega_f \), solve for \( \omega_f \):\[\omega_f = \frac{L_{total}}{I_f} = \frac{2695.536}{967.46525} = 2.787 \, \text{rad/s}\]Convert \( \omega_f \) to revolutions per second by dividing by \( 2\pi \):\[\omega_f = \frac{2.787}{2\pi} = 0.444 \, \text{rev/s}\]

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Key Concepts

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

Moment of Inertia
The moment of inertia is like the rotational equivalent of mass in linear motion. It determines how difficult it is to change the angular motion of an object. For a disk, the formula for moment of inertia is:
  • \(I = \frac{1}{2}MR^2\)
Here, \(M\) is the mass of the object, and \(R\) is its radius. This formula helps us calculate how much resistance the merry-go-round offers to changes in its angular state. When the mass of the object is distributed further from the axis of rotation, the moment of inertia is larger.

In the exercise, the merry-go-round, being disk-shaped, has its own moment of inertia, which we computed using the provided mass and radius. Additionally, once the person jumps onto the rim, we also consider their contribution to the total moment of inertia as a point mass, because the person is concentrated at one location on the rim. This is denoted by adding \(mr^2\) to the initial moment of inertia from the merry-go-round.
Angular Momentum
Angular momentum, often symbolized as \(L\), is a measure of how much rotation an object has, taking into account its moment of inertia and angular velocity. The principle of the conservation of angular momentum tells us that the total angular momentum of a system remains constant, as long as no external torques are acting on it. This is a key part of understanding rotational dynamics.
  • Angular momentum can be calculated as \(L = I\omega\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.
In the problem, both the initial angular momentum from the merry-go-round and the angular momentum brought by the person jumping onto it are considered. These individual angular momenta are added together to find the total initial angular momentum of the system. By the conservation of angular momentum, this total will equal the final angular momentum after the person jumps onto the merry-go-round.

In essence, even when the person alters the distribution of mass (and thus the moment of inertia), the system's overall rotation maintains the same angular momentum throughout the interaction.
Angular Velocity
Angular velocity describes how fast an object spins around a central point (e.g., an axis). It's measured in radians per second or revolutions per second, similar to how we use speed for linear movement. Think of angular velocity as the pace at which an object rotates. In this exercise, angular velocity changes when the person jumps onto the merry-go-round.
  • Initially, we calculated the angular velocity of the merry-go-round alone. Then, the effect of the person’s movement needed to be incorporated.
With the conservation of angular momentum, once we know the total and final moments of inertia, we can find the final angular velocity. During this process, the person's tangential speed contributes additional angular momentum. Ultimately, solving for the final angular velocity involves dividing the total angular momentum by the new, increased moment of inertia of the system. Conversion into revolutions per second helps relate this change to more familiar terms of spinning objects.

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Most popular questions from this chapter

A 0.34-kg meterstick balances at its center. If a necklace is suspended from one end of the stick, the balance point moves \(9.5 \mathrm{cm}\) toward that end. (a) Is the mass of the necklace more than, less than, or the same as that of the meterstick? Explain. (b) Find the mass of the necklace.

A torque of \(0.97 \mathrm{N} \cdot \mathrm{m}\) is applied to a bicycle wheel of radius \(35 \mathrm{cm}\) and mass \(0.75 \mathrm{kg}\). Treating the wheel as a hoop, find its angular acceleration.

\(\mathrm{A} 47.0-\mathrm{kg}\) uniform rod \(4.25 \mathrm{m}\) long is attached to a wall with a hinge at one end. The rod is held in a horizontal position by a wire attached to its other end. The wire makes an angle of \(30.0^{\circ}\) with the horizontal, and is bolted to the wall directly above the hinge. If the wire can withstand a maximum tension of \(1450 \mathrm{N}\) before breaking, how far from the wall can a \(68.0-\mathrm{kg}\) person sit without breaking the wire?

A mass \(M\) is attached to a rope that passes over a diskshaped pulley of mass \(m\) and radius \(r .\) The mass hangs to the left side of the pulley. On the right side of the pulley, the rope is pulled downward with a force \(F .\) Find \((a)\) the acceleration of the mass, (b) the tension in the rope on the left side of the pulley, and \((c)\) the tension in the rope on the right side of the pulley. (d) Check your results in the limits \(m \rightarrow 0\) and \(m \rightarrow \infty\)

Calculate the angular momentum of the Earth about its own axis, due to its daily rotation. Assume that the Earth is a uniform sphere.

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