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A large horizontal disk is rotating on a vertical axis through its center. Its moment of inertia is \(I=4000 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The disk is revolving freely at a rate of \(0.150 \mathrm{rev} / \mathrm{s}\) when a \(90.0-\mathrm{kg}\) person drops straight down onto it from an overhanging tree limb. The person lands and remains at a distance of \(3.00 \mathrm{~m}\) from the axis of rotation. What will be the rate of rotation after the person has landed?

Short Answer

Expert verified
The final rate of rotation is approximately 0.125 revolutions per second.

Step by step solution

01

Understand the conservation of angular momentum

In this scenario, the principle of conservation of angular momentum applies. The total angular momentum before the person lands on the disk must equal the total angular momentum after the person lands. Mathematically, this is represented as:\[ L_{initial} = L_{final} \] where \( L \) is the angular momentum.
02

Calculate the initial angular momentum

The initial angular momentum \( L_{initial} \) is given by the product of the moment of inertia of the disk \( I \) and its initial angular velocity \( \omega_{initial} \):\[ L_{initial} = I \cdot \omega_{initial} \]Convert the initial angular velocity from revolutions per second to radians per second:\[ \omega_{initial} = 0.150 \times 2\pi \approx 0.942 \text{ rad/s} \]Now calculate:\[ L_{initial} = 4000 \cdot 0.942 \approx 3768 \text{ kg} \cdot \text{m}^2/\text{s} \]
03

Determine the change in the moment of inertia

When the person lands on the disk, they add to the system's moment of inertia. The moment of inertia due to the person \( I_{person} \) can be calculated as:\[ I_{person} = m \cdot r^2 \]where \( m = 90.0 \text{ kg} \) and \( r = 3.00 \text{ m} \).\[ I_{person} = 90.0 \times (3.00)^2 = 810 \text{ kg} \cdot \text{m}^2 \]
04

Calculate the final moment of inertia

The total moment of inertia after the person lands on the disk \( I_{final} \) is the sum of the disk's moment of inertia and the person's contribution:\[ I_{final} = I + I_{person} = 4000 + 810 = 4810 \text{ kg} \cdot \text{m}^2 \]
05

Calculate the final angular velocity

Using the conservation of angular momentum:\[ L_{final} = I_{final} \cdot \omega_{final} \]\[ L_{initial} = L_{final} \]Thus,\[ 3768 = 4810 \cdot \omega_{final} \]Solve for \( \omega_{final} \):\[ \omega_{final} = \frac{3768}{4810} \approx 0.783 \text{ rad/s} \]
06

Convert the final angular velocity to revolutions per second

To convert \( \omega_{final} \) back to revolutions per second, divide by \( 2\pi \):\[ \text{Final rate of rotation} = \frac{0.783}{2\pi} \approx 0.125 \text{ rev/s} \]

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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 a fundamental concept in rotational dynamics that determines an object's resistance to changes in its rotational motion. It’s analogous to mass in linear motion. In this exercise, the disk's moment of inertia is given as \( I = 4000 \, \text{kg} \cdot \text{m}^2 \). This tells us how much torque is needed to achieve a certain angular acceleration. When the person lands on the disk, the moment of inertia changes. We calculate the additional moment of inertia due to the person as \( I_{\text{person}} = m \cdot r^2 \), where \( m = 90 \, \text{kg} \) and \( r = 3.0 \, \text{m} \). This shows how the person’s mass and the distance from the axis contribute to the new rotational inertia.The final moment of inertia becomes the sum \( I_{\text{final}} = I + I_{\text{person}} = 4810 \, \text{kg} \cdot \text{m}^2 \). This reveals how every component's placement affects the overall rotational motion.
Angular Velocity
Angular velocity \((\omega)\) is a measure of how quickly an object rotates or revolves relative to another point, usually the center of rotation. It is commonly expressed in radians per second (rad/s) or revolutions per second (rev/s). Initially, the disk rotates at \(0.150 \, \text{rev/s}\). For calculation ease in physics, we convert this to rad/s using the formula \( \omega = \text{revs} \times 2\pi \approx 0.942 \, \text{rad/s} \). This initial angular velocity is crucial for determining the rotational energy and momentum before any external interference. After the person lands on the disk, the rotational speed changes due to increased moment of inertia. The final angular velocity \( \omega_{\text{final}} \approx 0.783 \, \text{rad/s} \) shows how the shared system slows down as mass is added further from the center.
Rotational Dynamics
Rotational dynamics deals with the torques and angular motions influencing rotating objects. Understanding this concept is vital in solving physics problems about rotations. The conservation of angular momentum \( (L = I \times \omega) \) is a key principle here, stating that unless an external torque acts on a system, the total angular momentum remains constant.In the provided exercise, the initial angular momentum \( L_{\text{initial}} \approx 3768 \, \text{kg} \cdot \text{m}^2/\text{s} \) is solely due to the disk. The conservation principle implies this must equal the final angular momentum \( L_{\text{final}} \) of the system that now includes the person.This explains the reduced final angular velocity \( \omega_{\text{final}} \) even though the total angular momentum remains unchanged. Knowing these dynamics helps in anticipating the outcomes when conditions like mass distribution or force application change.
Physics Problem Solving
Physics problem-solving often involves breaking down complex physical situations into approachable steps, as shown in the exercise. This method involves identifying known quantities and applicable physical laws, like conservation laws, to find unknown properties.For instance:
  • Start by identifying the principle involved, like in this case, conservation of angular momentum.
  • Calculate initial known values such as \( L_{\text{initial}} \).
  • Determine contributions from any changes, like the added mass in this exercise, affecting the system.
  • Use these calculations to find unknowns such as \( \omega_{\text{final}} \).
Breaking down the problem into such logical steps can simplify complex calculations and enhance understanding, a strategy useful across various physics domains. This approach turns textbook exercises into insightful physics practice.

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Most popular questions from this chapter

The wheel on a grinder is a homogeneous \(0.90-\mathrm{kg}\) disk with a \(8.0-\mathrm{cm}\) radius. It coasts uniformly to rest from 1400 rpm in a time of \(35 \mathrm{~s}\). How large a frictional torque slows its motion? Let's first find \(\alpha\) from the change in \(\omega ;\) then we can use \(\tau=I \alpha\) to find \(\tau\). We know that \(f=1400 \mathrm{rev} / \mathrm{min}\) \(=23.3 \mathrm{rev} / \mathrm{s}\), and since \(\omega=2 \pi f, \omega_{i}=146 \mathrm{rad} / \mathrm{s}\) and \(\omega_{f}=0 .\) Therefore, $$ \alpha=\frac{\omega_{f}-\omega_{i}}{t}=\frac{-146 \mathrm{rad} / \mathrm{s}}{35 \mathrm{~s}}=-4.2 \mathrm{rad} / \mathrm{s}^{2} $$ We also need \(I\). For a uniform disk, $$ I=\frac{1}{2} M r^{2}=\frac{1}{2}(0.90 \mathrm{~kg})(0.080 \mathrm{~m})^{2}=2.9 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2} $$ Then \(\tau=I \alpha=\left(0.0029 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)\left(-4.2 \mathrm{rad} / \mathrm{s}^{2}\right)=-1.2 \times 10^{-2} \mathrm{~N} \cdot \mathrm{m}\)

As shown in Fig. \(10-3\), a mass \(m=400 \mathrm{~g}\) hangs from the rim of a frictionless pulley of radius \(r=15 \mathrm{~cm}\). When released from rest, the mass falls \(2.0 \mathrm{~m}\) in \(6.5 \mathrm{~s}\). Find the moment of inertia of the wheel. The hanging mass linearly accelerates downward due to its weight, and the pulley angularly accelerates clockwise due to the torque produced by the rope). The two motions are linked by the fact that \(a_{T}=r \alpha\). Consequently we will need to determine \(a_{T}\), and then \(\alpha\), and then \(F_{T}\), and then \(\tau\), and then \(I\). Remember that Newton's Second Law is central here (i.e., \(\tau=I \alpha\) for the wheel and \(F=m a\) for the mass). First we find \(a\) using \(y=v_{i} t+\frac{1}{2} a t^{2}\), since the mass accelerates down uniformly: $$ 2.0 \mathrm{~m}=0+\frac{1}{2} a(6.5 \mathrm{~s})^{2} $$ which yields \(a=0.095 \mathrm{~m} / \mathrm{s}^{2}\), and that equals the tangential acceleration \(\left(a_{T}\right)\) of a point on the rim of the pulley, which equals the acceleration \(a\) of the rope. Then, from \(a_{T}=\alpha r\), $$ \alpha=\frac{a_{T}}{r}=\frac{0.095 \mathrm{~m} / \mathrm{s}^{2}}{0.15 \mathrm{~m}}=0.63 \mathrm{rad} / \mathrm{s}^{2} $$ The net force on the mass \(m\) is \(m g-F_{T}\) and so \(F=m a\) becomes $$ \begin{aligned} m g-F_{T} &=m a_{T} \\ (0.40 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)-F_{T} &=(0.40 \mathrm{~kg})\left(0.095 \mathrm{~m} / \mathrm{s}^{2}\right) \end{aligned} $$ from which it follows that \(F_{T}=3.88 \mathrm{~N}\). Now \(\tau=I \alpha\) for the wheel: $$ \left(F_{T}\right)(r)=I \alpha \quad \text { or } \quad(3.88 \mathrm{~N})(0.15 \mathrm{~m})=I\left(0.63 \mathrm{rad} / \mathrm{s}^{2}\right) $$ from which we get \(I=0.92 \mathrm{~kg} \cdot \mathrm{m}^{2}\)

An force of \(200 \mathrm{~N}\) acts tangentially on the rim of a wheel \(25 \mathrm{~cm}\) in radius. \((a)\) Find the torque. \((b)\) Repeat if the force makes an angle of \(40^{\circ}\) to a spoke of the wheel.

Determine the constant torque that must be applied to a \(50-\mathrm{kg}\) flywheel, with radius of gyration \(40 \mathrm{~cm}\), to give it a frequency of 300 rpm in \(10 \mathrm{~s}\) if it's initially at rest.

A 90 -kg person stands at the edge of a stationary children's merry-go-round (essentially a disk) at a distance of \(5.0 \mathrm{~m}\) from its center. The person starts to walk around the perimeter of the disk at a speed of \(0.80 \mathrm{~m} / \mathrm{s}\) relative to the ground. What rotation rate does this motion impart to the disk if \(I_{\text {disk }}=20000 \mathrm{~kg} \cdot \mathrm{m}^{2} ?\left[\right.\) Hint \(:\) For the person, \(\left.I=m r^{2} .\right]\)

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