/*! 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 98 A 55-kg runner runs around the e... [FREE SOLUTION] | 91Ó°ÊÓ

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A 55-kg runner runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the earth has magnitude 2.8 \(\mathrm{m} / \mathrm{s}\) . The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.20 rad/s relative to the earth. The radius of the turntable is \(3.0 \mathrm{m},\) and its moment of inertia about the axis of rotation is 80 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can model the runner as a particle.)

Short Answer

Expert verified
The final angular velocity of the system is approximately 0.775 rad/s.

Step by step solution

01

Define Angular Momentum Conservation

Since there are no external torques acting on the system, the total angular momentum must be conserved. This means the initial angular momentum of the runner and the turntable will equal the final angular momentum when the runner comes to rest relative to the turntable. Therefore, \( L_{initial} = L_{final} \).
02

Calculate Initial Angular Momentum

Calculate the initial angular momentum of the system. The angular momentum of the turntable is \( L_{turntable} = I_{turntable} \times \omega_{turntable} \), where \( I_{turntable} = 80 \, \mathrm{kg \cdot m^2} \) and \( \omega_{turntable} = 0.20 \, \mathrm{rad/s} \). So, \( L_{turntable} = 80 \, \mathrm{kg \cdot m^2} \times 0.20 \, \mathrm{rad/s} = 16 \, \mathrm{kg \cdot m^2/s} \). Now, the angular momentum of the runner is \( L_{runner} = m_{runner} \times v_{runner} \times r \), where \( m_{runner} = 55 \, \mathrm{kg} \), \( v_{runner} = 2.8 \, \mathrm{m/s} \), and \( r = 3.0 \, \mathrm{m} \). So, \( L_{runner} = 55 \, \mathrm{kg} \times 2.8 \, \mathrm{m/s} \times 3.0 \, \mathrm{m} = 462 \, \mathrm{kg \cdot m^2/s} \). The initial total angular momentum \( L_{initial} = L_{runner} - L_{turntable} = 462 - 16 = 446 \, \mathrm{kg \cdot m^2/s} \).
03

Find Final Angular Velocity

In the final state, the runner comes to rest relative to the turntable, so they have the same angular velocity denoted as \( \omega_{final}\). Therefore, the final angular momentum is given by \( L_{final} = (I_{turntable} + m_{runner} \times r^2) \times \omega_{final} \). With \( r = 3.0 \, \mathrm{m} \), \( I_{runner} = 55 \times 3.0^2 = 495 \, \mathrm{kg \cdot m^2} \). Consequently, \( L_{final} = (80 + 495) \times \omega_{final} = 575 \times \omega_{final} \). Equating initial and final angular momentum, we have \( 446 = 575 \times \omega_{final} \), which gives \( \omega_{final} = \frac{446}{575} \approx 0.775 \mathrm{rad/s} \).

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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 (often symbolized as \( I \)) is a measure of how an object's mass is distributed with respect to an axis of rotation. This property determines how much torque is required for a given angular acceleration about that axis. Imagine trying to spin a barbell versus a single dumbbell; the barbell has more of its mass far from the rotational axis, leading to a higher moment of inertia and making it harder to spin. The moment of inertia for a point mass, like the runner in this exercise, about a given axis is calculated as the product of the mass and the square of the perpendicular distance from the axis, expressed as \( I = m imes r^2 \). In our exercise, since the runner is modeled as a particle, this simplifies our calculation and helps in easy understanding.Understanding moment of inertia is key when examining rotational systems where different components may contribute to the system's overall resistance to changes in rotation.
  • If the mass is concentrated further from the axis, the moment of inertia increases.
  • If the mass is closer to the axis, the moment of inertia decreases.
In our case, knowing the moment of inertia of both the turntable and the runner helps in calculating the total angular momentum, which governs the conservation laws applied here.
Angular Velocity
Angular velocity is a vector quantity that represents how fast an object rotates around an axis. It is like the rotational counterpart to linear velocity. Instead of moving in a straight line, an object under angular motion sweeps an angle over time, and its angular velocity, typically denoted by \( \omega \), tells us how quickly this angle is changing.Angular velocity is measured in radians per second (rad/s), showing how many radians the object rotates in one second. In the given exercise, the runner initially has a perpendicular linear velocity, which contributes to their angular velocity when considering their motion relative to the turntable.
  • For the turntable, we already know its initial angular velocity as 0.20 rad/s, rotating opposite to the runner's direction.
  • When the runner comes to rest with respect to the turntable, their final shared angular velocity \( \omega_{final} \) needs to be calculated by balancing out the initial angular momentum.
Connecting these dynamic changes to the concept of angular momentum conservation helps determine the new angular velocity of the entire system when states change.
Rotational Kinematics
Rotational kinematics deals with the description of the rotation of a rigid body without considering the forces or torques that cause the motion. It is directly analogous to linear kinematics but focuses on angular quantities. This includes concepts like angular displacement, angular velocity, and angular acceleration, similar to their linear counterparts but involved in circular paths or rotational motion. In the exercise, understanding rotational kinematics can help to conceptualize how the angular velocity of the system changes when the external conditions such as the runner's motion change. Since no external torques are applied, we rely on conservation of angular momentum to predict the final state of the system. The interplay of the runner's kinematics with the motion of the turntable ultimately determines the combined system's final angular velocity. The key takeaway: - When a system undergoes changes in rotation and involves different mass elements, rotational kinematics provides a detailed quantitative description of how these changes manifest. - The principles of rotational kinematics simplify complex interactions by employing concepts already familiar from linear motion, merely applied pivotally around an axis. This ability to translate between linear and angular descriptions is what allows us to solve complex rotational problems like the one presented here.

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Most popular questions from this chapter

A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley, similar to what is shown in Fig. \(10.10 .\) The pulley is a uniform disk with mass 10.0 \(\mathrm{kg}\) and radius 50.0 \(\mathrm{cm}\) and turns on frictionless bearings. You measure that the stone travels 12.6 \(\mathrm{m}\) in the first 3.00 s starting from rest. Find (a) the mass of the stone and (b) the tension in the wire.

A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel 0.650 \(\mathrm{m}\) in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects 0.200 \(\mathrm{m}\) at each side of the wheel, and a woman holds the ends of the shaft in her hands. The mass of the system is 8.00 \(\mathrm{kg}\) ; its entire mass may be assumed to be located at its rim. The shaft is horizontal, and the wheel is spinning about the shaft at 5.00 \(\mathrm{rev} / \mathrm{s}\) . Find the magnitude and direction of the force each hand exerts on the shaft (a) when the shaft is at rest; (b) when the shaft is rotating in a horizontal plane about its center at \(0.050 \mathrm{rev} / \mathrm{s} ;\) (c) when the shaft is rotating in a horizontal plane about its center at 0.300 \(\mathrm{rev} / \mathrm{s}\) . (d) At what rate must the shaft rotate in order that it may be supported at one end only?

The solid wood door of a gymnasium is 1.00 \(\mathrm{m}\) wide and 2.00 \(\mathrm{m}\) high, has total mass \(35.0 \mathrm{kg},\) and is hinged along one side. The door is open and at rest when a stray basketball hits the center of the door head-on, applying an average force of 1500 \(\mathrm{N}\) to the door for 8.00 \(\mathrm{ms} .\) Find the angular speed of the door after the impact. [Hint: Integrating Eq. \((10.29)\) yields \(\Delta L_{z}=\int_{t_{1}}^{t_{2}}\left(\Sigma \tau_{z}\right) d t=\left(\sum \tau_{z}\right)_{\mathrm{av}} \Delta t .\) The quantity \(\int_{t_{1}}^{t_{2}}\left(\Sigma \tau_{z}\right) d t\) is called the angular impulse.]

A uniform, \(0.0300-\mathrm{kg}\) rod of length 0.400 \(\mathrm{m}\) rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.0200 \(\mathrm{kg}\) , are mounted so that they can slide along the rod. They are initially held by catches at positions 0.0500 \(\mathrm{m}\) on each side of the center of the rod, and the system is rotating at 30.0 rev/min. With no other changes in the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. (a) What is the angular speed of the system at the instant when the rings reach the ends of the rod? (b) What is the angular speed of the rod after the rings leave it?

A bicycle racer is going downhill at 11.0 \(\mathrm{m} / \mathrm{s}\) when, to his horror, one of his \(2.25-\mathrm{kg}\) wheels comes off as he is 75.0 \(\mathrm{m}\) above the foot of the hill. We can model the wheel as a thin- walled cylinder 85.0 \(\mathrm{cm}\) in diameter and neglect the small mass of the spokes. (a) How fast is the wheel moving when it reaches the foot of the hill if it rolled without slipping all the way down? (b) How much total kinetic energy does the wheel have when it reaches the bottomof the hill?

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