/*! 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 87 A \(55 \mathrm{~kg}\) runner run... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(55 \mathrm{~kg}\) 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 \(0.20 \mathrm{rad} / \mathrm{s}\) 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
By calculating the initial and final angular momentum of the system and setting them equal (due to the law of conservation of angular momentum), we can solve for the final angular velocity \(\omega_{f}\) using the given values and algebraic manipulation.

Step by step solution

01

Calculating initial angular momentum of the system

The total initial angular momentum of the system is the sum of the runner's angular momentum and the turntable's angular momentum. The angular momentum of the runner \(L_{r}\) can be calculated with the formula \(L_{r} = m*r*v\) where m is her mass, r is the distance from the center (which equals the radius of the turntable), and v is her speed. Similarily the turntable's angular momentum \(L_{t}\) can be calculated through \(L_{t} = I*\omega\) where I is the moment of inertia and \( \omega\) is the angular velocity. Summing these up will give the total initial angular momentum \(L_{i} = L_{r} + L_{t}\).
02

Calculating final angular momentum of the system

When the runner comes to rest relative to the turntable, she and the turntable should be considered as one object and hence they share the same final angular velocity. The total final angular momentum of the system \(L_{f}\) is thus: \(L_{f} = (I + m*r^{2})*\omega_{f}\). Since the angular momentum is conserved, \(L_{i} = L_{f}\). The only unknown in this equation is the final angular velocity \(\omega_{f}\), which we can solve for.
03

Solving for the final angular velocity

We subtitute the values we've calculated in step 1 and step 2 into the equation \(L_{i} = L_{f}\) and solve for the final angular velocity of the system. Therefore, the final angular velocity of the turntable can be calculated by algebraically manipulating this equation to isolate \(\omega_{f}\).

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Key Concepts

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

Angular Velocity
Understanding angular velocity is essential in physics, particularly when analyzing rotating systems. Angular velocity, symbolized by \(\omega\), measures the rate of rotation about an axis and is usually expressed in radians per second (rad/s). In the context of our textbook problem, the turntable initially rotates with an angular velocity of magnitude \(0.20 \: \text{rad/s}\) in a direction opposite to the runner's movement. This concept is crucial: it helps us comprehend how quickly objects are spinning and is a key component in calculating angular momentum, which plays the leading role in solving our exercise.

When the runner comes to rest on the turntable, they both share a common angular velocity; understanding how to merge these separate speeds into one cohesive measure is what makes this problem such a fascinating study in the conservation of angular momentum. To reach the final angular velocity, we rely on the unchanging nature of angular momentum when no external torques act on the system, leading us to a comprehensive solution as we equate initial and final states.
Moment of Inertia
The moment of inertia, often symbolized by \(I\), is a measure of an object's resistance to changes in its rotation rate. It's akin to mass in linear motion, but for rotating systems. Moment of inertia depends on the distribution of mass relative to the axis of rotation: the farther the mass is from the axis, the higher the moment of inertia.

In the exercise provided, the turntable has a moment of inertia of \(80 \: \text{kg} \cdot \text{m}^2\), indicating how much resistance it offers against changes in its angular velocity. This moment of inertia, combined with the runner's mass and the radius of the turntable, directly influences calculations for the final angular velocity of the system. By considering the runner as a particle and applying the parallel axis theorem, we enhance the moment of inertia to account for the motion of the runner, a crucial step for accurate problem-solving in rotational dynamics.
Physics Problem Solving
Solving physics problems often involves a systematic approach that includes understanding the concepts, identifying known and unknown variables, and applying the appropriate physical laws. In our scenario, the law of conservation of angular momentum is the guiding principle.

By breaking the problem into steps, we first identify the initial and final conditions—this paves the way for the establishment of relationships between these states. It's about dissecting complex scenarios: we calculated the initial angular momentum by summing the runner's and turntable's contributions, and then deduced the system's final state under the assumption of angular momentum conservation.

Effective problem-solving in physics typically involves algebraic manipulation, unit consistency, and critical thinking. By combining these elements, students can navigate through challenging problems and arrive at solutions that encapsulate the behavior of physical systems. The key lies in sequential reasoning, which transcends mere memorization of formulas and pushes towards a deeper comprehension of the underlying principles, as is the case in preserving angular momentum while finding the system's final angular velocity.

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Most popular questions from this chapter

A wheel with radius \(0.0600 \mathrm{~m}\) rotates about a horizontal frictionless axle at its center. The moment of inertia of the wheel about the axle is \(2.50 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The wheel is initially at rest. Then at \(t=0 \mathrm{a}\) force \(F=(5.00 \mathrm{~N} / \mathrm{s}) t\) is applied tangentially to the wheel and the wheel starts to rotate. What is the magnitude of the force at the instant when the wheel has turned through 8.00 revolutions?

A large wooden turntable in the shape of a flat uniform disk has a radius of \(2.00 \mathrm{~m}\) and a total mass of \(120 \mathrm{~kg}\). The turntable is initially rotating at \(3.00 \mathrm{rad} / \mathrm{s}\) about a vertical axis through its center. Suddenly, a \(70.0 \mathrm{~kg}\) parachutist makes a soft landing on the turntable at a point near the outer edge. (a) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.) (b) Compute the kinetic energy of the system before and after the parachutist lands. Why are these kinetic energies not equal?

A local ice hockey team has asked you to design an apparatus for measuring the speed of the hockey puck after a slap shot. Your design is a 2.00 -m-long, uniform rod pivoted about one end so that it is free to rotate horizontally on the ice without friction. The \(0.800 \mathrm{~kg}\) rod has a light basket at the other end to catch the \(0.163 \mathrm{~kg}\) puck. The puck slides across the ice with velocity \(\overrightarrow{\boldsymbol{v}}\) (perpendicular to the rod), hits the basket, and is caught. After the collision, the rod rotates. If the rod makes one revolution every 0.736 s after the puck is caught, what was the puck's speed just before it hit the rod?

Stabilization of the Hubble Space Telescope. The Hubble Space Telescope is stabilized to within an angle of about 2 -millionths of a degree by means of a series of gyroscopes that spin at 19,200 rpm. Although the structure of these gyroscopes is actually quite complex, we can model each of the gyroscopes as a thin-walled cylinder of mass \(2.0 \mathrm{~kg}\) and diameter \(5.0 \mathrm{~cm},\) spinning about its central axis. How large a torque would it take to cause these gyroscopes to precess through an angle of \(1.0 \times 10^{-6}\) degree during a 5.0 hour exposure of a galaxy?

A uniform solid cylinder with mass \(M\) and radius \(2 R\) rests on a horizontal tabletop. A string is attached by a yoke to a frictionless axle through the center of the cylinder so that the cylinder can rotate about the axle. The string runs over a disk-shaped pulley with mass \(M\) and radius \(R\) that is mounted on a frictionless axle through its center. A block of mass \(M\) is suspended from the free end of the string (Fig. \(\mathbf{P 1 0 . 7 9}\) ). The string doesn't slip over the pulley surface, and the cylinder rolls without slipping on the tabletop. Find the magnitude of the acceleration of the block after the system is released from rest.

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