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Two skaters, each of mass \(50 \mathrm{~kg}\), approach each other along parallel paths separated by \(3.0\) \(\mathrm{m}\). They have opposite velocities of \(1.4 \mathrm{~m} / \mathrm{s}\) each. One skater carries one end of a long pole of negligible mass, and the other skater grabs the other end as she passes. The skaters then rotate around the center of the pole. Assume that the friction between skates and ice is negligible. What are (a) the radius of the circle, (b) the angular speed of the skaters, and (c) the kinetic energy of the two-skater system? Next, the skaters pull along the pole until they are separated by \(1.0 \mathrm{~m}\). What then are (d) their angular speed and (e) the kinetic energy of the system? (f) What provided the energy for the increased kinetic energy?

Short Answer

Expert verified
(a) Radius: 1.5 m, (b) Angular speed: 0.933 rad/s, (c) KE: 98 J, (d) New angular speed: 8.4 rad/s, (e) New KE: 882 J, (f) Work done by skaters.

Step by step solution

01

Determine the Radius of Rotation

The skaters are separated by 3.0 m initially, so the distance between each skater and the center of the pole is half of that. Therefore, the radius of the circle, which is the swing distance for each skater, is:\[ r = \frac{3.0 \, \text{m}}{2} = 1.5 \, \text{m} \]
02

Calculate the Angular Speed

Conservation of linear momentum in a system where they turn into circular motion implies that the initial linear momentum equals the angular momentum. Each skater's velocity component contributing to angular momentum is their perpendicular distance from the center at that instance:Initial angular momentum, \(L = mvr \theta\), sets rotational inertia (\(I\)) and angular velocity (\(\omega_1\)):\[ I = mr^2 \] \[ L = 2 \times \left(50 \, \text{kg} \times 1.4 \, \text{m/s} \times 1.5 \, \text{m}\right) = I \omega_1 \] Substitute \(I\):\[ 2 \times 50 \times 1.4 \times 1.5 = 2 \times 50 \times 1.5^2 \times \omega_1 \]Solving gives \[ \omega_1 \approx 0.933 \, \text{rad/s} \].
03

Calculate the Initial Kinetic Energy

The initial kinetic energy is given by the sum of the kinetic energies of both skaters:\[KE_i = 2 \times \frac{1}{2} \times 50 \, \text{kg} \times (1.4 \, \text{m/s})^2 = 98 \, \text{J} \]Because the skaters maintain the same speed initially, this applies as is.
04

Calculate the Angular Speed with Smaller Radius

When the distance between skaters reduces to 1.0 m, the distance from each to the center is 0.5 m:New rotational inertia:\[ I' = m \left(\frac{1.0}{2}\right)^2 \]Because initial angular momentum (previous step) equals final angular momentum:\[ 2 \times 50 \times 1.4 \times 1.5 = 2 \times 50 \times 0.5^2 \times \omega_2 \]Solve for \(\omega_2\):\[ \omega_2 \approx 8.4 \, \text{rad/s} \].
05

Calculate the Final Kinetic Energy

The new kinetic energy involves this new angular speed:\[KE_f = \frac{1}{2} I' \omega_2^2 \]Substitute values:\[KE_f = 2 \times \frac{1}{2} \times 50 \times 0.5^2 \times (8.4)^2 \, \text{J} = 882 \text{ J}\]
06

Analyze Energy Source for Increased Kinetic Energy

The energy comes from the work done by the skaters pulling inward on the pole, which decreases the radius and increases rotational speed, converting this work to rotational kinetic energy, due to conservation of angular momentum.

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Key Concepts

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

Conservation of Angular Momentum
Imagine two skaters twirling around each other with a consistent pattern. Their performance can be explained by the principle known as the Conservation of Angular Momentum. In simple terms, angular momentum is a quantity that remains constant if no external torque is acting on a system. It is the "spin" version of linear momentum.

Angular momentum (\(L\) ) is given by the product of rotational inertia (\(I\) ) and angular velocity (\(\omega\) ). When the skaters interlink and begin to rotate, they exhibit this conservation law practically. Initially, they have a certain linear momentum which then translates into angular momentum due to the act of rotation. In our exercise, changes in the distance between skaters influence their angular speed. As they pull closer, the skaters spin faster to maintain the same amount of angular momentum, like a figure skater pulling their arms in during a spin to rotate faster.
Rotational Kinetic Energy
Rotational Kinetic Energy is the energy possessed by a rotating object. The faster the rotation, the more kinetic energy the object holds. For objects in circular motion, this energy depends on both the angular speed and the moment of inertia.

The formula for rotational kinetic energy is given by:\[KE_{rot} = \frac{1}{2} I \omega^2\]Where \(I\) is the moment of inertia, and \(\omega\) is the angular velocity. In the problem, the skaters' initial and final kinetic energies are calculated by considering their speeds and the radius at which they circle one another. Notice that when the radius decreases and the angular velocity increases, the rotational kinetic energy also increases. This transition is a fascinating interplay between moment of inertia and angular speed.
Circular Motion
Circular motion is a path that leads an object to rotate along the circumference of a circle. In simpler terms, it is the movement of an object in a circle or along a circular path.

For our skaters, once they grab both ends of the pole, they begin rotating in circular motion around a central axis. The radius of their motion initially is half the distance between them. The circular movement creates centripetal force, which is necessary to keep the skaters moving in a circle. This force is directed towards the center of the circle and is crucial for maintaining their path. In the exercise, changes in radius affect the circumferential speed, but the continued circular path is facilitated by the constant interaction between their forces and motion.
Physics Problem Solving
Problem-solving in physics begins with understanding the key concepts that govern a situation. Breaking down problems into understandable pieces is essential.

To solve problems like our skaters’ exercise, it is helpful to:
  • Identify the principles involved, such as conservation laws or energy balances.
  • Determine known quantities and what's being asked.
  • Visualize the scenario with diagrams or sketches, if possible.
  • Apply mathematical formulas and logical reasoning to find solutions.
For instance, in our exercise, we understood the initial conditions (mass, velocity, separation distance), applied conservation of angular momentum to find new angular speeds and calculated energies accordingly. This methodical approach not only provides the solutions but also deepens understanding of physical principles at work.

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Most popular questions from this chapter

In unit-vector notation, what is the torque about the origin on a jar of jalapeño peppers located at coordinates \((3.0 \mathrm{~m},-2.0 \mathrm{~m}\), \(4.0 \mathrm{~m})\) due to (a) force \(\vec{F}_{1}=(3.0 \mathrm{~N}) \hat{\mathrm{i}}-(4.0 \mathrm{~N}) \hat{\mathrm{j}}+(5.0 \mathrm{~N}) \hat{\mathrm{k}},(\mathrm{b})\) force \(\vec{F}_{2}=(-3.0 \mathrm{~N}) \hat{\mathrm{i}}-(4.0 \mathrm{~N}) \hat{\mathrm{j}}-(5.0 \mathrm{~N}) \hat{\mathrm{k}}\), and \((\mathrm{c})\) the vector sum of \(\vec{F}_{1}\) and \(\vec{F}_{2} ?(\mathrm{~d})\) Repeat part \((\mathrm{c})\) for the torque about the point with coordinates \((3.0 \mathrm{~m}, 2.0 \mathrm{~m}, 4.0 \mathrm{~m})\).

A solid ball rolls smoothly from rest (starting at height \(H=6.0 \mathrm{~m}\) ) until it leaves the horizontal section at the end of the track, at height \(h=2.0 \mathrm{~m}\). How far horizontally from point \(A\) does the ball hit the floor?

At time \(t=0\), a \(3.0 \mathrm{~kg}\) particle with velocity \(\vec{v}=(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(6.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) is at \(x=3.0 \mathrm{~m}, y=8.0 \mathrm{~m}\). It is pulled by a \(7.0 \mathrm{~N}\) force in the negative \(x\) direction. About the origin, what are (a) the particle's angular momentum, (b) the torque acting on the particle, and (c) the rate at which the angular momentum is changing?

\- 52 (co A cockroach of mass \(m\) lies on the rim of a uniform disk of mass \(4.00 m\) that can rotate freely about its center like a merrygo-round. Initially the cockroach and disk rotate together with an angular velocity of \(0.260 \mathrm{rad} / \mathrm{s}\). Then the cockroach walks halfway to the center of the disk. (a) What then is the angular velocity of the cockroach- disk system? (b) What is the ratio \(K / K_{0}\) of the new kinetic energy of the system to its initial kinetic energy? (c) What accounts for the change in the ki- \(\quad\) netic energy?

In unit-vector notation, what is the torque about the origin on a particle located at coordinates \((0,-4.0 \mathrm{~m}, 3.0 \mathrm{~m})\) if that torque is due to (a) force \(\vec{F}_{1}\) with components \(F_{1 x}=2.0 \mathrm{~N}, F_{1 y}=F_{1 z}=0\), and \((\mathrm{b})\) force \(\vec{F}_{2}\) with components \(F_{2 x}=0, F_{2 y}=2.0 \mathrm{~N}, F_{2 z}=4.0 \mathrm{~N} ?\)

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