/*! 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 113 One December identical twins Jen... [FREE SOLUTION] | 91Ó°ÊÓ

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One December identical twins Jena and Jackie are playing on a large merry-go- round (a disk mounted parallel to the ground, on a vertical axle through its center) in their school playground in northern Minnesota. Each twin has mass 30.0 \(\mathrm{kg}\) . The icy coating on the merry-go-round surface makes it frictionless. The merry-go-round revolves at a constant rate as the twins ride on it. Jena, sitting 1.80 \(\mathrm{m}\) from the center of the merry-go-round, must hold on to one of the metal posts attached to the merry-go-round with a horizontal force of 60.0 \(\mathrm{N}\) to keep from sliding off. Jackie is sitting at the edge, 3.60 \(\mathrm{m}\) from the center. (a) With what horizontal force must Jackie hold on to keep from falling off? (b) If Jackie falls off, what will be her horizontal velocity when she becomes airborne?

Short Answer

Expert verified
(a) 120 N; (b) 7.2 m/s

Step by step solution

01

Calculate Centripetal Force for Jena

To find the centripetal force exerted on Jena, use the formula:\[F_c = m \cdot a_c\]where \(F_c\) is the centripetal force, \(m\) is the mass, and \(a_c\) is the centripetal acceleration. Jena needs 60 N force to stay on the merry-go-round. Therefore, the centripetal acceleration \(a_c\) can be calculated using:\[a_c = \frac{F_c}{m} = \frac{60 \text{ N}}{30.0 \text{ kg}} = 2 \text{ m/s}^2\]
02

Calculate Angular Velocity

Since centripetal acceleration \(a_c\) also equals \(r \cdot \omega^2\) where \(r\) is the radius (distance from center) and \(\omega\) is the angular velocity, we use Jena's position:\[a_c = r \cdot \omega^2\]Substituting for \(a_c\) and \(r\) for Jena:\[2 = 1.80 \cdot \omega^2 \Rightarrow \omega^2 = \frac{2}{1.80} \Rightarrow \omega = \sqrt{\frac{2}{1.80}}\]
03

Calculate Centripetal Force for Jackie

Jackie is sitting at 3.60 m from the center. Using \(a_c = r \cdot \omega^2\), and knowing \(\omega\) from Jena's calculation:\[a_c = 3.60 \times \frac{2}{1.80} = 4 \text{ m/s}^2\]The force required for Jackie:\[F_{c, Jackie} = m \cdot a_c = 30.0 \times 4 = 120 \text{ N}\]
04

Calculate Horizontal Velocity for Jackie when Airborne

When Jackie falls off, she continues moving with the linear velocity given by \(v = \omega \cdot r\). From previous steps, \(\omega^2 = \frac{2}{1.80}\), hence:\[v = \sqrt{\frac{2}{1.80}} \times 3.60\]Simplifying gives:\[v = \sqrt{\frac{4}{1.80}} \times 3.60 = 2.0 \times 3.60 = 7.2 \text{ m/s}\]

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how quickly an object rotates around a central point or axis. It describes the rate of rotation and is usually represented by the Greek letter \( \omega \). Angular velocity is different from linear velocity because it specifically measures rotational motion.
In the problem, we use the concept of angular velocity to determine how the merry-go-round spins as the twins hold on. For Jena, sitting at a radius of 1.80 meters, her centripetal acceleration \( a_c \) is linked to the angular velocity using the formula:
  • \( a_c = r \cdot \omega^2 \)
We calculated Jena’s centripetal acceleration as 2 m/s², and inserted it into the formula to find \( \omega \). Thus, \( \omega = \sqrt{\frac{2}{1.80}} \).
This relationship helps us determine the spinning rate essential to find the forces acting on both Jena and Jackie as they cling to the merry-go-round.
Centripetal Acceleration
Centripetal acceleration is the rate at which an object's velocity changes as it moves along a circular path. Even at a constant speed, as in our merry-go-round problem, a changing direction equals acceleration. This type of acceleration is always directed towards the center of the circle in which the object is moving.
For both Jena and Jackie, knowing the centripetal acceleration is critical to assess the force required to keep them on the merry-go-round. Jena's centripetal acceleration is calculated using the force she exerts to hold on:
  • \( a_c = \frac{F_c}{m} \)
Where \( F_c = 60 \text{ N} \) and \( m = 30.0 \text{ kg} \), resulting in \( a_c = 2 \text{ m/s}^2 \).
With Jena's centripetal acceleration known, we are able to calculate the angular velocity which is crucial for solving Jackie's problem.
Physics Problem Solving
Physics problems, like the merry-go-round scenario, often require applying concepts systematically to find the desired results. In our case, we used step-by-step reasoning to solve for forces and velocities. We began with what we know and applied relevant formulas to solve for what we didn’t know.
Effective problem-solving in physics involves:
  • Understanding the situation and visualizing the physical setup.
  • Identifying known and unknown variables from the problem statement.
  • Using appropriate physics equations and formulas to link the knowns to the unknowns.
  • Applying logical mathematical steps to derive solutions.
In our exercise, identifying the forces and acceleration helped calculate the centripetal force Jackie needs to resist and the velocity she would have if she became airborne.

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Most popular questions from this chapter

A block with mass \(m_{1}\) is placed on an inclined plane with slope angle \(\alpha\) and is connected to a second hanging block with mass \(m_{2}\) by a cord passing over a small, frictionless pulley (Fig. P5.68). The coefficient of static friction is \(\mu_{\mathrm{s}}\) and the coefficient of kinetic friction is \(\mu_{\mathrm{k}}\) (a) Find the mass \(m_{2}\) for which block \(m_{1}\) moves up the plane at constant speed once it is set in motion. (b) Find the mass \(m_{2}\) for which block \(m_{1}\) moves down the plane at constant speed once it is set in motion. (c) For what range of values of \(m_{2}\) will the blocks remain at rest if they are released from rest?

A 45.0 -kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that the crate just begins to move when your force exceeds 313 \(\mathrm{N}\) . After that you must reduce your push to 208 \(\mathrm{N}\) to keep it moving at a steady 25.0 \(\mathrm{cm} / \mathrm{s}\) (a) What are the coefficients of static and kinetic friction between the crate and the floor? (b) What push must you exert to give it an acceleration of 1.10 \(\mathrm{m} / \mathrm{s}^{2} ?\) (c) Suppose you were performing the same experiment on this crate but were doing it on the moon instead, where the acceleration due to gravity is 1.62 \(\mathrm{m} / \mathrm{s}^{2}\) . (i) What magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)?

A rock with mass \(m=3.00\) kg falls from rest in a viscous medium. The rock is acted on by a net constant downward force of 18.0 \(\mathrm{N}\) (a combination of gravity and the buoyant force exerted by the medium) and by a fluid resistance force \(f=k v\) where \(v\) is the speed in \(\mathrm{m} / \mathrm{s}\) and \(k=2.20 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}(\) see Section 5.3\() .\) (a) Find the initial acceleration \(a_{0 .}\) (b) Find the acceleration when the speed is 3.00 \(\mathrm{m} / \mathrm{s} .\) (c) Find the speed when the acceleration equals 0.1\(a_{0}\) (d) Find the terminal speed \(v_{\mathrm{t}}\) . (e) Find the coordinate, speed, and acceleration 2.00 s after the start of the motion. (f) Find the time required to reach a speed of 0.9\(v_{t} .\)

A 15.0 -kg load of bricks hangs from one end of a rope that passes over a small, frictionless pulley. A 28.0 kg counterweight is suspended from the other end of the rope, as shown in Fig. E5.15. The system is released from rest. (a) Draw two free-body diagrams, one for the load of bricks and one for the counterweight. (b) What is the magnitude of the upward acceleration of the load of bricks? (c) What is the tension in the rope while the load is moving? How does the tension compare to the weight of the load of bricks? To the weight of the counterweight?

A flat (unbanked) curve on a highway has a radius of 220.0 \(\mathrm{m} .\) A car rounds the curve at a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) . (a) What is the minimum coefficient of friction that will prevent sliding? (b) Suppose the highway is icy and the coefficient of friction between the tires and pavement is only one-third what you found in part (a). What should be the maximum speed of the car so it can round the curve safely?

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