/*! 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 60 A small 0.500 -kg object moves o... [FREE SOLUTION] | 91Ó°ÊÓ

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A small 0.500 -kg object moves on a frictionless horizontal table in a circular path of radius \(1.00 \mathrm{~m} .\) The angular speed is \(6.28 \mathrm{rad} / \mathrm{s} .\) The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than \(105 \mathrm{~N}\), what is the radius of the smallest possible circle on which the object can move?

Short Answer

Expert verified
The smallest possible radius is approximately 5.32 meters.

Step by step solution

01

Identify the centripetal force and its relation to tension

The tension in the string provides the centripetal force required to keep the object moving in a circular path. Thus, we can write the relation:\[T = F_c = m \cdot \omega^2 \cdot r\]where:- \(T\) is the tension,- \(m\) is the mass of the object,- \(\omega\) is the angular speed,- \(r\) is the radius of the circle.
02

Substitute known values into the equation

Given:- \(m = 0.500\, \text{kg}\)- \(\omega = 6.28\, \text{rad/s}\)- Maximum \(T = 105\, \text{N}\)Replace these values into the equation for tension:\[105 = 0.500 \cdot (6.28)^2 \cdot r \]
03

Solve for radius \(r\)

Rearrange the equation to find \(r\):\[r = \frac{105}{0.500 \cdot (6.28)^2}\]Calculate the value:\[r = \frac{105}{0.500 \cdot 39.4384} \]
04

Calculate the result

Continue by calculating the division:\[r = \frac{105}{19.7192} \approx 5.32 \, \text{meters}\]Thus, the smallest radius at which the object can move without exceeding the tension limit is approximately \(5.32 \, \text{m}\).

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Key Concepts

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

circular motion
Circular motion is the movement of an object along the circumference of a circle. This type of motion is characterized by a constant change in the direction of the object, even if its speed remains constant. This is what makes circular motion unique and often more complex to understand compared to linear motion.
When an object moves in a circle, it experiences continuous centripetal acceleration, which always points towards the center of the circle. This means that to keep an object moving in a circular path, a centripetal force is required to push or pull it inward. Without this force, the object would simply move off in a straight line, according to Newton's First Law of Motion.
In the context of the original exercise, the tension in the string provides this necessary centripetal force to keep the object in circular motion. As you imagine pulling the string tighter and decreasing the circle's radius, the object needs more inward force to maintain the same angular speed, highlighting the relationship between centripetal force and curved motion. Circular motion might seem straightforward, but it involves a delicate balance of forces and acceleration to keep the motion steady.
tension in physics
Tension in physics refers to the force exerted along a string, rope, cable, or similar object. It is a pulling force, exerted by each end of the string and transmitted between connected objects. Understanding tension is crucial when analyzing objects in motion especially in systems where strings or ropes are used to exert forces.
In our exercise scenario, the string connecting the rotating object to the center of the circle is under tension. This tension is not just a static force; it actually provides the centripetal force needed to keep the object moving in circular motion. The tension needs to be strong enough to maintain motion, but it must not exceed the string's strength limit, which in this case is 105 N.
Let’s think of it this way: as the object moves faster or the circle becomes smaller, the tension in the string increases. Imagine if the object is rotating so fast that the force acting on the string surpasses its limit. The string will break, and the object will fly off in a tangential direction. Tension here acts as a safety factor dictating the maximum feasible conditions for circular motion.
angular speed
Angular speed, sometimes referred to as angular velocity, describes how fast an object rotates or revolves around a central point. It is measured in radians per second (rad/s) and is a crucial descriptor of an object's rotation in physics.
In circular motion, angular speed tells us how quickly an object is moving around the circle. It's worth noting that unlike linear speed, which measures how fast an object moves along a straight path, angular speed is concerned with the path along a circle. It does not change whether the circle's diameter is large or small, as long as the change in angular position per unit time remains constant.
In the problem given, the angular speed is 6.28 rad/s, indicating a specific rate of rotation for the object as it moves along its circular path. Angular speed remains consistent as the radius varies unless acted upon by external forces. The relationship between linear speed \(v\), angular speed \(\omega\), and radius \(r\) is expressed by the formula \(v = \omega \cdot r\). This highlights that as the radius of the circle decreases, but angular speed remains the same, the linear speed also changes, affecting the tension in the string. Understanding angular speed helps us grasp how rotational dynamics work in practical situations like this exercise.

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Most popular questions from this chapter

In outer space two identical space modules are joined together by a massless cable. These modules are rotating about their center of mass, which is at the center of the cable, because the modules are identical (see the drawing). In each module, the cable is connected to a motor, so that the modules can pull each other together. The initial tangential speed of each module is \(v_{0}-17 \mathrm{~m} / \mathrm{s}\). Then they pull together until the distance between them is reduced by a factor of two. Determine the final tangential speed \(v_{\mathrm{f}}\) for each module.

Example 14 provides useful background for this problem. A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of \(125 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) When one person is standing at a distance of \(1.50 \mathrm{~m}\) from the center, the carousel has an angular velocity of \(0.600 \mathrm{rad} / \mathrm{s}\). However, as this person moves inward to a point located \(0.750 \mathrm{~m}\) from the center, the angular velocity increases to \(0.800 \mathrm{rad} / \mathrm{s}\). What is the person's mass?

A uniform door \((0.81 \mathrm{~m}\) wide and \(2.1 \mathrm{~m}\) high \()\) weighs \(140 \mathrm{~N}\) and is hung on two hinges that fasten the long left side of the door to a vertical wall. The hinges are \(2.1 \mathrm{~m}\) apart. Assume that the lower hinge bears all the weight of the door. Find the magnitude and direction of the horizontal component of the force applied to the door by (a) the upper hinge and (b) the lower hinge. Determine the magnitude and direction of the force applied by the door to (c) the upper hinge and (d) the lower hinge.

Concept Questions Two thin rods of length \(L\) are rotating with the same angular speed \(\omega\) (in \(\mathrm{rad} / \mathrm{s}\) ) about axes that pass perpendicularly through one end. \(\operatorname{Rod} \mathrm{A}\) is massless but has a particle of mass \(0.66 \mathrm{~kg}\) attached to its free end. Rod \(\mathrm{B}\) has a mass \(0.66 \mathrm{~kg}\), which is distributed uniformly along its length. (a) Which has the greater moment of inertia-rod A with its attached particle or rod B? (b) Which has the greater rotational kinetic energy? Account for your answers. Problem The length of each rod is \(0.75 \mathrm{~m}\), and the angular speed is \(4.2 \mathrm{rad} / \mathrm{s}\). Find the kinetic energies of rod A with its attached particle and of rod B. Make sure your answers are consistent with your answers to the Concept Questions.

A solid disk rotates in the horizontal plane at an angular velocity of \(0.067 \mathrm{rad} / \mathrm{s}\) with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is \(0.10 \mathrm{~kg} \cdot \mathrm{m}^{2}\). From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of \(0.40 \mathrm{~m}\) from the axis. The sand in the ring has a mass of \(0.50 \mathrm{~kg}\). After all the sand is in place, what is the angular velocity of the disk?

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