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A small \(0.500-\mathrm{kg}\) 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 5.33 meters.

Step by step solution

01

Determine the Centripetal Force

The centripetal force \( F_c \) required to keep the object moving in a circle can be calculated using the formula: \[ F_c = m r \omega^2 \] where \( m = 0.500 \, \text{kg} \), \( r = 1.00 \, \text{m} \), and \( \omega = 6.28 \, \text{rad/s} \). Calculate \( F_c \):\[ F_c = 0.500 \, \text{kg} \times 1.00 \, \text{m} \times (6.28 \, \text{rad/s})^2 \] \[ F_c = 0.500 \, \text{kg} \times 1.00 \, \text{m} \times 39.4384 \, \text{rad}^2/\text{s}^2 \] \[ F_c = 19.7192 \, \text{N} \] This is the force required to maintain the current circular motion.
02

Establish the Relationship of Forces and Radius

To ensure the object's motion remains radial without breaking the string, we need to express the maximum allowed tension \( T \) in terms of the radius \( r \): \[ T = m r' \omega^2 \] where \( r' \) is the new radius after pulling the string. Set \( T = 105 \, \text{N} \), which is the maximum allowable tension. Rearrange the equation to solve for \( r' \): \[ r' = \frac{T}{m \omega^2} \]
03

Calculate the Smallest Possible Radius

Substitute the known values into the equation for \( r' \):\[ r' = \frac{105 \, \text{N}}{0.500 \, \text{kg} \times (6.28 \, \text{rad/s})^2} \]\[ r' = \frac{105 \, \text{N}}{0.500 \, \text{kg} \times 39.4384 \, \text{rad}^2/\text{s}^2} \]\[ r' = \frac{105 \, \text{N}}{19.7192 \, \text{N}} \]\[ r' = 5.3263 \, \text{m} \] This is the radius where the tension reaches its maximum allowable value.

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Key Concepts

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

Angular Speed
Angular speed is a measure of how quickly an object moves along a circular path. In this exercise, the object moves with an angular speed of \(6.28\,\text{rad/s}\). Angular speed is different from linear speed, which is how fast an object moves in a straight line.
The formula to calculate angular speed \(\omega\) is:
  • \(\omega = \frac{\theta}{t} \)
where \(\theta\) is the angle in radians that the object covers in time \(t\). This means angular speed tells you how many radians the object travels through in a unit of time.
Angular speed helps determine various aspects of circular motion, like the centripetal force needed to maintain the object’s path in a circle. For instance, if you increase the angular speed while keeping the radius constant, you'll need more force to keep the object moving in its circular path.
Tension in Physics
Tension is the force exerted by a string or rope pulled tight by forces acting from opposite ends. In our exercise, the tension in the string must not exceed \(105\,\text{N}\). This tension is what keeps the object moving in its circular path and prevents it from flying outward.
When we talk about tension, we’re often concerned with the maximum value that the string can handle without breaking. The tension in this scenario is derived from the centripetal force formula:
  • \( T = m r \omega^2 \)
Where \(T\) is the tension, \(m\) is the mass of the object, \(r\) is the radius, and \(\omega\) is the angular speed. As the radius decreases, the tension must increase to provide the centripetal force needed for smaller circles.
It is important to calculate this accurately in real-world scenarios, because exceeding the tension limit can result in the string breaking, which could cause the object to fly off its intended path.
Circular Motion
Circular motion refers to the movement of an object along the circumference of a circle or rotation along a circular path. For the object in this exercise, circular motion is maintained by a force directed toward the center of the circle, known as the centripetal force.
As the object moves in its path, several forces act upon it:
  • The centripetal force pulls the object inward to maintain its path.
  • The tension in the string provides the necessary centripetal force.
Without these forces, the object would not continue its circular path and instead move off in a straight line due to inertia. It is crucial in circular motion to keep track of both the forces and the speed to ensure that the object can safely maintain its course.
Radius Calculation
The radius calculation is key in understanding the limits of circular motion, especially when maximum tension is a constraint, as it is in this example. Given:
  • The maximum tension, \(T = 105\,\text{N}\)
  • The object’s mass, \(m = 0.500\,\text{kg}\)
  • The angular speed, \(\omega = 6.28\,\text{rad/s}\)
You can calculate the smallest radius \(r'\) the object can maintain without breaking the string by rearranging the tension formula:\\[ r' = \frac{T}{m \omega^2} \] Calculating further, substituting in the values:
  • \( r' = \frac{105\,\text{N}}{0.500\,\text{kg} \times (6.28\,\text{rad/s})^2} \)
  • \( r' = 5.3263\,\text{m} \)
This radius represents the smallest circle the object can move in without the string breaking due to reached tension limits. This understanding aids in safety and design in practical scenarios involving rotations and swings.

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Most popular questions from this chapter

Concept Questions The drawing shows two identical systems of objects; each consists of three small balls (masses \(m_{1}, m_{2}\), and \(m_{3}\) ) connected by massless rods. In both systems the axis is perpendicular to the page, but it is located at a different place, as shown. (a) Do the systems necessarily have the same moments of inertia? If not, why not? (b) The same force of magnitude \(F\) is applied to the same ball in each system (see the drawing). Is the magnitude of the torque created by the applied force greater for system A or for system B? Or is the magnitude the same in the two cases? Explain. (c) The two systems start from rest. Will system A or system B have the greater angular speed at the same later time? Or will they have the same angular speeds? Justify your answer. Problem The masses of the balls are \(m_{1}=9.00 \mathrm{~kg}, m_{2}=6.00 \mathrm{~kg}\), and \(m_{3}=7.00 \mathrm{~kg}\). The magnitude of the force is \(F=424 \mathrm{~N}\). (a) For each of the two systems, determine the moment of inertia about the given axis of rotation. (b) Calculate the torque (magnitude and direction) acting on each system. (c) Both systems start from rest, and the direction of the force moves with the system and always points along the \(4.00-\mathrm{m}\) rod. What is the angular velocity of each system

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