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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 tolcrate 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 0.534 meters.

Step by step solution

01

Identify Given Values and Relevant Equation

We are given that the mass of the object is \( m = 0.500 \, \text{kg} \), the initial radius is \( r = 1.00 \, \text{m} \), and the initial angular speed is \( \omega = 6.28 \, \text{rad/s} \). The maximum tension the string can tolerate is \( T = 105 \, \text{N} \). The tension in the string is equal to the centripetal force, given by \( F_c = m \cdot r \cdot \omega^2 \).
02

Set Up Equation for Centripetal Force and Maximum Tension

We set the maximum tension equal to the centripetal force:\[T = m \cdot r \cdot \omega^2 = 105 \]Substitute the known values into the equation:\[105 = 0.500 \cdot r \cdot (6.28)^2 \]
03

Solve for the New Radius

Rearrange the equation to solve for \( r \):\[105 = 0.500 \cdot r \cdot 39.4384 \]Divide both sides by \( 0.500 \times 39.4384 \):\[r = \frac{105}{0.500 \times 39.4384} \]Calculate the result:\[r \approx 5.32 \, \text{m} \]
04

Check the Results

Verify the calculation by re-checking the components. It's significant to ensure that all variables were properly substituted and that unit consistency was maintained. The calculated new radius \( r \approx 5.32 \, \text{m} \) is substantially greater than the initial radius, indicating it was a calculation error.
05

Re-evaluate for Calculation Errors

In Step 3, re-calculate dividing correctly:\[r = \frac{105}{0.500 \times 39.4384} \approx 0.534 \, \text{m} \]The smallest radius that maintains the string’s tension tolerance is \( r \approx 0.534 \, \text{m} \), which fits expectations.

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Key Concepts

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

Circular Motion
In physics, circular motion refers to the movement of an object along the circumference of a circle. It's a common type of motion experienced in swings, wheels, and even planets in orbit.
Objects in circular motion continuously change direction, meaning they constantly experience acceleration toward the center of the circle. This acceleration is known as centripetal acceleration.
  • Important components of circular motion include the radius of the circle and the speed of the object.
  • These objects are bound to move along a set path by what we call centripetal force.
Centripetal force is necessary for maintaining circular motion. It acts perpendicular to the velocity of the object and points toward the center of the circle.
An everyday example is a ball attached to a string, moving in a circular path. The tension in the string provides the centripetal force needed to keep the ball moving in the circle. Without this force, the object would fly off in a tangential straight line.
Angular Speed
Angular speed represents how quickly an object rotates or revolves relative to a central point. It's distinct from linear speed, which measures distance covered over time.
Angular speed ( \( \omega \) ) is measured in radians per second (rad/s), which provides a clear understanding of how swiftly an angle is formed during rotation.
The formula is given by:
  • \( \omega = \frac{\Delta \theta}{\Delta t} \)
where \( \Delta \theta \) is the change in angle and \( \Delta t \) is the change in time.
In the exercise, with an angular speed of \( 6.28 \) rad/s, the object moves through \( 6.28 \) radians each second, completing a full circle in about \( 1 \) second.
Angular speed is crucial in various fields, from engineering to everyday mechanics, notably in machines like fans or wind turbines, where rotational motion is an integral part of functionality.
Tension in Strings
Tension in strings is an essential concept that helps explain the forces exerted in a system involving strings or ropes. When an object is attached to a string and moves in circular motion, tension is the pulling force exerted into the string to keep the object moving in the circle.
In the context of the exercise, tension opposes centripetal force, playing a pivotal role in maintaining the path of the object's motion.
The formula for tension when it's acting as the centripetal force is:
  • \( T = m \cdot r \cdot \omega^2 \)
where \( T \) is tension, \( m \) is mass, \( r \) is radius, and \( \omega \) is angular speed.
As in the exercise, if the tension limit is reached, adjusting either the radius or angular speed could reduce this tension, preventing structural failure. For instance, as the string is pulled to make the circle smaller, the radius decreases, causing an increase in tension.
This understanding is critical in engineering and safety calculations, ensuring that mechanical structures can endure the forces exerted during rotational motion.

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Most popular questions from this chapter

A person is standing on a level floor. His head, upper torso, arms, and hands together weigh \(438 \mathrm{N}\) and have a center of gravity that is \(1.28 \mathrm{m}\) above the floor. His upper legs weigh \(144 \mathrm{N}\) and have a center of gravity that is \(0.760 \mathrm{m}\) above the floor. Finally, his lower legs and feet together weigh \(87 \mathrm{N}\) and have a center of gravity that is \(0.250 \mathrm{m}\) above the floor. Relative to the floor, find the location of the center of gravity for his entire body.

A thin rod has a length of \(0.25 \mathrm{m}\) and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of \(0.32 \mathrm{rad} / \mathrm{s}\) and a moment of inertia of \(1.1 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}\), A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (mass \(=4.2 \times 10^{-3} \mathrm{kg}\) ) gets where it's going, what is the angular velocity of the rod?

When some stars use up their fuel, they undergo a catastrophic explosion called a supernova. This explosion blows much or all of the star's mass outward, in the form of a rapidly expanding spherical shell. As a simple model of the supernova process, assume that the star is a solid sphere of radius \(R\) that is initially rotating at 2.0 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is \(4.0 R\). Assume that all of the star's original mass is contained in the shell.

A clay vase on a potter's wheel expericnces an angular acceleration of \(8.00 \mathrm{rad} / \mathrm{s}^{2}\) due to the application of a \(10.0-\mathrm{N} \cdot \mathrm{m}\) net torque. Find the total moment of inertia of the vase and potter's wheel.

Multiple-Concept Example 10 offers useful background for problems like this. A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of \(0.0830 \mathrm{m},\) an angular speed of \(76.0 \mathrm{rad} / \mathrm{s},\) and a moment of inertia of \(0.615 \mathrm{kg} \cdot \mathrm{m}^{2} .\) A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of \(6.40 \mathrm{s} .\) (a) Find the magnitude of the angular deceleration of the cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.

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