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A \(1.0 \mathrm{kg}\) ball and a \(2.0 \mathrm{kg}\) ball are connected by a \(1.0-\mathrm{m}\) -long rigid, massless rod. The rod and balls are rotating clockwise about their center of gravity at 20 rpm. What torque will bring the balls to a halt in \(5.0 \mathrm{s} ?\)

Short Answer

Expert verified
Using the formulas for moment of inertia, angular velocity, and torque, we calculate the required torque to bring the rotating balls to a halt in 5.0 s. The detailed calculations are shown in the step-by-step solution above.

Step by step solution

01

Find Moment of Inertia for the System

To find the moment of inertia \(I\) for the system, note that the moment of inertia for a point mass \(m\) at a distance \(r\) from the rotation axis is given by \(I=mr^2\). The total moment of inertia of the system is the sum of the moments of inertia of the two balls. The mass of the first ball is 1.0 kg and the distance from the center of mass is 0.5 m. The mass of the second ball is 2.0 kg and the distance from the center of mass is also 0.5 m. Thus \(I = (1.0 \mathrm{kg}) * (0.5 \mathrm{m})^2 + (2.0 \mathrm{kg}) * (0.5 \mathrm{m})^2\).
02

Convert Rotations Per Minute to Rad/s

The angular velocity \( \omega \) of the system is initially given to be 20 rpm. Convert to rad/s by using the conversion factor of \( \frac{2\pi \mathrm{ rad }}{60 \mathrm{s}} \): \(\omega = 20 \frac{\mathrm{rotations}}{\mathrm{minute}} * \frac{2\pi \mathrm{rad}}{60 \mathrm{s}}\).\nThe final angular velocity is 0 rad/s, as the balls come to halt.
03

Calculate Angular Acceleration

The angular acceleration \( \alpha \) is the rate of change of angular velocity. It can be found using the formula \( \alpha = \frac{\omega_f - \omega_i}{t} \), where \( \omega_f \) is the final angular velocity, \( \omega_i \) is the initial angular velocity and \( t \) is 5.0 s, the time taken to halt the balls.
04

Calculate Required Torque

Now we can find the torque required to halt the balls using the equation \( \tau = I \cdot \alpha \), where \( I \) is the moment of inertia which we calculated in the first step and \( \alpha \) is the angular acceleration which we calculated in the third step.

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Key Concepts

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

Moment of Inertia
Moment of inertia, symbolized as \( I \), is a measure of an object's resistance to changes to its rotation. It can be thought of as the rotational equivalent of mass in linear motion. The moment of inertia depends on both the mass of the object and the distribution of that mass relative to the axis of rotation.

For point masses, the moment of inertia is given by the formula \( I = mr^2 \), where \( m \) is the mass of the object and \( r \) is the distance from the rotation axis. In the case of the exercise involving the balls and rod, the moment of inertia is calculated by adding the moments of inertia of the two balls, as the rod is considered massless. Each ball contributes differently to the rotational inertia because of their respective masses.

To assist in understanding, one can imagine moment of inertia as how difficult it would be to spin a wheel by pushing its rim; the further from the center you apply your force, or the heavier the rim, the harder it is to get the wheel spinning. This is because the moment of inertia is higher, so more torque is required to change its angular state.
Angular Velocity
Angular velocity, denoted by \( \omega \) (the Greek letter omega), is a vector quantity that represents the rate of rotation or angular displacement over time. The standard unit for angular velocity is radians per second (rad/s).

In our example, the system starts with an angular velocity of 20 rotations per minute (rpm). To use \( \omega \) in calculations, we first need to convert this to the standard unit. This is done using the conversion factor of \( \frac{2\pi \text{ rad }}{60 \text{ s}} \), since there are \( 2\pi \) radians in a complete rotation (360 degrees).

Understanding angular velocity is crucial when tackling problems related to rotational motion. It helps in predicting how fast an object will rotate in a certain amount of time. For instance, when you spin a top, you give it a certain angular velocity, determining how fast it spins. The conversion of units in the problem from rpm to rad/s is an essential step to solve for other quantities like angular acceleration and torque.
Torque
Torque, symbolized as \( \tau \), is a measure of the force that causes an object to rotate about an axis. The magnitude of torque depends on the force applied, the distance from the axis at which the force is applied, and the angle between the force and the lever arm.

The formula to calculate torque when a force is applied perpendicularly to the lever arm is \( \tau = rF \), where \( r \) is the lever arm, or the perpendicular distance from the axis of rotation to the line of action of the force, and \( F \) is the magnitude of the force. In the context of the rotating balls and rod, there's no specific force applied, but we can still calculate the torque required to bring the balls to a halt by using the relationship \( \tau = I \cdot \alpha \).

This equation directly ties torque to moment of inertia and angular acceleration (\( \alpha \)), demonstrating how torque, angular velocity, and moment of inertia are interconnected. Just as a greater force is required to stop a heavier moving object in a shorter time in linear dynamics, a greater torque must be applied to rapidly change the rotational motion of an object with greater moment of inertia.

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Most popular questions from this chapter

A bowling ball is far from uniform. Lightweight bowling balls are made of a relatively low-density core surrounded by a thin shell with much higher density. A 7.0 lb \((3.2 \mathrm{kg})\) bowling ball has a diameter of \(0.216 \mathrm{m} ; 0.196 \mathrm{m}\) of this is a \(1.6 \mathrm{kg}\) core, surrounded by a \(1.6 \mathrm{kg}\) shell. This composition gives the ball a higher moment of inertia than it would have if it were made of a uniform material. Given the importance of the angular motion of the ball as it moves down the alley, this has real consequences for the game. a. Model a real bowling ball as a \(0.196-\mathrm{m}\) -diameter core with mass \(1.6 \mathrm{kg}\) plus a thin \(1.6 \mathrm{kg}\) shell with diameter \(0.206 \mathrm{m}\) (the average of the inner and outer diameters). What is the total moment of inertia? b. How does your answer in part a compare to the moment of inertia of a uniform \(3.2 \mathrm{kg}\) ball with diameter \(0.216 \mathrm{m} ?\)

If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.72 rev/s, friction in the bearings causes the wheel to stop in just 12 s. If the moment of inertia of the wheel about its axle is \(0.30 \mathrm{kg} \cdot \mathrm{m}^{2},\) what is the magnitude of the frictional torque?

A \(1.5 \mathrm{kg}\) block and a \(2.5 \mathrm{kg}\) block are attached to opposite ends of a light rope. The rope hangs over a solid, frictionless pulley that is \(30 \mathrm{cm}\) in diameter and has a mass of \(0.75 \mathrm{kg}\). When the blocks are released, what is the acceleration of the lighter block?

To throw a discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant angular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about \(1.8 \mathrm{m}\). If the thrower takes \(1.0 \mathrm{s}\) to complete one revolution, starting from rest, what will be the speed of the discus at release?

U.S. nickels have a mass of \(5.00 \mathrm{g}\) and are \(1.95 \mathrm{mm}\) thick. If you stack 3 nickels on a table, how far above the table is their center of gravity?

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