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Chapter 10: Problem 99

A small ball with mass \(1.30 \mathrm{~kg}\) is mounted on one end of a rod \(0.780 \mathrm{~m}\) long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at 5010 rev/min. (a) Calculate the rotational inertia of the system about the axis of rotation. (b) There is an air drag of \(2.30 \times 10^{-2} \mathrm{~N}\) on the ball, directed opposite its motion. What torque must be applied to the system to keep it rotating at constant speed?

Short Answer

Expert verified
(a) Rotational inertia \(I = 0.78948\, \text{kg} \cdot \text{m}^2\). (b) Torque needed \(\tau = 0.01794 \, \text{N} \cdot \text{m}\).

Step by step solution

01

Calculate the angular velocity

First, we need to calculate the angular velocity in radians per second. Given that the system rotates at 5010 revolutions per minute (rev/min), we need to convert this to radians per second (rad/s).We know:- 1 revolution equals \(2\pi\) radians- 60 seconds equals 1 minuteTherefore,\[\omega = 5010 \text{ revolutions/min} \times \frac{2\pi \text{ radians}}{1 \text{ revolution}} \times \frac{1 \text{ min}}{60 \text{ seconds}}\]Calculate \(\omega\) to find the angular velocity.
02

Determine the rotational inertia

The rotational inertia \(I\) of the system about the axis of rotation, for a point mass, is given by:\[I = mr^2\]where:- \(m = 1.30 \text{ kg}\) is the mass of the ball- \(r = 0.780 \text{ m}\) is the length of the rod (the distance from the axis of rotation to the point mass)Substitute these values into the formula to calculate \(I\).
03

Calculate torque due to air drag

The torque \(\tau\) needed to counteract air drag is calculated using the relation:\[\tau = F \cdot r\]where:- \(F = 2.30 \times 10^{-2} \text{ N}\) is the air drag force- \(r = 0.780 \text{ m}\)Substitute the given values to find the required torque.

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Key Concepts

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

Understanding Angular Velocity
Angular velocity is a measure of how fast an object rotates or spins around an axis. Unlike linear velocity, which tells us how fast an object is moving in a straight line, angular velocity focuses on rotational speed.
It's measured in radians per second (rad/s), a unit that describes how many radians an object covers in one second. Radians are a natural unit for measuring angles, where one complete circle equals approximately 6.283 radians (or exactly 2蟺 radians).
  • 1 revolution equates to 2蟺 radians.
  • Given: 5010 revolutions per minute (rev/min).
  • To convert it to rad/s, multiply by 2蟺 and divide by 60.
This conversion helps us understand the system鈥檚 rotation in terms of time, making calculations like rotational inertia more manageable.
The Role of Torque in Rotational Motion
Torque is akin to force in linear motion, but in the world of rotation. Imagine opening a door: when you push farther from the hinges, the door swings open more easily. This is torque at work 鈥� the effectiveness of a force to rotate an object around an axis.
In mathematical terms, torque (\( \tau \)) is defined as the product of force and the distance from the axis of rotation:\[\tau = F \cdot r\]
  • Here, \( F \) is the applied force, which can oppose motion, just like air drag.
  • \( r \) is the distance from the axis of rotation.
In this context, the air drag force must be countered by applying torque to maintain constant rotational speed. So, when a ball rotates on a rod, any external force like air resistance needs a compensating torque. This ensures that the rotational speed remains unchanged, highlighting how torque sustains or alters an object's state of rotation.
Impact of Air Drag on Rotational Systems
Air drag, or air resistance, is a type of friction that acts against the motion of objects moving through the air. It plays a significant role in determining how much torque is needed to maintain rotational velocity in systems like the one described in our exercise.
Air drag depends on several factors, such as:
  • Shape and cross-sectional area of the object.
  • Speed of the object moving through the air.
  • Air density.
In the given system, as the ball rotates, it encounters a constant air drag force. To keep the ball spinning at a constant angular velocity, the applied torque must be equal and opposite to the air drag's torque. Understanding air drag helps in designing systems that either minimize energy use by reducing drag or efficiently counteract it using just the right amount of torque. As a result, learning how to calculate and manage air drag is essential for achieving and maintaining desired rotational motions.

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Most popular questions from this chapter

A disk rotates at constant angular acceleration, from angular position \(\theta_{1}=10.0 \mathrm{rad}\) to angular position \(\theta_{2}=70.0 \mathrm{rad}\) in \(6.00 \mathrm{~s}\). Its angular velocity at \(\theta_{2}\) is \(15.0 \mathrm{rad} / \mathrm{s}\). (a) What was its angular velocity at \(\theta_{1} ?\) (b) What is the angular acceleration? (c) At what angular position was the disk initially at rest? (d) Graph \(\theta\) versus time \(t\) and angular speed \(\omega\) versus \(t\) for the disk, from the beginning of the motion \((\operatorname{let} t=0\) then \()\).

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The masses and coordinates of four particles are as follows: \(50 \mathrm{~g}, x=2.0 \mathrm{~cm}, y=2.0 \mathrm{~cm} ; 25 \mathrm{~g}, x=0, y=4.0 \mathrm{~cm} ; 25 \mathrm{~g}, x=-3.0\) \(\mathrm{cm}, y=-3.0 \mathrm{~cm} ; 30 \mathrm{~g}, x=-2.0 \mathrm{~cm}, y=4.0 \mathrm{~cm} .\) What are the rota- tional inertias of this collection about the (a) \(x\), (b) \(y\), and (c) \(z\) axes? (d) Suppose the answers to (a) and (b) are \(A\) and \(B\), respectively. \(\quad\) Rotation Then what is the answer to (c) in axis terms of \(A\) and \(B\) ?c

Two uniform solid spheres have the same mass of \(1.65 \mathrm{~kg}\), but one has a radius of \(0.226 \mathrm{~m}\) and the other has a radius of \(0.854 \mathrm{~m}\). Each can rotate about an axis through its center. (a) What is the magnitude \(\tau\) of the torque required to bring the smaller sphere from rest to an angular speed of \(317 \mathrm{rad} / \mathrm{s}\) in \(15.5 \mathrm{~s} ?(\mathrm{~b})\) What is the magnitude \(F\) of the force that must be applied tangentially at the sphere's equator to give that torque? What are the corresponding values of (c) \(\tau\) and (d) \(F\) for the larger sphere?

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