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Chapter 9: Problem 63

A uniform wheel in the shape of a solid disk is mounted on a frictionless axle at its center. The wheel has mass \(5.00 \mathrm{~kg}\) and radius \(0.800 \mathrm{~m} .\) A thin rope is wrapped around the wheel, and a block is suspended from the free end of the rope. The system is released from rest and the block moves downward. What is the mass of the block if the wheel turns through 8.00 revolutions in the first \(5.00 \mathrm{~s}\) after the block is released?

Short Answer

Expert verified
The mass of the block is approximately \(2.04 \) kg.

Step by step solution

01

Calculation of angular velocity

First of all, we need to calculate the angular velocity of the wheel in rad/s. Given that the wheel turns through 8.00 revolutions in 5.00 s, we convert revolutions to radians as \(1 \) revolution \(\ = 2\pi \) rad. Therefore, angular velocity \( \omega = \dfrac{2\pi \times n}{t} = \dfrac{2\pi \times 8}{5} = 10.05 \) rad/s.
02

Calculation of angular acceleration

The wheel starts from rest so the initial angular velocity \( \omega_i = 0 \) rad/s. We can assume the angular acceleration \( \alpha \) to be constant, so it can be calculated using the formula \( \omega = \omega_i + \alpha \times t \). After rearranging and plugging values, we get \( \alpha = \dfrac{\omega - \omega_i}{t} = \dfrac{10.05}{5} = 2.01 \) rad/s^2.
03

Calculation of linear acceleration

The linear acceleration of the block \( a \) is related to the angular acceleration of the wheel \( \alpha \) through the equation \( a = \alpha \times r \). Substituting the known values, we get \( a = 2.01 \times 0.8 = 1.608 \) m/s^2.
04

Calculation of the mass of the block

We can now calculate the mass of the block. It only has two forces acting on it: the tension in the rope \( T \) and the weight due to gravity \( W = m \times g \), where \( g \approx 9.81 \) m/s^2 is the acceleration due to gravity. These two forces create a net force that provides the acceleration for the block: \( T - W = m \times a \). However, we know that the tension in the rope \( T \) is also providing the torque \( \tau = I \times \alpha \) for the wheel, where \( I = \dfrac{1}{2} M r^2 \) is the moment of inertia for the wheel (modeled as a solid disk), with \( M = 5.00 \) kg the mass of the wheel and \( r = 0.800 \) m the radius. Therefore, we also have \( T = \dfrac{1}{2} M r \alpha \). Substituting this into the equation for the net force and solving for \( m \), we get \( m = \dfrac{1}{2} M r \alpha / g + a \) kg. After substitution of known values, we obtain \( m = \dfrac{1}{2} \times 5.00 \times 0.800 \times 2.01 / 9.81 + 1.608 \approx 2.04 \) kg.

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Key Concepts

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

Understanding Angular Velocity
Imagine a figure skater spinning on ice - as the skater draws arms closer to the body, the spin rate increases. This is a dramatic display of angular velocity, which in the context of physics, refers to the rate of change of angular position of an object. It's the rotational equivalent of linear velocity, but instead of meters per second, we measure it in radians per second (rad/s).

In the exercise with the rotating wheel, calculating angular velocity helps us understand how fast the wheel is spinning. This understanding is crucial in solving an array of problems, from the simple motion of a playground merry-go-round to the more complex behavior of celestial bodies in astronomy.

Conversion Factor

To establish the connection between linear and rotational motion, remember that one complete revolution is equal to \(2\pi\) radians. Utilizing this conversion allows us to translate an object's rotational motion to a numerical value that can be used in subsequent calculations, such as the determination of angular acceleration or the effects of forces acting at a distance from the pivot point.
The Role of Angular Acceleration
When the wheel's speed changes, we encounter angular acceleration - the rate at which the angular velocity changes over time. It is measured in radians per second squared (rad/s\textsuperscript{2}). Just like the way a car's speedometer indicates the increasing speed, angular acceleration gives us a measure of how quickly an object is speeding up or slowing down its spin.

In our exercise, to find the angular acceleration, we need to know the final angular velocity and the time over which the change occurs. With this, you can use the formula \(\alpha = (\omega - \omega_i) / t\), where \(\omega_i\) is the initial angular velocity, which, for an object starting from rest, is zero. Understanding the connection between angular acceleration and forces will help students tackle problems involving rotational dynamics.

Practical Application

Angular acceleration is not just theoretical. It's pivotal in designing transportation systems, understanding geophysical phenomena, and even in robotics where robotic arms often mimic the acceleration and deceleration patterns of human arms.
Moment of Inertia Explained
Now let's explore the moment of inertia, which is the rotational equivalent of mass in linear motion. It measures an object's resistance to changes in its rotational motion. Think about trying to push a merry-go-round - it's harder to get moving if it has heavy objects on it, right? That's because it has a greater moment of inertia.

In the exercise, the moment of inertia for a solid disk is given by the equation \(I = \frac{1}{2} M r^2\), where \(M\) is the mass of the wheel, and \(r\) is its radius. It's like distribution of mass in linear motion but for rotation. Different shapes have different moments of inertia, even if their masses are the same, because the distribution of mass relative to the axis of rotation matters.

Application in Calculations

The concept of moment of inertia is fundamental when analyzing the rotational motion of objects. It's used in calculus-based physics to solve complex problems involving torques, angular momentum, and kinetic energy in rotating systems. Understanding its application allows for the analysis of how different shapes and mass distributions affect rotational motion.

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Most popular questions from this chapter

A physics student of mass \(43.0 \mathrm{~kg}\) is standing at the edge of the flat roof of a building, \(12.0 \mathrm{~m}\) above the sidewalk. An unfriendly \(\operatorname{dog}\) is running across the roof toward her. Next to her is a large wheel mounted on a horizontal axle at its center. The wheel, used to lift objects from the ground to the roof, has a light crank attached to it and a light rope wrapped around it; the free end of the rope hangs over the edge of the roof. The student grabs the end of the rope and steps off the roof. If the wheel has radius \(0.300 \mathrm{~m}\) and a moment of inertia of \(9.60 \mathrm{~kg} \cdot \mathrm{m}^{2}\) for rotation about the axle, how long does it take her to reach the sidewalk, and how fast will she be moving just before she lands? Ignore friction in the axle.

You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of \(0.200 \mathrm{rev} / \mathrm{s}^{2}\). The design specifications call for it to have a rotational kinetic energy of \(240 \mathrm{~J}\) after it has turned through 30.0 revolutions. What should be the moment of inertia of the flywheel about its rotation axis?

A uniform disk with radius \(R=0.400 \mathrm{~m}\) and mass \(30.0 \mathrm{~kg}\) rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to \(\theta(t)=(1.10 \mathrm{rad} / \mathrm{s}) t+\left(6.30 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}\) What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100 rev?

A uniform sphere with mass \(M\) and radius \(R\) is rotating with angular speed \(\omega_{1}\) about a frictionless axle along a diameter of the sphere. The sphere has rotational kinetic energy \(K_{1}\). A thin-walled hollow sphere has the same mass and radius as the uniform sphere. It is also rotating about a fixed axis along its diameter. In terms of \(\omega_{1},\) what angular speed must the hollow sphere have if its kinetic energy is also \(K_{1},\) the same as for the uniform sphere?

Two metal disks, one with radius \(R_{1}=2.50 \mathrm{~cm}\) and mass \(M_{1}=0.80 \mathrm{~kg}\) and the other with radius \(R_{2}=5.00 \mathrm{~cm}\) and mass \(M_{2}=1.60 \mathrm{~kg},\) are welded together and mounted on a frictionless axis through their common center (Fig. \(\mathbf{P 9 . 7 9}\) ). (a) What is the total moment of inertia of the two disks? (b) A light string is wrapped around the edge of the smaller disk, and a \(1.50 \mathrm{~kg}\) block is suspended from the free end of the string. If the block is released from rest at a distance of \(2.00 \mathrm{~m}\) above the floor, what is its speed just before it strikes the floor? (c) Repeat part (b), this time with the string wrapped around the edge of the larger disk. In which case is the final speed of the block greater? Explain.
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