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An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axle. If a constant net torque of \(5,00 \mathrm{~N} \cdot \mathrm{m}\) is applied to the tire for \(2.00 \mathrm{~s}\), the angular speed of the tire increases from 0 to 100 rev \(/\) min. The external torque is then removed, and the wheel is brought to rest in \(125 \mathrm{~s}\) by friction in its bearings. Compute (a) the moment of inertia of the wheel about the axis of rotation, (b) the friction torque, and (c) the total number of revolutions made by the wheel in the \(125 \mathrm{~s}\) time interval.

Short Answer

Expert verified
(a) \(\frac{15}{\pi} \text{ kg} \cdot \text{m}^2\), (b) \(-0.4 \text{ N} \cdot \text{m}\), (c) 156.25 revolutions.

Step by step solution

01

Convert Angular Speed from RPM to Radians per Second

First, convert the final angular speed from revolutions per minute (rev/min) to radians per second (rad/s). We know that 1 revolution is \(2\pi\) radians and 1 minute is 60 seconds. So, the conversion is:\[\omega = 100 \text{ rev/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{200\pi}{60} \text{ rad/s} = \frac{10\pi}{3} \text{ rad/s}\]
02

Calculate the Angular Acceleration

Using the formula \(\tau = I\alpha\) where \(\tau\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration, find \(\alpha\) first since it's needed for later calculations:\[\alpha = \frac{\Delta \omega}{\Delta t} = \frac{\frac{10\pi}{3} \text{ rad/s} - 0}{2.00 \text{ s}} = \frac{5\pi}{3} \text{ rad/s}^2\]
03

Calculate the Moment of Inertia

Rearrange the formula \(\tau = I\alpha\) to solve for \(I\), the moment of inertia:\[I = \frac{\tau}{\alpha} = \frac{5.00 \text{ N} \cdot \text{m}}{\frac{5\pi}{3} \text{ rad/s}^2} = \frac{15}{\pi} \text{ kg} \cdot \text{m}^2\]
04

Calculate the Friction Torque

After the initial acceleration period, the wheel is brought to rest by friction in its bearings over a duration of 125s. Use the deceleration formula: \(-\alpha_{friction} = \frac{-\omega}{t}\) to find the frictional angular deceleration, then use it in \(\tau_{friction} = I \times \alpha_{friction}\).\[\alpha_{friction} = -\frac{\frac{10\pi}{3} \text{ rad/s}}{125 \text{ s}} = -\frac{2\pi}{75} \text{ rad/s}^2\]Using this to find \(\tau_{friction}\):\[\tau_{friction} = I \alpha_{friction} = \left(\frac{15}{\pi}\right)\left(-\frac{2\pi}{75}\right) = -0.4 \text{ N} \cdot \text{m}\]
05

Calculate the Total Number of Revolutions

During deceleration, calculate the total angular displacement using:\[\theta = \omega t + \frac{1}{2}\alpha t^2 = \left(\frac{10\pi}{3}\right)(125) + \frac{1}{2}\left(-\frac{2\pi}{75}\right)(125)^2\]Simplify to get:\[\theta = \frac{1250\pi}{3} - \frac{625\pi}{6} = \frac{2500\pi - 625\pi}{6} = \frac{1875\pi}{6} \text{ radians}\]Convert radians to revolutions:\[\frac{1875\pi}{6} \times \frac{1 \text{ rev}}{2\pi \text{ rad}} = \frac{1875}{12} = 156.25 \text{ revolutions}\]

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Key Concepts

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

Angular Acceleration
Angular acceleration is a key concept when analyzing rotational motion. Just as linear acceleration measures how quickly an object's velocity changes, angular acceleration measures how quickly an object's rotational speed changes. Specifically, it is the change of angular velocity over time and is denoted by the symbol \( \alpha \). This concept is crucial in determining how fast a spinning object, like a bicycle wheel, reaches its desired rotational speed.
When you apply a constant net torque \( \tau \) to an object, it causes angular acceleration according to the formula \( \tau = I \alpha \), where \( I \) is the moment of inertia. The moment of inertia is like the rotational equivalent of mass in linear motion—it's a measure of how much an object resists changes to its rotational speed.
In the given exercise, the bicycle wheel initially accelerates from rest to a certain angular speed. Using the net torque (\( 5.00 \, \mathrm{N} \cdot \mathrm{m} \)) and time applied (2.00 seconds), we can find the angular acceleration using the rearranged formula \( \alpha = \frac{\Delta \omega}{\Delta t} \), where \( \Delta \omega \) is the change in angular velocity. This helps us understand how much rotational force was needed to achieve the final speed of the wheel.
Friction Torque
Friction torque is the rotational equivalent of friction that acts against the motion of an object, eventually bringing it to a halt. When considering how an object like a bicycle wheel slows down, friction torque becomes very important. This is because friction in the bearings or other parts of the system creates a resisting torque that decreases the wheel's speed over time.
Friction torque can be determined through the formula \( \tau_{\text{friction}} = I \times \alpha_{\text{friction}} \), where \( \alpha_{\text{friction}} \) is the angular deceleration or the decrease in angular velocity over time as the wheel slows down due to friction. In the exercise, after the applied torque is removed, the friction torque within 125 seconds halts the wheel’s motion, illustrating the effect of friction.
  • To calculate \( \alpha_{\text{friction}} \), use \( -\alpha_{\text{friction}} = \frac{-\omega}{t} \), where \( \omega \) is the initial angular speed.
  • This offers insight into how the system dissipates energy until rest, highlighting the importance of friction in realistic scenarios where rotation is involved.
Understanding these concepts helps students apply real-world physics to technical systems like bike wheels.
Angular Displacement
Angular displacement measures how much an object has rotated during a certain period and it is usually measured in radians. It is different from angular speed or velocity because it focuses solely on the extent or degree of the rotation rather than how fast the rotation occurs.
In the context of the exercise, once the external torque is removed, understanding the total number of revolutions the wheel completes before coming to a stop is imperative. This is where angular displacement comes into play. You use the formula \( \theta = \omega t + \frac{1}{2} \alpha t^2 \) to calculate it, where \( \omega \) is the angular velocity and \( \alpha \) is the angular acceleration or deceleration. This formula gives you the total radians through which the wheel has rotated during its slowing down phase.
To convert this angular displacement from radians to actual revolutions—a more practical unit for something like a bicycle wheel—you divide by \( 2\pi \) since there are \( 2\pi \) radians in one full revolution.
  • This measured displacement not only provides data on the wheel's performance but is also useful for students to translate theoretical physics into mechanical applications.
  • It also serves as a bridge between linear and rotational motion concepts, fostering a deeper understanding of how these dynamics interact.

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Most popular questions from this chapter

Atwood's machine. Figure 10.77 illustrates an Atwood's machine. Find the linear accelerations of blocks \(A\) and \(B,\) the angular acceleration of the wheel \(C\), and the tension in each side of the cord if there is no slipping between the cord and the surface of the wheel. Let the masses of blocks \(A\) and \(B\) be \(4.00 \mathrm{~kg}\) and \(2.00 \mathrm{~kg}\). respectively. the moment of inertia of the wheel about its axis be \(0.300 \mathrm{~kg} \cdot \mathrm{m}^{2},\) and the radius of the wheel be \(0.120 \mathrm{~m}\).

A large wooden turntable in the shape of a flat disk has a radius of \(2.00 \mathrm{~m}\) and a total mass of \(120 \mathrm{~kg}\). The turntable is initially rotating at \(3.00 \mathrm{rad} / \mathrm{s}\) about a vertical axis through its center. Suddenly, a \(70.0 \mathrm{~kg}\) parachutist makes a soft landing on the tumtable at a point on its outer edge. Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.)

A solid uniform sphere of mass \(5 \mathrm{~kg}\) and radius \(0.1 \mathrm{~m}\) rolls down an inclined plane without slipping. If the sphere's center of mass moves through a vertical distance of \(1.5 \mathrm{~m}\), what is the rotational work done on the sphere by the static-friction force?

A thin. light string is wrapped around the rim of a \(4.00 \mathrm{~kg}\) solid uniform disk that is \(30.0 \mathrm{~cm}\) in diameter. A person pulls on the string with a constant force of \(100.0 \mathrm{~N}\) tangent to the disk, as shown in Figure \(10.53 .\) The disk is not attached to arything and is free to move and hum. (a) Find the angular acceleration of the disk about its center of mass and the linear acceleration of its center of mass. (b) If the disk is replaced by a hollow thin-walled cylinder of the same mass and diameter, what will be the accclerations in part (a)?

A cord is wrapped around the rim of a wheel \(0.250 \mathrm{~m}\) in radius, and a steady pull of \(40.0 \mathrm{~N}\) is exerted on the cord. The wheel is mounted on frictionless bearings on a horizontal shaft through its center. The moment of incrtia of the wheel about this shaft is \(5.00 \mathrm{~kg} \cdot \mathrm{m}^{2}\). Compute the angular acceleration of the wheel.

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