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Chapter 9: Problem 25

A flywheel with a radius of 0.300 \(\mathrm{m}\) starts from rest and accelerates with a constant angular acceleration of 0.600 \(\mathrm{rad} / \mathrm{s}^{2}\) . Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim (a) at the start; (b) after it has turned through \(60.0^{\circ} ;\) (c) after it has turned through \(120.0^{\circ} .\)

Short Answer

Expert verified
At the start, tangential acceleration is 0.180 m/s虏, and the resultant is the same. After 60掳, radial and resultant accelerations are 0.376 m/s虏 and 0.416 m/s虏. After 120掳, they are 0.758 m/s虏 and 0.780 m/s虏.

Step by step solution

01

Understanding Tangential Acceleration

The tangential acceleration (\(a_t\)) of a point on the rim of the flywheel is the product of the angular acceleration (\(\alpha\)) and the radius (\(r\)). It is given by the formula: \[a_t = \alpha \times r\]Substitute \(\alpha = 0.600 \, \mathrm{rad/s^2}\) and \(r = 0.300 \, \mathrm{m}\) to calculate \(a_t\).
02

Computation of Tangential Acceleration

Using the formula for tangential acceleration: \[a_t = 0.600 \, \mathrm{rad/s^2} \times 0.300 \, \mathrm{m} = 0.180 \, \mathrm{m/s^2}\]This is the tangential acceleration for all sections of the exercise since the angular acceleration is constant.
03

Determine Radial Acceleration at the Start

Radial acceleration (\(a_r\)) is calculated using the formula: \[a_r = \omega^2 \times r\]At the start, the angular velocity (\(\omega\)) is zero because the wheel starts from rest, so \(a_r = 0\).
04

Determine Resultant Acceleration at the Start

The resultant acceleration (\(a_{res}\)) is the vector sum of tangential and radial accelerations. Initially, since \(a_r = 0\), the resultant acceleration is equal to the tangential acceleration:\[a_{res} = a_t = 0.180 \, \mathrm{m/s^2}\]
05

Calculate Radial Acceleration after 60 Degrees

First, convert 60 degrees to radians: \[\theta = 60^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{\pi}{3} \, \mathrm{rad}\]Find the angular velocity using \[\omega^2 = \omega_0^2 + 2\alpha\theta\]With \(\omega_0 = 0\), \[\omega^2 = 2 \times 0.600 \, \mathrm{rad/s^2} \times \frac{\pi}{3}\approx 1.2566\]And hence, \[\omega \approx \sqrt{1.2566} \approx 1.12 \, \mathrm{rad/s}\]Using \(a_r = \omega^2 \times r\), we find:\[a_r = (1.12 \, \mathrm{rad/s})^2 \times 0.300 \, \mathrm{m} \approx 0.376 \, \mathrm{m/s^2}\]
06

Compute Resultant Acceleration after 60 Degrees

Use the formula for resultant acceleration: \[a_{res} = \sqrt{a_t^2 + a_r^2}\]Substitute the values found earlier: \[a_{res} = \sqrt{(0.180)^2 + (0.376)^2} \approx 0.416 \, \mathrm{m/s^2}\]
07

Calculate Radial Acceleration after 120 Degrees

Convert 120 degrees to radians: \[\theta = 120^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{2\pi}{3} \, \mathrm{rad}\]Find the angular velocity using \[\omega^2 = 2 \times 0.600 \, \mathrm{rad/s^2} \times \frac{2\pi}{3} \approx 2.5132\]And hence, \[\omega \approx \sqrt{2.5132} \approx 1.59 \, \mathrm{rad/s}\]Using \(a_r = \omega^2 \times r\), we find:\[a_r = (1.59 \, \mathrm{rad/s})^2 \times 0.300 \, \mathrm{m} \approx 0.758 \, \mathrm{m/s^2}\]
08

Compute Resultant Acceleration after 120 Degrees

Finally, use the formula for resultant acceleration:\[a_{res} = \sqrt{a_t^2 + a_r^2}\]Substitute: \[a_{res} = \sqrt{(0.180)^2 + (0.758)^2} \approx 0.780 \, \mathrm{m/s^2}\]

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Key Concepts

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

Understanding Angular Acceleration
Angular acceleration (\( \alpha \)) refers to the rate of change of angular velocity over time. It plays a significant role in rotational dynamics and is analogous to linear acceleration in linear motion. In our problem, a flywheel starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s虏.
This means that the angular velocity increases by 0.600 rad/s every second. Angular acceleration is a vector quantity, having both magnitude and direction. For commonly used rotational systems, usually only the magnitude is of concern.
  • Angular acceleration is typically measured in radians per second squared (rad/s虏).
  • Constant angular acceleration simplifies calculations as it allows the use of linear kinematic equations adapted for rotational motion.
  • It's instrumental in calculating other rotational quantities like tangential and radial accelerations.
Tangential Acceleration Explained
Tangential acceleration (\( a_t \)) is the linear acceleration experienced by a point on the rim of a rotating object. It is directly related to the angular acceleration. For instance, in our flywheel problem, the formula \( a_t = \alpha \times r \) helps to convert angular acceleration into linear terms. Here, \( r \) represents the radius of the flywheel.
Using the given angular acceleration (\( \alpha = 0.600 \, \mathrm{rad/s^2} \)) and radius (\( r = 0.300 \, \mathrm{m} \)), we found that \( a_t = 0.180 \, \mathrm{m/s^2} \), which remains constant for all states because the angular acceleration is constant.
  • This acceleration occurs along the edge of the circle created by the revolving point.
  • A critical aspect is its perpendicularity to the radial acceleration, accounting for the linear speed change as the point moves in a circle.
  • Tangential acceleration is vital for determining how quickly an object speeds up or slows down its rotation.
Understanding Radial Acceleration
Radial acceleration (\( a_r \)), also known as centripetal acceleration, acts towards the center of the circular path and is essential for maintaining the circular motion. Unlike tangential acceleration, radial acceleration depends on the square of the angular velocity (\( \omega \)), calculated with \( a_r = \omega^2 \times r \).
Initially, since the wheel begins from rest, the radial acceleration equals zero because \( \omega \) is zero. However, as the flywheel turns, such as at 60掳 or 120掳, the angular velocity increases, resulting in radial acceleration values like 0.376 m/s虏 and 0.758 m/s虏, respectively.
  • Radial acceleration ensures that the rotating point changes direction, keeping it in a circular path.
  • It highlights the relationship between rotation speed and the necessary centripetal force.
  • If a point weren't subject to radial acceleration, it would deviate from its circular path.
Exploring Resultant Acceleration
Resultant acceleration (\( a_{res} \)) combines both tangential and radial accelerations, determining the net acceleration at any point on the rotating body. We used the formula \( a_{res} = \sqrt{a_t^2 + a_r^2} \) to find this value, which varies based on the system's angular velocity.
Initially, since radial acceleration is zero, resultant acceleration only includes the tangential component (0.180 m/s虏). However, as the flywheel continues to rotate, resultant acceleration changes, being 0.416 m/s虏 after turning 60掳 and 0.780 m/s虏 after 120掳.
  • Resultant acceleration provides a comprehensive view of how the object moves in circular motion.
  • It accounts for both speed changes and direction changes, providing crucial insight into real-world applications.
  • Considering both accelerations is essential for accurately predicting the motion of rotating systems.

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Most popular questions from this chapter

At \(t=0\) the current to a de electric motor is reversed, resulting in an angular displacement of the motor shaft given by \(\theta(t)=(250 \mathrm{rad} / \mathrm{s}) t-\left(20.0 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}-\left(1.50 \mathrm{rad} / \mathrm{s}^{3}\right) t^{3} .(\mathrm{a}) \mathrm{At}\) what time is the angular velocity of the motor shaft zero? (b) Calculate the angular acceleration at the instant that the motor shaft has zero angular velocity. (c) How many revolutions does the motor shaft turn through between the time when the current is reversed and the instant when the angular velocity is zero? (d) How fast was the motor shaft rotating at \(t=0,\) when the current was reversed? ( c) Calculate the average angular velocity for the time period from \(t=0\) to the time calculated in part (a).

An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev \(/ \min\) to 200 rev \(/ \min\) in 4.00 s. (a) Find the angular acceleration in rev/s' and the number of revolutions made by the motor in the \(4.00-\) - interval. (b) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part (a)?

A thin uniform rod of mass \(M\) and length \(L\) is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.

A cylinder with radius \(R\) and mass \(M\) has density that increases linearly with distance \(r\) from the cylinder axis, \(\rho=\alpha r\) where \(\alpha\) is a positive constant. (a) Calculate the moment of inertia of the cylinder about a longitudinal axis through its center in terms of \(M\) and \(R .\) (b) Is your answer greater or smaller than the moment of inertia of a cylinder of the same mass and radius but of uniform density? Explain why this result makes qualitative sense.

A compound disk of outside diameter 140.0 \(\mathrm{cm}\) is made up of a uniform solid disk of radius 50.0 \(\mathrm{cm}\) and area density 3.00 \(\mathrm{g} / \mathrm{cm}^{2}\) surrounded by a concentric ring of inner radius 50.0 \(\mathrm{cm}\) , outer radius \(70.0 \mathrm{cm},\) and area density 2.00 \(\mathrm{g} / \mathrm{cm}^{2} .\) Find the moment of inertia of this object about an axis perpendicular to the plane of the object and passing through its center.
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