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Chapter 9: Problem 58

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?

Short Answer

Expert verified
The exact value of the linear acceleration will depend on the solution to the quadratic equation in step 2. After solving for \( t \) and substituting in the formula for linear acceleration, you will arrive at the acceleration of a point on the rim of the disk.

Step by step solution

01

Calculate Angular Acceleration

The first step is to calculate the angular acceleration, because it will be used in the equation for linear acceleration. Given the equation for \( \theta(t) \), we can find the angular acceleration \( \alpha \) by taking the second derivative: \( \alpha = d^2 \theta/dt^2 = 2 * 6.30 rad/s^2 = 12.6 rad/s^2 \). Hence, the angular acceleration is \( \alpha = 12.6 rad/s^2 \).
02

Calculate Angular Velocity at the Given Time

The rotational angle is given as 0.100 rev. Since 1 rev = \(2 \pi\) rad, so the time when the disk has turned through 0.100 rev is \( t = (0.100*2 \pi)/(1.10 rad/s + 6.30 rad/s^2 * t) \). We solve this quadratic equation for \( t \) and substitute in the formula for angular velocity \( \omega = (1.10 rad/s) t + (6.30 rad/s^2) t^2 \) to get the angular velocity at this time.
03

Calculate Linear Acceleration

Once we have calculated the angular acceleration \( \alpha \) and the angular velocity \( \omega \), we can compute the linear acceleration of a point on the rim of the disk using the formula \( a = r \alpha + (r \omega)^2 \). Substituting \( r = 0.4 m \), \( \alpha = 12.6 rad/s^2 \) and the calculated value for \( \omega \), this will give us the required linear acceleration.

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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 measure of how quickly the angular velocity of an object changes with time. In the context of a rotating disk, as found in the example problem, angular acceleration can be understood as how fast the disk is speeding up or slowing down its rotation. It's calculated by taking the second derivative of the angular position \theta(t), with respect to time.

For the given exercise, the angular acceleration \( \alpha \) is determined by differentiating the angular position function \( \theta(t) \) twice with respect to time. This leads us to find that \( \alpha = 12.6 \, \text{rad/s}^2 \). Understanding this concept is imperative, as it serves as a key component in determining linear acceleration, particularly in a disk rotating around a central axle.
Angular Velocity
Angular velocity, denoted by \( \omega \), represents the rate of change of the angular position of a rotating object, essentially describing how fast the object is spinning. It is analogous to linear velocity, but instead of the change in position, it considers the change in rotation angle over time.

To solve for angular velocity in our exercise, we need to determine the time at which the disk has turned through 0.100 revolutions. This involves converting revolutions to radians and then solving a quadratic equation based on the provided \( \theta(t) \) relationship. Once the correct time is found, \( \omega \) is calculated using this time in the expression for angular position, providing us the angular velocity at that instant. This angular velocity is crucial to calculating the linear acceleration experienced at the rim of the disk.
Uniform Circular Motion
Uniform circular motion describes the movement of an object in a circular path at constant speed. Even though the speed is constant, there is still a type of acceleration present, called centripetal acceleration, which is always directed towards the center of the circle. This is why an object in uniform motion can have a linear acceleration even without changing speed.

In the case of the disk in our exercise, uniform circular motion applies if we consider the constant component of angular velocity. However, since the disk's angular velocity also changes due to angular acceleration, the motion is not purely uniform. The total linear acceleration at any point on the rim is the vector sum of radial acceleration (due to angular acceleration) and centripetal acceleration (due to angular velocity). This relationship allows us to calculate linear acceleration which is of utmost interest in analyzing the dynamics of a rotating system like the given disk.

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Most popular questions from this chapter

An airplane propeller is \(2.08 \mathrm{~m}\) in length (from tip to tip) with mass \(117 \mathrm{~kg}\) and is rotating at 2400 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod. (a) What is its rotational kinetic energy? (b) Suppose that, due to weight constraints, you had to reduce the propeller's mass to \(75.0 \%\) of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?

An airplane propeller is rotating at 1900 rpm (rev/min). (a) Compute the propeller's angular velocity in rad/s. (b) How many seconds does it take for the propeller to turn through \(35^{\circ} ?\)

About what axis will a uniform, balsa-wood sphere have the same moment of inertia as does a thin-walled, hollow, lead sphere of the same mass and radius, with the axis along a diameter?

A uniform, solid disk with mass \(m\) and radius \(R\) is pivoted about a horizontal axis through its center. A small object of the same mass \(m\) is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

A wheel of diameter \(40.0 \mathrm{~cm}\) starts from rest and rotates with a constant angular acceleration of \(3.00 \mathrm{rad} / \mathrm{s}^{2}\). Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (a) \(a_{\mathrm{rad}}=\omega^{2} r\) and (b) \(a_{\mathrm{rad}}=v^{2} / r\)
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