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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

The rotating blade of a blender turns with constant angular acceleration \(1.50 \mathrm{rad} / \mathrm{s}^{2}\). (a) How much time does it take to reach an angular velocity of \(36.0 \mathrm{rad} / \mathrm{s},\) starting from rest? (b) Through how many revolutions does the blade turn in this time interval?

\(\mathrm{At} t=0\) a grinding wheel has an angular velocity of \(24.0 \mathrm{rad} / \mathrm{s}\) It has a constant angular acceleration of \(30.0 \mathrm{rad} / \mathrm{s}^{2}\) until a circuit breaker trips at \(t=2.00 \mathrm{~s}\). From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between \(t=0\) and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?

A thin, uniform rod is bent into a square of side length \(a\). If the total mass is \(M,\) find the moment of inertia about an axis through the center and perpendicular to the plane of the square. (Hint: Use the parallel-axis theorem.)

You are to design a rotating cylindrical axle to lift \(800 \mathrm{~N}\) buckets of cement from the ground to a rooftop \(78.0 \mathrm{~m}\) above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady \(2.00 \mathrm{~cm} / \mathrm{s}\) when it is turning at \(7.5 \mathrm{rpm} ?\) (b) If instead the axle must give the buckets an upward acceleration of \(0.400 \mathrm{~m} / \mathrm{s}^{2},\) what should the angular acceleration of the axle be?

Energy is to be stored in a \(70.0 \mathrm{~kg}\) flywheel in the shape of a uniform solid disk with radius \(R=1.20 \mathrm{~m}\). To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is \(3500 \mathrm{~m} / \mathrm{s}^{2}\). What is the maximum kinetic energy that can be stored in the flywheel?
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