/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 A uniform disk with radius \(R=0... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 62

A uniform disk with radius \(R=0.400 \mathrm{m}\) and mass 30.0 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 \(\theta(t)=\) \((1.10 \mathrm{rad} / \mathrm{s}) t+\left(8.60 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2} .\) 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 \(\mathrm{rev}\) ?

Short Answer

Expert verified
The resultant linear acceleration is the combination of tangential and centripetal accelerations.

Step by step solution

01

Convert Revolutions to Radians

The angle in radians is given by multiplying the number of revolutions by \( 2\pi \). So, if the disk has turned through 0.100 revolutions, the angle is \( 0.100 \times 2\pi = 0.20\pi \) radians.
02

Establish the Position Equation

The angle \( \theta(t) \) is given by \( \theta(t) = (1.10 \, \mathrm{rad/s}) \times t + (8.60 \, \mathrm{rad/s^2}) \times t^2 \). We need to find the time \( t \) when \( \theta = 0.20 \pi \) radians so, \((1.10 \, \mathrm{rad/s}) t + (8.60 \, \mathrm{rad/s^2}) t^2 = 0.20\pi \). This requires solving a quadratic equation.
03

Solve the Quadratic Equation for Time t

Rearrange the equation as \(8.60 t^2 + 1.10 t - 0.20 \pi = 0\). Use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 8.60\), \(b = 1.10\), and \(c = -0.20\pi\). Calculate the discriminant and solve for time \(t\).
04

Calculate Angular Velocity and Angular Acceleration

After finding \(t\), calculate the angular velocity \( \omega(t) \) by differentiating \( \theta(t) \), giving \( \omega(t) = 1.10 + 2 \times 8.60 \times t \). The angular acceleration \( \alpha \) is the derivative of \( \omega(t) \), which is constant and equal to \(2 \times 8.60 = 17.2 \, \mathrm{rad/s^2}\).
05

Determine Linear Acceleration Components

The tangential acceleration is given by \( a_t = R \cdot \alpha = 0.4 \times 17.2 = 6.88 \, \mathrm{m/s^2}\). The centripetal acceleration is given by \( a_c = R \cdot \omega(t)^2 \). Substitute the calculated \( \omega(t) \) into this equation.
06

Calculate and Combine Linear Accelerations

Combine the tangential and centripetal components using the Pythagorean theorem for vector addition: the resultant linear acceleration \( a = \sqrt{a_t^2 + a_c^2} \). Substitute the values calculated in Step 5 and solve for \( a \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Angular Velocity
In rotational motion, angular velocity represents how fast an object rotates or revolves relative to another point, typically around an axis. It's akin to linear velocity but on a circular path. Defined as the rate of change of angular displacement, angular velocity is a vector quantity, meaning it has both magnitude and direction.
  • The standard unit for angular velocity is radians per second (rad/s).
  • It is represented by the symbol \(\omega\).
In the problem, angular velocity is determined by differentiating the expression for \(\theta(t)\), which provides the instantaneous rate of rotation. This calculation helps us understand how swiftly the disk turns at any moment, which is crucial for determining other parameters like tangential and centripetal accelerations.
Tangential Acceleration
Tangential acceleration is directly linked to the increase or decrease in angular velocity over time. It measures how quickly the tangential (linear) speed of a point on a rotating object changes. This acceleration occurs along the edge of the circular path.
  • The formula for tangential acceleration is \(a_t = R \cdot \alpha\), where \(\alpha\) is the angular acceleration.
  • It reflects the linear aspect of rotational motion.
  • In this exercise, we've calculated it as \(6.88 \, \mathrm{m/s^2}\).
Tangential acceleration is vital because it not only affects the speed at which an object moves along its path but also contributes to the overall linear acceleration when combined with centripetal acceleration.
Centripetal Acceleration
Centripetal acceleration is essential for maintaining circular motion. It acts towards the center of the circle, holding the object on its curved path.
The equation used is \(a_c = R \cdot \omega^2\), where \(\omega\) is the angular velocity.
  • It is always directed towards the axis of rotation.
  • It's responsible for changing the direction of the velocity vector of the object but not its magnitude.
Without centripetal acceleration, an object would move off in a straight line rather than following a curved path. In the given problem, once we determine \(\omega (t)\), it becomes straightforward to compute \(a_c\), giving insight into how strongly the object is "pulled" towards the center of the circular path.
Quadratic Equation
The calculation for time \(t\) in this problem involves solving a quadratic equation. Quadratic equations are fundamental in physics for finding values at precise moments in the parabolic curve of motion.
The typical form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants.
  • The solution is often found using the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
  • The discriminant \(b^2 - 4ac\) determines the nature of the roots.
Quadratic equations help predict time when certain rotational conditions are met, essential for stepping into calculations of rotational dynamics like angular velocity and accelerations. In the problem, rearranging \(1.10t + 8.60t^2 = 0.20\pi\) into a standard quadratic form allows us to solve for the exact moment when the given angle in radians is achieved.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Measuring \(I .\) As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an axis through its center. Since you were a good physics student, you know what to do. You measure the diameter of the wheel to be 0.740 \(\mathrm{m}\) and find that it weighs 280 \(\mathrm{N.}\) You mount the wheel, using frictionless bearings, on a horizontal axis through the wheel's center. You wrap a light rope around the wheel and hang an 8.00 -kg mass from the free end of the rope, as shown in Fig. 9.17 . You release the mass from rest; the mass descends and the wheel turns as the rope unwinds. You find that the mass has speed 5.00 \(\mathrm{m} / \mathrm{s}\) after it has descended 2.00 \(\mathrm{m} .\) (a) What is the moment of inertia of the wheel for an axis perpendicular to the wheel at its center? (b) Your boss tells you that a larger \(I\) is needed. He asks you to design a wheel of the same mass and radius that has \(I=19.0 \mathrm{kg} \cdot \mathrm{m}^{2} .\) How do you reply?

Small blocks, each with mass \(m,\) are clamped at the ends and at the center of a rod of length \(L\) and negligible mass. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through (a) the center of the rod and (b) a point one-fourth of the length from one end.

While riding a multispeed bicycle, the rider can select the radius of the rear sprocket that is fixed to the rear axle. The front sprocket of a bicycle has radius 12.0 \(\mathrm{cm} .\) If the angular speed of the front sprocket is 0.600 rev/s, what is the radius of the rear sprocket for which the tangential speed of a point on the rim of the rear wheel will be 5.00 \(\mathrm{m} / \mathrm{s} ?\) The rear wheel has radius 0.330 \(\mathrm{m} .\)

A uniform sphere with mass 28.0 \(\mathrm{kg}\) and radius 0.380 \(\mathrm{m}\) is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere. If the kinetic energy of the sphere is \(176 \mathrm{J},\) what is the tangential velocity of a point on the rim of the sphere?

A thin, light wire is wrapped around the rim of a wheel, as shown in Fig. E9.49. The wheel rotates without friction about a stationary horizontal axis that passes through the center of the wheel. The wheel is a uniform disk with radius \(R=0.280 \mathrm{m}\) . An object of mass \(m=4.20 \mathrm{kg}\) is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration. If the suspended object moves downward a distance of 3.00 \(\mathrm{m}\) in \(2.00 \mathrm{s},\) what is the mass of the wheel?
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Particle Model of Matter

Read Explanation

Waves Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Energy Physics

Read Explanation

Medical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.