/*! 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 45 A wheel of radius \(R\) starts f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A wheel of radius \(R\) starts from rest and accelerates with a constant angular acceleration \(\alpha\) about a fixed axis. At what time t will the centripetal and tangential accelerations of a point on the rim have the same magnitude?

Short Answer

Expert verified
The time \(t\) is \(\frac{1}{\sqrt{\alpha}}\).

Step by step solution

01

Understanding Tangential and Centripetal Acceleration

The tangential acceleration, denoted as \(a_t\), is given by the formula \(a_t = R \alpha\). The centripetal acceleration, \(a_c\), is given by \(a_c = R \omega^2\), where \(\omega\) is the angular velocity.
02

Determine Angular Velocity from Angular Acceleration

The angular velocity \(\omega\) of the wheel as a function of time can be found using the equation \(\omega = \alpha t\), since the wheel starts from rest. This equation is derived from the basic kinematics equation \(\omega = \omega_0 + \alpha t\), where the initial angular velocity \(\omega_0 = 0\).
03

Equate Tangential and Centripetal Acceleration

Set the tangential acceleration equal to the centripetal acceleration: \(R \alpha = R (\alpha t)^2\). Simplify this equation by canceling \(R\) from both sides to get \(\alpha = \alpha^2 t^2\).
04

Solve for Time \(t\)

Solve \(\alpha = \alpha^2 t^2\) for \(t\). Divide both sides by \(\alpha\) to get \(1 = \alpha t^2\). Then solve for \(t\) by taking the square root of both sides: \(t = \frac{1}{\sqrt{\alpha}}\).

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 91Ó°ÊÓ!

Key Concepts

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

Understanding Angular Acceleration
Angular acceleration is a fundamental concept in rotational motion. It describes how quickly the angular velocity of an object changes with time. Think of angular acceleration, denoted as \( \alpha \), as the rotational equivalent of linear acceleration. This tells us how fast an object speeds up or slows down when spinning. Units involved are radians per second squared (rad/s²). If a wheel or any rotating object starts from rest and speeds up, angular acceleration is what governs this change. For instance, in our problem, the wheel starts with an angular velocity of zero and begins to accelerate with a constant angular acceleration \( \alpha \). It's crucial in converting the rotational effects into measurable accelerations that impact how we calculate motion parameters like tangential and centripetal accelerations.
Exploring Tangential Acceleration
Tangential acceleration \( a_t \) is related to the change in speed of an object along its path of motion. For rotational objects like a wheel, it's the acceleration experienced by a point on the rim in a direction tangent to its path. Mathematically, tangential acceleration is expressed as \( a_t = R \alpha \), where \( R \) is the radius of the wheel and \( \alpha \) is the angular acceleration. This shows the direct relationship with angular acceleration, allowing us to connect rotational motion to directional speed changes that are tangential to the wheel's path. This measurement helps determine how quickly a point on the wheel's rim accelerates as it rolls. It's also pivotal in solving our exercise, as we need it to be equal to the centripetal acceleration at a specific time \( t \).
Therefore, knowing about tangential acceleration aids in understanding how rotational effects translate into linear motion along the object's path.
Defining Centripetal Acceleration
Centripetal acceleration \( a_c \) is essential for any object moving in a circular path. It keeps the object moving in its circular trajectory by constantly changing the direction of its velocity. In our problem, centripetal acceleration is calculated using the formula \( a_c = R \omega^2 \), wherein \( R \) denotes the radius and \( \omega \) is the angular velocity. This acceleration is always directed towards the center of the circle, giving the object its curved path. It's an intrinsic part of rotational motion and is key in understanding how the motion is sustained in a circle. In our problem, angular velocity \( \omega \) can be expressed in terms of angular acceleration \( \alpha \) and time \( t \) using the relation \( \omega = \alpha t \).
This relationship helps in determining when the centripetal and tangential accelerations are equal, highlighting the intersection of rotational and linear motion principles.

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

Find the angular speed of the Earth as it spins about its axis. Give your result in rad/s.

A \(2.0-\mathrm{kg}\) solid cylinder (radius \(=0.10 \mathrm{m},\) length \(=0.50 \mathrm{m})\) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is \(0.75 \mathrm{m}\) high and \(5.0 \mathrm{m}\) long. When the cylinder reaches the bottom of the ramp, what are (a) its total kinetic energy, (b) its rotational kinetic energy, and (c) its translational kinetic energy?

An object at rest begins to rotate with a constant angular acceleration. If the angular speed of the object is \(\omega\) after the time \(t,\) what was its angular speed at the time \(t / 2 ?\)

A person rides on a 12-m-diameter Ferris wheel that rotates at the constant rate of 8.1 rpm. Calculate the magnitude and direction of the force that the seat exerts on a \(65-\mathrm{kg}\) person when he is (a) at the top of the wheel, (b) at the bottom of the wheel, and (c) halfway up the wheel.

Playing a CD The record in an old-fashioned record player always rotates at the same angular speed. With CDs, the situation is different. For a \(\mathrm{CD}\) to play properly, the point on the \(\mathrm{CD}\) where the laser beam shines must have a linear speed \(v_{t}=1.25 \mathrm{m} / \mathrm{s},\) as indicated in Figure \(10-27\) (a) As the CD plays from the center outward, does its angular speed increase, decrease, or stay the same? Explain. (b) Find the angular speed of a CD when the laser beam is \(2.50 \mathrm{cm}\) from its center. (c) Repeat part (b) for the laser beam \(6.00 \mathrm{cm}\) from the center. (d) If the CD plays for \(66.5 \mathrm{min},\) and the laser beam moves from \(2.50 \mathrm{cm}\) to \(6.00 \mathrm{cm}\) during this time, what is the CD's average angular acceleration?

See all solutions

Recommended explanations on Physics Textbooks

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.