   Chapter 12.3, Problem 48E

Chapter
Section
Textbook Problem

# Circular Motion In Exercises 47-50, consider a particle moving on a circular path of radius b described by r ( t ) = b cos ω t i + b sin ω t j where ω = d u / d t is the constant angular speed.Show that the magnitude or the acceleration vector is b ω 2

To determine

To prove: The magnitude of the acceleration vector is bω2 if a particle which is moving on a circular path of radius b is r(t)=bcosωti+bsinωtj where ω=du/dt.

Explanation

Given:

The provided particle path is r(t)=bcosωti+bsinωtj.

Formula used:

The acceleration vector is the derivative of the velocity vector,

a(t)=v(t).

Proof:

Consider the particle path is,

r(t)=bcosωti+bsinωtj

Differentiate position vector with respect to t gives a velocity vector,

v(t)= r(t)=(bωsinωti+bωcosωtj)

Differentiate velocity vector with respect to t gives an acceleration vector,

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