Consider a circular disc which is rotatingabout its geometric center with acircular motion such that the angularacceleration is a quadratic function intime such that the time derivative of theangular acceleration is proportional tothe time. For a particular angularacceleration motion at a time of t₁ =0.1047 second the angular accelerationis a₁ =-0.1568 meter per secondsquared and at a time of t2=0.514second the angular acceleration is a2=0.9995 meter per second squared. At atime of t3=1.1999 second the angularvelocity is w=9.9219 radians persecond. Calculate the following at timet4=2.2023 second and a point on thedisc where the radius is r=0.6351 meter:Question 1) the scalar magnitude of thevelocity at time 2.2023 secondQuestion 2) the scalar magnitude of thenormal component of the vector of theacceleration at time 2.2023 secondQuestion 3) the scalar magnitude of thetangential component of the vector ofthe acceleration at time 2.2023 second

Answered: Consider a circular disc which is…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Consider a circular disc which is rotating
about its geometric center with a
circular motion such that the angular
acceleration is a quadratic function in
time such that the time derivative of the
angular acceleration is proportional to
the time. For a particular angular
acceleration motion at a time of t₁ =
0.1047 second the angular acceleration
is a₁ =-0.1568 meter per second
squared and at a time of t2=0.514
second the angular acceleration is a2
=0.9995 meter per second squared. At a
time of t3=1.1999 second the angular
velocity is w=9.9219 radians per
second. Calculate the following at time
t4=2.2023 second and a point on the
disc where the radius is r=0.6351 meter:
Question 1) the scalar magnitude of the
velocity at time 2.2023 second
Question 2) the scalar magnitude of the
normal component of the vector of the
acceleration at time 2.2023 second
Question 3) the scalar magnitude of the
tangential component of the vector of
the acceleration at time 2.2023 second

Expert Solution

Step 1: Determine the magnitude of the scalar velocity at time 2.0512 seconds.

1) We may use the following equation to get the scalar magnitude of the velocity at time 2.0512 seconds:

$bold italic v bold space bold equals bold space bold italic omega bold space bold italic r$

where:

v = the scalar magnitude of the velocity

$ω$ = the angular velocity

r = the radius

Substituting the known values, we get:

$v space equals space left parenthesis space 8.5565 space r a d divided by s space space right parenthesis space left parenthesis space 0.9599 m space right parenthesis space equals space 8.21338 space m divided by s$

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