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=0.1949 second the angularacceleration is a₁ =-0.198 meter persecond squared and at a time of t2=0.6111 second the angularacceleration is a2=0.5872 meter persecond squared. At a time of t3=1.2799second the angular velocity is w=8.5565radians per second. Calculate thefollowing at time t4=2.0512 second anda point on the disc where the radius is r-0.9599 meter:Question 1) the scalar magnitude of thevelocity at time 2.0512 secondQuestion 2) the scalar magnitude of thenormal component of the vector of theacceleration at time 2.0512 secondQuestion 3) the scalar magnitude of thetangential component of the vector ofthe acceleration at time 2.0512 second

1) To calculate the scalar magnitude of the velocity at time 2.1407" data-mce-style="display:…

Answered: Question 1) the scalar magnitude of the…

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
=0.1949 second the angular
acceleration is a₁ =-0.198 meter per
second squared and at a time of t2
=0.6111 second the angular
acceleration is a2=0.5872 meter per
second squared. At a time of t3=1.2799
second the angular velocity is w=8.5565
radians per second. Calculate the
following at time t4=2.0512 second and
a point on the disc where the radius is r
-0.9599 meter:
Question 1) the scalar magnitude of the
velocity at time 2.0512 second
Question 2) the scalar magnitude of the
normal component of the vector of the
acceleration at time 2.0512 second
Question 3) the scalar magnitude of the
tangential component of the vector of
the acceleration at time 2.0512 second

Expert Solution

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

1) To calculate the scalar magnitude of the velocity at time $2.1407$ seconds, we can use the following equation:

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

where:

v is the scalar magnitude of the velocity

$ω$ is the angular velocity

r is 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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