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:…

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

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

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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