Answer to Question #230101 in Calculus for Unknown346307

Question #230101

Question:

A particle moves such that its vector is given by π‘Ÿ

 = cos πœ”π‘‘π‘– + sin πœ”π‘‘π‘— where πœ” is a constant. Show that:

(i)

Velocity, 𝑣 of the particle is perpendicular to r

(ii)

Acceleration, π‘Ž is directed towards the origin and has magnitude proportional to the 

distance from the origin

(iii) π‘Ÿ 𝑋 𝑣is a constant vector



1
Expert's answer
2021-08-30T16:09:38-0400

(i)


v=drdt=βˆ’Ο‰sin⁑ωti+Ο‰cos⁑ωtjv=\dfrac{dr}{dt}=-\omega\sin \omega ti+\omega\cos \omega tj

rβ‹…v=cos⁑ωt(βˆ’Ο‰sin⁑ωt)+sin⁑ωt(Ο‰cos⁑ωt)=0r\cdot v=\cos \omega t(-\omega\sin \omega t)+\sin \omega t(\omega \cos \omega t)=0

Hence rβŠ₯v.r \perp v.


(ii)


a=d2rdt2=dvdt=βˆ’Ο‰2cos⁑ωtiβˆ’Ο‰2sin⁑ωtja=\dfrac{d^2r}{dt^2}=\dfrac{dv}{dt}=-\omega^2\cos \omega ti-\omega^2\sin \omega tj

=βˆ’Ο‰2(cos⁑ωti+sin⁑ωtj)=βˆ’Ο‰2r=-\omega^2(\cos \omega ti+\sin\omega tj)=-\omega^2r

Hence π‘Ž is directed towards the origin and has magnitude proportional to the  distance from the origin


a=(βˆ’Ο‰2)ra=(-\omega^2)r

(iii)


rΓ—v=∣ijkcos⁑ωtsin⁑ωt0βˆ’Ο‰sin⁑ωtΟ‰cos⁑ωt0∣r\times v=\begin{vmatrix} i & j & k \\ \cos\omega t & \sin\omega t & 0\\ -\omega \sin\omega t & \omega\cos\omega t & 0\\ \end{vmatrix}


=k∣cos⁑ωtsin⁑ωtβˆ’Ο‰sin⁑ωtΟ‰cos⁑ωt∣=k\begin{vmatrix} \cos\omega t & \sin\omega t \\ -\omega \sin\omega t & \omega \cos\omega t \end{vmatrix}

=(Ο‰cos⁑2Ο‰t+Ο‰sin⁑2Ο‰t)k=(\omega\cos^2\omega t+\omega\sin^2\omega t)k

=Ο‰k=\omega k

Therefore rΓ—vr\times v is a constant vector (Ο‰=const\omega=const ).



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment