Answer:
Explanation:
(a) we can prove that the particle moves in a circle by taking the square of the norm of r(t)
[tex]|r(t)|=\sqrt{A^{2}[cos^{2}(\omega t)+sin^{2}(\omega t)]}\\cos^{2}(\omega t)+sin^{2}(\omega t)=1\\|r(t)|=A\\r=A[/tex]
the norm of the position vector does not depend of time, so |r| is constant and is a radius of a circle.
(b) the sped of the particle is the norm of the velocity v(t). Velocity is calculated by derivating r(t)
[tex]v(t)=\frac{dr(t)}{dt}=A(-\omega sin(\omega t)\hat{i}+\omega cos(\omega t)\hat{j})\\|v(t)|=\sqrt{A^{2}\omega^{2}(sin^{2}(\omega t)+cos^{2}(\omega t))}\\v=A\omega[/tex]
A and w are constant, hence the speed of the particle is constant.
(c) the acceleration is the derivative of the velocity
[tex]a(t)=\frac{dv(t)}{dt}=A(-\omega ^{2}cos(\omega t)\hat{i}-\omega^{2}sin(\omega t)\hat{j})\\ a(t)=-\omega^{2}r(t)\\|a(t)|=a=\omega^{2}r[/tex]
(d)
[tex]F_{c}=ma_{c}=m\frac{v^{2}}{r}=m\frac{A^{2}\omega ^{2}}{A}=mA\omega ^{2}[/tex]
I hope this is useful for you
regards
