Answer:
Acceleration of the particle = [tex]bw^{2}[/tex]
Step-by-step explanation:
We are given the position vector of a particle moving in a circle of radius b units.
r(t) = b cos(ωt)i + b sin(ωt)j
Velocity , v =[tex]\frac{dr}{dt}[/tex] = -bω sin(ωt)i + bω cos(ωt)j
The magnitude of velocity, v =[tex]\sqrt{v_x^{2} +v_y^{2} }[/tex]
Squaring both sides,
[tex]v^{2} = b^{2} w^{2}(sin^{2}(wt)+cos^{2}(wt))[/tex]
Since [tex]sin^{2}(wt)+cos^{2}(wt))[/tex] = 1
[tex]v^{2} = b^{2}w^{2}[/tex]
The acceleration towards the centre is called the centripetal acceleration and is given by
a = [tex]\frac{v^{2} }{r}[/tex]
a = [tex]\frac{b^{2}w^{2}}{b}[/tex]
a = [tex]bw^{2}[/tex]
