Question

In this exercise, consider a particle moving on a circular path of radius b described by r(t) = b cos(ωt)i + b sin(ωt)j, where ω = du/dt is the constant angular velocity. Find the acceleration vector and show that its direction is always toward the center of the circle.

Answer 1

Answer:

Acceleration of the particle = $bw^{2}$

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 =$\frac{dr}{dt}$ = -bω sin(ωt)i + bω cos(ωt)j

The magnitude of velocity, v =$\sqrt{v_x^{2} +v_y^{2} }$

Squaring both sides,

$v^{2} = b^{2} w^{2}(sin^{2}(wt)+cos^{2}(wt))$

Since $sin^{2}(wt)+cos^{2}(wt))$ = 1

$v^{2} = b^{2}w^{2}$

The acceleration towards the centre is called the centripetal acceleration and is given by

a = $\frac{v^{2} }{r}$

a = $\frac{b^{2}w^{2}}{b}$

a = $bw^{2}$