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

Acceleration of the particle =
`bw^(2)`

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 =
`(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 =
`(v^(2) )/(r)`

a =
`(b^(2)w^(2))/(b)`

a =
`bw^(2)`