tangential acceleration in nonuniform circular motion is const.if we apply eqn  s=ut+1/2 at^2,s represents distance covered along circumference or displacement?

2 Answers | Add Yours

valentin68's profile pic

valentin68 | College Teacher | (Level 3) Associate Educator

Posted on

For some reason the ds diferrential above does not appear like it should:

`ds =R*d(theta) =omega*R +epsilon*R*t*dt =v + a*t*dt`

valentin68's profile pic

valentin68 | College Teacher | (Level 3) Associate Educator

Posted on

If the tangential acceleration in the circular motion is constant in module (its direction is changing because the trajectory is circular) then the equation

`s = v*t +a*t^2/2`

represents the distance covered along the circle.

To demonstrate this take an infinitesimal displacement along the same circle

`ds =R*d(theta)`

If the circular motion has a tangential acceleration uniform in module, it means the angular acceleration `epsilon` is constant as a vector over time. For angles `theta` on the circle the following equation is valid

`theta =omega*t +epsilon*t^2/2`

By differentiating we get

`d(theta) = omega +epsilon*t*dt`

Multiplying with R we obtain

`ds =R*d(theta) = omega*R +epsion*R*t*dt =v +a*t*dt`

By integration from `s=0` to `s` (from angles `theta=0` to `theta` ) we obtain the total distance covered along the circle (not the displacement vector).

`s =R*theta = v*t +a*t^2/2`

`R*theta` (`theta` is measured in radians) represents the distance along the circle since `R*2pi` represents the circumference of the entire circle.

We’ve answered 318,989 questions. We can answer yours, too.

Ask a question