2 Answers | Add Yours
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`
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
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 319,210 questions. We can answer yours, too.Ask a question