Expression for Centripetal Acceleration | Physics Grade XI

Circular Motion

☰   Related Articles

Refraction Through Prisms

Expression for Centripetal Acceleration | Physics Grade XI

Expression for Centripetal Acceleration

Expression for Centripetal Acceleration
Consider a particle moving with a constant speed v in a circular path of radius r with center O. At a point A on its path, its velocity vA is along the tangent at A and at B, its velocity vB is along the tangent at B. since the direction of the velocity at A and B are different, an acceleration occurs from A to B.

Let ∆t be the time taken to move from A to B covering a small displacement Δl and angular displacement Δθ.

Expression for centripetal accelerationAs shown in figure vA and vB are redrawn where Δv represents the change in velocity, Δv = vB - vA, which form three sides of a triangle PQR. Two triangles OAB and PQR are similar since both are isosceles triangles and angles labelled Δθ are the same. So, the ratios of the corresponding sides of such triangles are equal.

    Δv/v = Δl/r
  or, Δv = Δl/r . Δl
The rate of change of velocity is
Δv/Δt = v/t . Δl/Δt
For Δt->0,
Limit Δt-> 0 Δv/Δt = a and,
Limit Δt-> 0 Δl/Δt = v
Then, the above equation is written as
a = v/r . v
or, a = v2/ r
Since v = r. ω, the centripetal acceleration can be written as
a = r. ω2

You may also like to read:

Join with us on social media to see our updates on your feed.
facebook logo twitter logo