Vf
∆r
rf
Vi
∆θ
ri
An object is moving along a circular path
of radius r with a velocity that changes
direction but has a constant magnitude so
that: rf = ri and vf = vi
∆V
Vi
∆θ
Vf
Since the velocity vectors are perpendicular to the position vectors the
angle between the velocity vectors is the same as the angle between the
position vectors as shown in the second figure. The triangles shown in
the figures are similar. If we let r = ri = rf and v=vi = vf we have:
∆v
∆r
=
v
r
∆v
∆r
=
v
r
Multiplying this equation by v and dividing by ∆t gives:
If we take the limit as ∆t
0 we obtain:
radial acceleration
dv v dr
=
dt r dt
∆v v ∆r
=
∆t r ∆t
⇒
ar =
v2
r
velocity
Direction of the Acceleration
dv
vf
vi
When the time interval ∆t is made very small the angle between vi
and vf is also very small and the angle between dv and vi (or vf) is 90
degrees. Since v is parallel to the tangent to the circle, dv points
towards the center of the circle. Since dv determines the direction of
ar, the radial acceleration points towards the center of the path.
1
v
Acceleration of an object in circular motion
with constant speed.
ar
v
at
total
acceleration
ar
Acceleration components of an object in circular
motion and speeding up. The component at is the
tangential acceleration component.
v
Acceleration components of an object in circular
motion and slowing down.
ar
total
acceleration
at
2