1. Uniform Circular Motion
A special form of 2D motion is circular motion.
Examples: ball on a string, cars exiting the highway, planets orbiting the sun
Circular motion is accelerated motion, even if the speed of the object
remains constant.
WHY?
The direction of the velocity changes at every point

2. Uniform Circular Motion
• an object in uniform circular motion experiences a centripetal acceleration of magnitude:
• centripetal acceleration always points toward the center of the circle.

3. Period and Frequency Uniform circular motion is periodic. Its
behavior repeats after one period, T.
The period, T, is the amount of time it takes for one complete revolution.
The frequency, f, of motion is the rate with which the motion repeats.
The frequency is the inverse of the period.
per second
seconds
Hertz

4. Period and Speed The period and frequency can be related to
the speed of an object in uniform circular
motion.
The object travels one circumference, 2πr,
every period.
Thus

5. The Dynamics of Circular Motionv
To remain in uniform
circular motion, an
object must be constantly
accelerating toward the
center of the circle.
To constantly be
center of the circle, the
object must experience a
net centripetal force. v

6. Horizontal circular motion
‘view from above’
Consider a box being whirled
in a horizontal circle on a frictionless table. Draw the forces acting on the box. FT
Fg and FN are not drawn since they point into and out of the paper respectively. (They are not important to the problem as they do not act into or opposite the direction of the acceleration)
Fc = FT
In what direction does this box accelerate?
ALWAYS towards the center

7. Vertical circular motion
‘view from the side’
Consider a box being whirled
in a vertical circle. Draw the forces acting on the box.
Note: At the top of
the circle, Fg
would be added to FT
Fg is now drawn since it is opposite to the direction the box accelerates. (There is no FN as it is not on a surface) FT Fg
Fc = FT - Fg

8. A person is flying a kg model airplane connected to a 10 m long string in a horizontal circle. The string exerts a force of 2.65 N on the plane. What is the plane’s speed?
Given:
‘view from above’
m = kg
r = 10 m
FT= 2.65 N
v = ? m/s
Fc = FT FT v2
2.65 = 0.088 10
v = 17.4 m/s

9. A pendulum 80 cm long is lifted above its equilibrium position and released. As the kg pendulum bob passes through its lowest position the tension in the pendulum cord is N, what is the pendulum's speed at this point?
Given:
‘view from the side’
m = kg
Fc = FT - Fg
r = 80 cm = 0.80 m
FT = 0.735N
v = ? m/s v2
0.735 – 0.49 = 0.050
0.80 FT ac
v = 2.0 m/s Fg

10. A special case – just completing a vertical loop
Unlike horizontal circular motion, in vertical circular motion the
speed, as well as the direction of the object, is constantly
changing. Gravity is constantly either speeding up the object
as it falls, or slowing the object down as it rises.
If we wanted to calculate the minimum or
critical velocity needed for the block to
just be able to pass through the top of the
circle without the rope sagging then
we would start by setting the tension in the
rope to ____________.
zero *
ΣFc = Fg
*only valid for the minimum velocity at the top to just complete a vertical loop.