Download Circular Motion

Forces and
Motion in Two
Dimensions
Circular Motion
Objectives
 Explain
the acceleration of an object
moving in a circle at constant speed
 Describe how centripetal acceleration
depends upon the object’s speed and
the radius of the circle
 Recognize the direction of the force that
causes centripetal acceleration
 Explain how the rate of circular motion is
changed by exerting torque on it
Describing Circular Motion
 The
position of the object relative to the
center of the circle is the position vector R.
 During its path, the length of the vector
doesn’t change, but the direction does.
v1
r1
r2
v2
Describing Circular Motion
 To
find velocity, first find the
displacement vector (∆r) or (r2-r1)
over a time interval (∆t.)
¯v = Δr / Δt
A
moving object’s average velocity
is ∆d / ∆t. But for an object in circular
motion, the average velocity is ∆r /
∆t.
Describing Circular Motion
 The
velocity is at
right angles to
the position
vector and
tangent to the
circular path.
 ∆V is found by
subtracting the
vectors V2 and
V1.
Centripetal Acceleration
 Centripetal
acceleration is the
acceleration of an
object in uniform
circular motion and
is always pointed
toward the center
of the circle.
Centripetal Acceleration
 Acceleration
is ∆V/ ∆t and is directed
toward the center.
ac = v2/r
Centripetal Acceleration
 Period
(t) = the time needed for an object
to make a complete revolution, the
circumference of the circle, 2πr.

The object’s speed is represented by:
V = 2πr/t
 So,
substitute 2πr/t for V and the
centripetal acceleration is:
Ac = (2πr/t)2 / r
or
4π2r / t2
Torque
 In
relation to uniform circular
motion, you have considered
objects such as a person on
a merry-go-round
 These can be considered
point masses
 Now, consider rigid rotating
objects


A mass that rotates around its
own axis
Ex: doors
Torque
 To
open a door, you push at a distance
from the hinges (axis of rotation) and in a
direction perpendicular to the door
 The product of the force applied to an
object, the distance from the rotation
point, and the direction of the force
applied is called torque

Unit is for torque is Newton meters (Nm)
Torque
 Solve
for torque using the following equation:
T = r FsinΘ
T = torque
r= radius
F= force applied
Θ= direction