2. Circular Motion
Constant force
is toward
center.
velocity is
tangent to path.
v
Fc
Circular Motion is motion along a circular
path in which the direction of its velocity is
always changing. Thus, it is accelerating.
3. Initial quantities involved in
circular motion:
𝑡 − period for N revolutions
𝑓 − frequency of N revolutions
𝑣 =
2𝜋𝑅
𝑡
= 2𝜋𝑅𝑓 𝑓 =
1
𝑡
6. Dynamics of Circular Motion
The question of an outward force can be
resolved by asking what happens when the
string breaks!
When central force is
removed, ball continues in
straight line.
v
Ball moves tangent to
path, NOT outward as
might be expected.
Centripetal force is needed to change direction.
7. Car Example
There is an outward force, but it does not act
ON you. It is the reaction force exerted BY you
ON the door. It affects only the door.
The centripetal
force is exerted
BY the door ON
you. (Centrally)
Fc
F’
Reaction
8. From Newton’s Second Law: 𝐹𝑛𝑒𝑡 = 𝑚𝑎
𝐹𝑛𝑒𝑡 = 𝐹𝑐
In circular motion, the net force is the
centripetal force!
𝐹𝑛𝑒𝑡 = 𝐹𝑐 = 𝑚
𝑣2
𝑅
9. 1. Horizontal circular motion or Uniform
Circular Motion - motion along a circular
path in which there is no change in speed,
only a change in direction.
2. Vertical circular motion – non-uniform
circular motion
Kinds of Circular Motion
10. Car Negotiating a Flat Turn
R
v
Is there also an outward force
acting ON the car?
Ans. No, but the car does exert an
outward reaction force ON the road.
Fc
Applications: Circular Motion
11. Car Negotiating a Flat Turn
The centripetal force Fc is
that of static friction fs:
The central force FC and the friction force fs
are not two different forces that are equal.
The nature of this central force is static
friction.
Fc = fsR
v
m
Fc
N
mg
fs
R
12. Finding the maximum speed for
negotiating a turn without slipping.
The car is on the verge of slipping when FC
is equal to the maximum force of static
friction fs.
R
v
m
Fc
Fc = fs
n
mg
fs
R
13. q
Optimum Banking Angle
By banking a curve at the
optimum angle, the normal
force N can provide the
necessary centripetal force
without the need for a friction
force.
optimum
N
fs = 0
w
R
v
m
Fc
14. Free-body Diagram
N
mg
q
q
Acceleration a is toward the
center. Set x axis along the
direction of ac , i. e.,
horizontal (left to right).
N
mg
q
N sin q
N cos q
+ ac
q
N
mg
x
15. Motion in a Vertical Circle
Consider the forces on a
ball attached to a string as
it moves in a vertical loop.
Note also that the positive
direction is always along
acceleration, i.e., toward
the center of the circle.
Note changes in positions.
+
T
mg
v
Bottom
Maximum
tension T, W
opposes Fc
+
v
T
mg
Top Right
Weight has no
effect on T
+
T
mg
v
Top Right
Weight causes
small decrease
in tension T
v
T
mg
+
Left Side
Weight has no
effect on T
+
T
mg
v
Bottom
v
T
mg
Top of Path
Tension is
minimum as
weight helps
Fc force
+
16. R
v
v
As an exercise, assume
that a central force of
Fc = 40 N is required to
maintain circular motion
of a ball and W = 10 N.
The tension T must
adjust so that central
resultant is 40 N.
At top: 10 N + T = 40 N
Bottom: T – 10 N = 40 N T = __?___T = 50 N
T = 30 NT = _?_
T
10 N
+
+
T
10 N
17. Homework:
1. In the banking of curves at optimum angle 𝜃, an
object must move with a certain speed so that it
can move in the curve without friction. Prove that
this speed is 𝑣 = 𝑔𝑅 sin 𝜃.
2. A 3.0-kg rock swings in a horizontal circle of radius
2.0 m. If it takes 2.0 s to complete one revolution,
what is its tangential speed? What is its
centripetal acceleration? What is the tension of
the string used? Draw the FBD of the rock.
3. In the frictionless loop-the-loop apparatus, derive
the equation of the normal force at the top of the
loop in terms of the mass of the cart m, the radius
of the loop R, the tangential speed v, and the
acceleration of gravity g. Note: Draw FBD.
R
v
v
