Download Chapter 5, Part II

Newton’s Laws + Circular Motion
Sect. 5-2: Uniform Circular Motion; Dynamics
A particle moving in uniform circular
motion at radius r, speed v = constant:
• Centripetal acceleration:
aR = (v2/r) , aR  v always!!
aR is radially inward always!
• Newton’s 1st Law:
There must be a force acting!
For an object to be in uniform
nd
• Newton’s 2 Law:
circular motion, there must be
∑F = ma = maR
a net force acting on it. We
know the acceleration, so we
= m(v2/r) (magnitude)
can immediately write the force:
Direction: The total force must
be radially inward always!
A force is required to keep
an object moving in a circle.
If the speed is constant, the force is
directed toward the center of the
circle. The direction of the force is
continually changing so that it is
always pointed toward the center
of the circle.
Example: A ball twirled on the end of a string. In that case,
the force is the tension in the string.
• A particle moving in uniform circular motion, radius r
(speed v = constant)
• The acceleration: aR = (v2/r), aR  v always!!
aR is radially inward always!
• Newton’s 1st Law: There must be a force acting!
• Newton’s 2nd Law: ∑F = ma = maR= m(v2/r)
The total force ∑F must be radially inward always!
 The force which enters N’s 2nd Law  Centripetal Force
(Center directed force)
• NOT a new kind of force! Could be string tension,
gravity, etc. It’s the right side of ∑F = ma, not the left side!
(It’s the form of “ma”, for circular motion!)
Centripetal Force
You can understand that the
centripetal force must be
inward by thinking about the
ball on a string. Strings only
pull; they never push.
MISCONCEPTION!!
The force on the ball is NEVER
Outward (“Centrifugal”). It is
ALWAYS inward (Centripetal) !!
An outward (“centrifugal”)
force ON THE BALL is
NOT
F
a valid concept!
The string tension force
ON THE BALL
is inward (centripetal).
The string tension force
F
on the ball is INWARD toward
the center of the circle!
What happens if the cord on the
ball is broken or released?
For the ball to move in a circle, there must be
an inward (Centripetal) force pointed
towards the circle center so that the natural
tendency of the object to move in a straight line
(Newton’s 1st Law!) will be overcome.
If the centripetal force goes to zero, the ball
will fly off in a direction tangent to the circle
(Newton’s 1st Law again!)
There is no
centrifugal force pointing outward on the ball!
Example
Sparks fly in straight line tangentially from the
edge of a rotating grinding wheel
Example 5-3 (Estimate)
Estimate the force a person must exert on a string attached to a
0.15 kg ball to make the ball revolve in a horizontal circle of radius
0.6 m. The ball makes 2 rev/s.
m = 0.15 kg, r = 0.6 m, f = 2 rev/s  T = 0.5 s
Assumption: Circular path is  in horizontal plane, so
φ  0  cos(φ)  1
Newton’s 2nd Law: ∑F = ma  FTx = max= maR = m(v2/r)
v = (2πr/T) = 7.54 m/s
So, the tension is (approximately)
FTx  14 N
Example 5-4: Revolving ball (vertical circle)
A ball, mass m = 0.15 kg on the end of
a (massless) cord of length r = 1.1 m
cord is swung in a vertical circle.
Calculate:
a. The minimum speed the ball must
have at the top of its arc so that the
ball continues moving in a circle.
b. The tension in the cord at the
bottom of the arc, assuming that
there the ball is moving at twice the
speed found in part a.
Hint: The minimum speed at the top will
happen for the minimum tension FT1
Note: Here, string tension &
gravity are acting together (both
enter Newton’s 2nd Law!) to
produce centripetal acceleration.
Problem 7
r = 0.72 m, v = 4 m/s, m = 0.3 kg
Use: ∑F = maR
• At the top of the circle:
Vertical forces: (down is positive!)
FT1 + mg = m(v2/r)
FT1 = 3.73 N
• At the bottom of the circle:
Vertical forces: (up is positive!)
FT2 - mg = m(v2/r)
FT2 = 9.61 N
Example: Exercise C, p. 111
A Ferris wheel rider moves in a
vertical circle of radius r at
constant speed v. Is the normal
force FN1 that the seat exerts on
the rider at the top of the wheel
a. less than, b. more than, or
c. equal to the normal force FN2
that the seat exerts on the rider at
the bottom of the wheel?
Use
Newton’s 2nd Law: ∑F = maR
at top & bottom, solve for normal
force & compare.
1
2
Conceptual Example 5-5
A tether ball is hit so that it revolves
around a pole in a circle of radius r at
constant speed v. In what direction is
the acceleration? What force causes it?
Newton’s 2nd Law: ∑F = ma
x:
∑Fx = max
 FTx = maR = m(v2/r)
y: ∑Fy = may = 0
 FTy - mg = 0, FTy = mg