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Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
Lecture 9
Chapter 5
Uniform Circular Motion
02.19.2014Physics I
Course website:http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI
Lecture Capture: http://echo360.uml.edu/danylov2013/physics1spring.html
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
Outline
Chapter 5. Sections 5.2 – 5.4
Circular Motion Dynamics Terminal Velocity
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
Motion in a circular pathInstantaneous velocity is tangent to the path
Velocity is a vector (magnitude/direction) Magnitude of the velocity stays
the same (constant) but Direction of the velocity changes
So there must be accelerationthat causes that
Let’s find, first, a direction of this acceleration.
Uniform Circular Motion
v
v
v
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
Centripetal acceleration
The instantaneous acceleration is in the direction of ,and is given by:12 vvv
1v
r1 2r 2v
1v
2v v
Average acceleration v
tv2
v1
t2 t1
a limt 0
v
t dv
dt
Ra
Let’s derive it:
This acceleration is called centripetal, or radial, acceleration, and it points toward the center of the circle.
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
Look at change in velocity for infinitesimally small time interval
Centripetal Acceleration: derivation
vv l
r
aR v2
r
aR vr
limt 0
lt
aR lim
t 0vrlt
aR
limt 0
vt
v vrl
v
Similar triangles
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
Centripetal Acceleration
m v
v
v
mm
Ra
Ra
Ra
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
For an object to be in uniform circular motion, there must be a net force acting on it radially inwards.
This can be any one of the forces we have already encountered: tension, gravity, normal force, friction, …
Centripetal Acceleration: Dynamics
ConcepTest 1 Missing Link
Once the ball leaves the tube, there is no longer
a force to keep it going in a circle. Therefore, it
simply continues in a straight line, as Newton’s
First Law requires!
A Ping-Pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the Ping-Pong ball leaves the track, which path will it follow?
Follow-up: What physical force provides the centripetal acceleration?
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
NF
Examples of centripetal acceleration (Wall of death)
RmvFN
2
v –velocity of the motorbikeR- radius of the circle
FN
Ffr
mg
Bike going in a circle: the wall exerts an inward force (normal force) on a bike to make it move in a circle.
Normal force provides the centripetal acceleration
maF
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
Examples of centripetal acceleration (II).
TF
Tension provides the centripetal acceleration
RmvFT
2
A hammer going in a circle: the cord exerts an inward force (tension) on a hammer to make it move in a circle.
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
mg
Ty
Tx
aR
Example: Conical pendulum
T
y
x
)1(0 mgTF yy
)2(2
RmvSinTTx
xxx maTF
X component of tension provides the centripetal acceleration
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
Examples of centripetal acceleration (I).
Car going in a circle: the road exerts an inward force (friction) on a car to make it move
in a circle.
frF
NF
gm
Rvar
2
RmvFfr
2
v –velocity of the carR- radius of the circle
Friction provides the centripetal accelerationhttp://phys23p.sl.psu.edu/phys_anim/mech/car_Fc_new.avi
The friction force between tires and road provides the centripetal force that keeps the car moving in a circle. If this force is too small, the car continues in a straight line!
A) car’s engine is not strong enough to keep the car from being pushed out
B) friction between tires and road is not strong enough to keep car in a circle
C) car is too heavy to make the turnD) a deer caused you to skidE) none of the above
You drive your car too fast around a curve and the car starts to skid. What is the correct description of this situation?
ConcepTest 2 Around the Curve
Follow-up: What could be done to the road or car to prevent skidding?
RmvFfr
2
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
Banked curve
But sometimes, friction force is not enough to keep a car on a circular road.
Banking the curve can help to keep cars from skidding.
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
mg
FNcos
FNsin
Fy = FNcos mg = 0
Fx = FNsin = maxax= v2/R
FN
FNsin mv2/R
tan v2/gRIndependent of object mass !!!
Banked Curves
X component of normal force provides the centripetal acceleration
http://phys23p.sl.psu.edu/phys_anim/mech/car_banked_new.avi
x
y FNcos mg
y
x
Take a ratio
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
Loop the Loop
http://phys23p.sl.psu.edu/phys_anim/mech/car_vert_bare.avi
http://phys23p.sl.psu.edu/phys_anim/mech/car_vert_fail.avi
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
Loop the LoopTo make the loop-the-loop at a constant speed, what minimum speed does the motorcycle need?
RvmmgFN
2
gm
v
R
v
NF
gm
NF
1
2
y
y
2 yy maF Rvaa Ry
2
mgFmRv N
When FN=0 (feels like no weight), then speed is minimum
gRv http://phys23p.sl.psu.edu/phys_anim/mech/car_vert_bare.avi
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
Loop the LoopApparent weight at the bottom, point 1?
RvmmgFN
2
R
v
gm
NF
1
y
1 yy maF Rvaa Ry
2
Thus, FN>mg. You would feel heavier (similar to an elevator)
mgRvmFN
2
http://phys23p.sl.psu.edu/phys_anim/mech/car_vert_bare.avi
ConcepTest 3 Going in Circles
A) FN remains equal to mg
B) FN is smaller than mg
C) FN is larger than mg
D) none of the above
You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force FN exerted by your seat is equal to your weight mg. How does FN change at the top of the Ferris wheel when you are in motion?
You are in circular motion, so there has to
be a centripetal force pointing inward. At
the top, the only two forces are mg (down)
and N (up), so N must be smaller than mg.
Follow-up: Where is FN larger than mg?
RvmFmg N
2
NFRvmmg
2
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
Drag Force Concepts
Force on object as it moves through Fluid or GasVelocity dependent forcesFormulae complexGood approximations:
vbFD
2bvFD
Department of Physics and Applied Physics95.141, Spring 2014, Lecture 9
If FD bv
Terminal Velocity
yy maF maFmg D
mabvmg
When mg=bv, then a=0 and vT=const (Terminal velocity)
R
vFc points toward the center of the circle (i.e., downward in this case). The weight vectorpoints down and the normal force (exerted by the hill) points up. The magnitude of the net force, therefore, isFc = mg – N.
mg N
A skier goes over a small round hill with radius R. Because she is in circular motion, there has to be a centripetal force. At the top of the hill, what is Fc of the skier equal to?
ConcepTest 4 Going in Circles II
Follow-up: What happens when the skier goes into a small dip?
A) Fc = N + mg
B) Fc = mg – N
C) Fc = T + N – mg
D) Fc = N
E) Fc = mg