Uniform Circular Motion

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


Outline

Chapter 5. Sections 5.2 – 5.4

Circular Motion Dynamics Terminal Velocity


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


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.


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


Centripetal Acceleration

m v

v

v

mm

Ra

Ra

Ra


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?


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


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.


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


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


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.


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


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


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


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


Drag Force Concepts

Force on object as it moves through Fluid or GasVelocity dependent forcesFormulae complexGood approximations:

vbFD

2bvFD


If FD bv

Terminal Velocity

yy maF maFmg D

mabvmg

When mg=bv, then a=0 and vT=const (Terminal velocity)


Thank youSee you on Monday

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

Uniform Circular Motion

Documents

Transcript of Uniform Circular Motion