Uniform Circular Motion Gravity - UMass Lowellfaculty.uml.edu/Andriy_Danylov/Teaching/documents/... · wheel when you are in motion? You are in circular motion, so there has to be

Department of Physics and Applied PhysicsPHYS.1410 Lecture 12 Danylov

Lecture 12

Chapter 8

Uniform Circular MotionGravity

Physics I

Course website:http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI


Today we are going to discuss:

Chapter 8:

Uniform Circular Motion: Section 8.2 Circular Orbits: Section 8.3 Reasoning about Circular Motion: Section 8.4

IN THIS CHAPTER, you will learn to solve problems about motion in two dimensions.


Examples. 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

Ncos

Nsin

Fz = Ncos mg = 0

Fr = Nsin = marar= v2/R

N

Nsin mv2/R

tan v2/gRIndependent of object mass !!!

Banked Curves (solution)

r component of normal force provides the centripetal accelerationhttp://phys23p.sl.psu.edu/phys_anim/mech/car_banked_new.avi

r

z

Ncos mg

z

r

Take a ratio


Example: 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?

RvmmgN

2

gm

v

R

v

N

gm

N

1

2

r

r

2 rr maF Rvar

2

mgNmRv

When N=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?

RvmmgN

2

R

v

gm

N

1

r

1 rr maF Rvar

2

Thus, N > mg. You would feel heavier (similar to an elevator)

mgRvmN

2

http://phys23p.sl.psu.edu/phys_anim/mech/car_vert_bare.avi

ConcepTest Going in Circles

A) N remains equal to mg

B) N is smaller than mg

C) N 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 N exerted by your seat is equal to your weight mg. How does N 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 N larger than mg?

RvmNmg

2

NRvmmg

2

N

RvmmgN

2

RvmmgN

2

Bottom


The motion of Satellites

Section 6.3Section 8.3


universal gravitational constant

M

m

F

Gravitational force is central, attractive, proportional to masses, and inversely proportional to the square of the distance

2rmMGF

Newton’s Law of Universal Gravitation

G 6.671011 N m2 kg2

A gravitational force of attraction exists between all objects that have mass


Gravity near the Earth’s SurfaceWhat should we get for the acceleration due to gravity at the Earth’s surface?

MEARTH 5.981024 kg REARTH 6.38106 m G 6.671011 Nm2

kg2

mg

g GME

RE2 9.80 m/s2

2rmGMF E

2E

E

RmGMF

ME

m Let’s apply N.2nd law for gravitational force

maF

2rmGMma E

Denote the acceleration due to gravity as 2)(

rGMrga E

If r=RE (on the surface)

F


Newton’s Cannon on a Mountain

http://waowen.screaming.net/revision/force&motion/ncananim.htm


Example:


Weighlessness (Explanation)


The “weightlessness” in orbit is exactly the same, everything is accelerating equally independent of mass.

The satellite and all its contents are in free fall, except with a huge tangential velocity. So, there is no normal force. This is what leads to the experience of weightlessness.

“Weightlessness”

They do have a gravitational force acting on them, though!


Thank youSee you on Wednesday


mg

Tz

Tr

ar

Example: Conical pendulum

T

z

r

)1(0 mgTF zz

)2(2

RmvSinTTr

rrr maTF

r component of tension provides the centripetal acceleration

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


Variations in “g” on the surface

The acceleration due to gravityvaries over the Earth’s surface dueto altitude, local geology, and theshape of the Earth, which is notquite spherical.

Documents

Uniform Circular Motion Gravity - UMass Lowellfaculty.uml.edu/Andriy_Danylov/Teaching/documents/... · wheel when you are in motion? You are in circular motion, so there has to be