### Centripetal Force & Newton`s Universal Law of

```Flashback


An object experiencing circular
motion has an angular speed of 10
rad/s and is traveling 5 m away from
its axis of rotation. Find the object’s
centripetal acceleration.
Hint: ac = rω2
Announcements



Units & Symbols Quiz on Wednesday
Circular Motion Quiz on Thursday
My contact information:
• Email:
[email protected]
• Cell: 859 466 6402
Centripetal Force &
Newton’s Universal Law
of Gravitation
Circular Motion
Physics
Mr. McCallister
Recall:



ac = vt2/r
ac = rω2
Newton’s Second Law: F = ma
Centripetal Force

Centripetal Force is the force
responsible for circular motion
• Symbol: Fc
• Unit: N



Recall: F = ma
Fc = mvt2/r
Fc = mrω2
Centripetal Force


Centripetal force is not a new kind of
force.
It is the name given to any kind of
force responsible for keeping an
object in circular motion.
• Ex: Car turning = friction is Fc
David’s sling = tension is Fc
Practice 7H p 261 #2-4

#1
• A girl sits in a tire that is attached to an
overhanging tree limb by a rope 2.10 m
in length. The girl’s father pushes her to
a tangential speed of 2.50 m/s. If the
magnitude of the force that maintains
her circular motion is 88.0 N, what is
the girl’s mass?
• Formula: Fc = mvt2/r
• Substitute: 88 = m(2.50)2 / 2.10
• Solve: m = 29.6 kg
Fundamental forces

All forces, at their most basic
fundamental level, are field forces!
• Ex: Atoms in hand repelling atoms in
wall

All mass in the universe attracts all
other mass in the universe with a
gravitational force. The size of the
force depends on:
• the amount of mass the objects have
• how far apart the objects are.
Gravitational Force

Gravitational force: an attractive
force between all mass in the
universe. The size of the force
depends on:
• the amount of mass the objects have
• how far apart the objects are.


Symbol: Fg
Unit: N
Newton’s Universal Law of
Gravitation

Fg = G m1 m2 / r2
• Where



m1 & m2 are masses in kg
r is the distance the masses are separated
in m. r is measured from the center of mass
of large objects (ex: planet’s core)
G is the constant of universal gravitation
• 6.673 x 10-11 N m2 / kg2
Stop to Think…


Compare the gravitational force on
the moon from the earth to the
gravitational force on the earth from
the moon.
They are equal! (Newton’s Third
Law)
Calculating g.



Let m1 = person’s mass, m2 =
Earth’s mass
Recall weight = mass x g, where g =
9.81 m/s2
So Fg = mg and Fg = G m1 m2 / r2
• mg = G m mE / rE2

The person’s mass cancels…
• g = G mE / rE2
Calculating g.


g = G mE / rE2
• mE = 5.98 x 1024 kg
• rE = 6.37 x 106 m
• G = 6.673 x 10-11 N m2/kg2
Substitute and Solve:
• g = 6.673 x 10-11 x 5.98 x 1024
(6.37 x 106)2
• g = 9.83 m/s2
Practice 7I pg 265 #2-4

#1
• If the mass of each ball in Sample
Problem 7I is 0.800 kg, at what distance
between the balls will the gravitational
force between the balls have the same
magnitude as that in Sample Problem
7I? How does the change in mass affect
the magnitude of the gravitational
force?
Practice 7I pg 265 #2-4

#1
• Formula:

Fg = G m1 m2 / r2
• Substitution:

8.92 x 10-11 = 6.673 x 10-11 x (.8 x .8)/r2
• Solve:

r = 0.692 m
Defying Gravity

Imagine a cannon on a tall mountain:
vesc, Earth = 11,200 m/s
Black Holes

In 1916, Karl Schwarzschild
predicted an object so massive and
dense that something travelling at
even the speed of light near the
dense object could not escape its
gravity.
Black Holes



The radius at which light cannot
escape the object’s gravity is called
The physical space at the
event horizon.
The physical space inside the event
horizon is called a black hole.
Gravitational Lensing

Massive objects can bend light, just
like a lens.
Homework due Tomorrow:

By end of class tomorrow:
• p. 261 7H #2-4
• p. 265 7I #2-3
• p. 265 Section Review #1-5
```