Centripetal Acceleration
– acceleration towards the center of a circle.
– a.k.a.
Radial Acceleration (a
R
)

a
R a
R v v
Ball rolling in a straight line (inertia) v
Same ball, hooked to a string a
R
= v 2 r

If you are on a carousel at constant speed, are you experiencing acceleration?

If you twirl a yo-yo and let go of the string, what way will it fly?

Period (T)
– Time for one complete (360 o ) revolution
– seconds
Frequency
– Number of revolutions per second
– rev/s or Hertz (Hz)
T = 1 f

a
R
= v 2 r
T = 1 f
v = 2 p r
T v = 2 p rf

A 150-g ball is twirled at the end of a 0.600 m string. It makes 2.00 revolutions per second. Find the period, velocity, and acceleration.
(0.500 s , 7.54 m/s, 94.8 m/s 2 )

The moon has a radius with the earth of about
384,000 km and a period of 27.3 days. a. Calculate the acceleration of the moon toward the earth.
b. Convert the answer to g’s.
(Ans: 2.72 X 10 -3 m/s 2 , 2.78 X 10 -4 g )

Jupiter is about 778 X 10 6 km from the sun. It takes 4332.6 days to orbit the sun.
a) Calculate the circumference of Jupiter’s . orbit.
(4.89 X 10 12 m) b) Calculate Jupiter’s period in seconds. (3.74 X
10 8 s) c) Calculate Jupiter’s orbital speed. (1.30 X 10 4 m/s) d) Calculate Jupiter’s centripetal acceleration towards the sun. (2.18 X 10 -4 m/s 2 )

Centripetal Force
– “center seeking” force that pulls an object in a circular path.
– Yo-yo
– Planets
– Merry-go-round
– Car rounding a curve?

• Doesn’t exist
• “apparent outward force”
• When you let the string go, the ball will continue in a straight line path. No new acceleration involved.
• Water in swinging cup example

Direction water wants to go
Centripetal
Force of string

S
F = ma
R
= mv 2 r
A 0.150 kg yo-yo is attached to a 0.600 m string and twirled at 2 revolutions per second. a) Calculate the circumference of the circle (3.77 m) b) Calculate the linear speed (7.54 m/s) c) Calculate the centripetal force in the string (14.2
N)

An electron orbits the nucleus with a radius of 0.5 X
10 -10 m. The electron has a mass of 9 X 10 -31 kg and a speed of 2.3 X 10 6 m/s.
a. Calculate the centripetal force on the electron.
(9.52 X 10 -8 N) b. Calculate the frequency of an electron. (7.32 X
10 15 Hz) c. Convert the velocity to miles/s. (1429 miles/s) d. What provides the centripetal force?

Thor’s Hammer
(mjolnir) has a mass of 10 kg and the handle and loop have a length of 50 cm. If he can swing the hammer at a speed of
3 m/s, what force is exerted on Thor’s hands?
(Ans: 180 N)

Can Thor swing his hammer so that it is perfectly parallel to the ground?
F
H

mg
What angle will the hammer take with the horizontal?
Mass = 10 kg r = 50 cm v = 3 m/s q

mg
What angle will the hammer take with the horizontal?
q
Let’s resolve the F
R vector into it’s components:
F
Hx
= F h cos q
F
Hx
= mv 2 /r
F
H cos q
= mv 2 /r
S
F y
= 0 (the hammer is not rising or falling)
0 = F
H sin q – mg
Two Equations: Two Unknowns

F
H cos q
= mv 2 /r
F
H
= mv 2 rcos q mv 2 sin q
= mg r cos q sin q
= gr cos q v 2
F
H sin tan q
= gr v 2 q
= mg q
= 28.6
o

A 0.0050 kg walnut is swung in a radius of 50.0 cm. The walnut makes 2.50 revolutions per second.
a) Calculate the linear speed of the walnut. (7.85 m/s) b) Draw a free-body diagram of the walnut.
c) Calculate the centripetal force needed to keep it in a circle. (0.616 N) d) Calculate the force of tension and the angle the string makes with the horizontal. (0.618 N,
4.5
o )

A 0.150 kg ball is swung on a 1.10-m string in a vertical circle. What minimum speed must it have at the top of the circle to keep moving in a circle?
mg F
T
At the top of the circle, both the weight and the tension in the string contribute to the centripetal force
S
F = F
T
+ mg v = 3.28 m/s

What is the tension in the cord at the bottom of the arc if the ball moves at the minimum speed? (v = 3.28 m/s)
F
T
= 2.94 N
F
T mg

A rollercoaster vertical loop has a radius of
20.0 m. Assume the coaster train has a mass of 3,000 kg.
a) Calculate the minimum speed the coaster needs to have to make the loop. (14.0 m/s) b) Calculate the normal force the tracks provide to the train at the bottom of the curve if the train is travelling at 25.0 m/s.
(123,150 N) c) Calculate the normal force the tracks provide at the top of the curve if the train is travelling at 25.0 m/s. (64, 350 N)

The ferris wheel at Knoebels has a radius of
16.8 m and travels at a speed of 3.52 m/s.
a) Calculate the frequency and period (0.033 Hz,
30 s) b) Calculate the normal force that the seat provides to a 56.0 kg rider at the top. (507 N) c) Calculate the normal force that the seat provides to a 56.0 kg rider at the bottom. (590
N)

A car travels over a round hill (radius = 50.0 m).
a) Calculate the maximum speed at which the car can take the hill. (22.0 m/s) b) Calculate the normal force on a 1000.0 kg car if it is travelling over the hill at 10.0 m/s. (7.80 X
10 3 N)

• Friction provides centripetal force
• Use ( m s
). Wheels are turning, not sliding, across the surface
• Wheel lock = kinetic friction takes over. m k is always less than m s
, so the car is much more likely to skid.

A 1000-kg car rounds a curve (r=50 m) at a speed of
14 m/s.
F
N mg
F fr
= F c a) Calculate the centripetal force needed to keep the car on the road b) Calculate if the car will skid if the road is dry and m s
= 0.60
c) Calculate if the car will skid if the road is icy and m s
= 0.25

A 15,000-kg truck can safely round a 150 m curve at a speed of 20 m/s. a) Calculate the centripetal force needed to keep the car on the road (40,000 N ) b) Calculate the coefficient of static friction (0.27) c) Calculate the maximum speed a 1000 kg Cube car can take the turn. (20 m/s)

The Rotor at an amusement park has a radius of
7.0 m and makes 30 rev/min.
a) Calculate the speed of the rotor. ( 22.0 m/s) b) Draw a free body diagram of a person in the rotor. What causes the F
N
?
c) Calculate the coefficient of static friction between the person and the wall. (0.14)

• Banked to reduce the reliance on friction
• Part of the Normal Force now contributes to the centripetal force

F
C
= F fr
+ F
N sin q
(ideally, we bank the road so that no friction is required: F fr
= 0)

A 1000-kg car rounds a 50 m radius turn at 14 m/s. What angle should the road be banked so that no friction is required?
F
N q mg q

Now we will simply work with the Normal
Force to find the component that points to the center of the circle
F
N cos q q q mg
F
N
F
N sin q q
First consider the y forces.
S
F y
= F
N cos q
- mg
Since the car does not move up or down:
S
F y
= 0
0 = F
N cos q – mg
F
N cos q
= mg
F
N
= mg/cos q

mv 2 = F
N sin q r mv 2 = mgsin q r v 2 = gtan q cos q r v 2 = gtan q r v 2 = tan q gr

tan q
= (14 m/s) 2
(50 m)(9.8m/s 2 )
= 0.40
q
= 22 o

A 2,000-kg Nascar car rounds a 300 m radius turn at 200 miles/hr. a. Convert the speed to m/s. (89.4 m/s) b. What angle should the road be banked so that no friction is required? (70 o ) c. Suppose a track is only banked at 35.0
o , calculate the maximum speed that a car can take the turn. (45.3 m/s, 101 mph) d. Looking at the formula for banking angle, how could a track designer decrease that angle?

• True weightlessness exists only very far from planets
• “Apparent weightlessness” can be achieved on earth

a= 0
S
F = F
N
0 = F
N
– mg
– mg
F
N
= mg
Suppose Chewbacca has a mass of 102 kg:
F
N
= mg = (102kg)(9.8m/s 2 )
F
N
= 1000 N
F
N mg a is zero

a = 4.9 m/s 2
S
F = F
N
– mg ma = F
N
– mg
F
N
= ma + mg
F
N
= m(a + g)
F
N
=(102kg)(4.9m/s 2 +9.8 m/s 2 )
F
N
= 1500 N
F
N mg a is upward

a = 4.9 m/s 2
S
F = mg - F
N ma = mg - F
N
F
N
= mg - ma
F
N
= m(g - a)
F
N
=(102kg)(9.8m/s 2 – 4.9 m/s 2 )
F
N
= 500 N
F
N mg a is down

At what acceleration will he feel weightless?
F
N
= 0
S
F = mg - F
N ma = mg -F
N ma = mg - 0 ma = mg a = 9.8 m/s 2
Apparent weightlessness occurs if a > g
F
N mg

Calculate the apparent weight of a 56.0 kg man in an elevator if the elevator is: a) Accelerating upwards at 2.00 m/s 2 . (661 N) b) Accelerating downwards at 2.00 m/s 2 (436 N) c) Accelerating downwards at 9.80 m/s 2 (0 N) d) Accelerating sideways at 9.80 m/s 2 . (549 N)

Even when you are running, you fell weightless between strides.

Why don’t satellites fall back onto the earth?
• Speed
• They are “falling” due to the pull of gravity
• Can feel
“weightless” (just like in the elevator)

Is gravity caused by the earth’s rotation?
Will a man down here fall off if the earth stops rotating?

Newton’s Law of Universal Gravitation
1. Every object in the universe is attracted to every other object. (based on mass)
2. The force drops off with the distance squared.
(As distance increases, the force of gravity drops very quickly)

F= Gm
1 m
2 r 2
G = 6.67 X 10 -11 N-m 2 /kg 2 m
1
= mass of one object m
2
= mass of second object r = distance from center of objects

Cavendish proves the law in
1798

Everything in the solar system pulls on everything else.
Sun pulls on Earth All the other planets also pull on the
Earth
Some comets/meteors are actually from outside our solar system and were captured by our sun’s gravity.

What is the force of gravity between two 60.0 kg (132 lbs) people who standing 2.00 m apart?
F= Gm
1 m
2 r 2
= (6.67 X 10 -11 N-m 2 /kg 2 )(60kg)(60kg)
(2.00m) 2
F = 6.00 X 10 -8 N
Such a force is so small that it is almost impossible to measure.

What is the force of gravity between a 60 kg person and the earth? Assume the earth has a mass of
5.98 X 10 24 kg and a radius of 6,400,000 m
(4,000 miles).
F= Gm
1 m
2
= (6.67 X 10 -11 N m 2 /kg 2 )(60kg)(5.98 X 10 24 kg) r 2 (6,400,00 m) 2
F = 584 N

A 2000-kg satellite orbits the earth at an altitude of
6380 km (the radius of the earth) above the earth’s surface.
What is the force of gravity on the satellite?
F= Gm
1 m
2
= (6.67 X 10 -11 N m 2 /kg 2 )(2000kg)(5.98 X 10 24 kg) r 2 (6,380,00 m + 6,380,00 m) 2
F = 4900 N

What is the net force on the moon when it is at a right angle with the sun and the earth?
Relevant Data:
M
M
= 7.35 X 10 22 kg
M
E
= 5.98 X 10 24 kg
M
S
= 1.99 X 10 30 kg r
MS
= 1.50 X 10 11 m r
ME
= 3.84 X 10 8 m

Calculate each force separately:
F
ME
= 1.99 X 10 20 N
F
MS
= 4.34 X 10 20 N
F
R
2 = F
ME
2 + F
MS
2
F
R
= 4.77 X 10 20 N tan q
= opp = F
ME q
= 24.6
o adj F
MS
F
ME
F
MS q
F
R
Earth
Sun

Calculate the mass of the earth knowing that it has a radius of 6.38 X 10 6 m. Start using the weight formula.

g = Gm
E r
E
2 g = (6.67 X 10 -11 N-m 2 /kg 2 )(5.98 X 10 24 kg)
(6.38 X 10 6 m) 2 g = 9.80 m/s 2

Calculate the value of g at the top of Mt.
Everest, 8848 m above the earth’s surface.
g = Gm
E r 2 g = (6.67 X 10 -11 N-m 2 /kg 2 )(5.98 X 10 24 kg)
(6.38 X 10 6 m + 8848 m) 2 g = 9.77 m/s 2

g varies with:
• Altitude
• Location
– Earth is not a perfect sphere
– Different mineral deposits can change density
– “salt domes” are low density salt regions near petroleum deposits

Objects weigh about 1/6 their weight on Earth on the Moon. Calculate the mass of the moon, knowing that the radius of the moon is 1734 km.

An object weighs 200 N on earth.
a) Calculate the acceleration of gravity on
Mars (3.71 m/s 2 ) b) Calculate its weight on Mars (75.5 N)
R e
= 6370 km
R m
= 3440 km
M e
= 5.98 X 10 24 kg
M m
= 0.11M
e

Three 5.00 kg bowling balls are placed at the corners of an equilateral triangle whose sides are
1.50 m long. Calculate the magnitude and direction of the gravitational force on the top ball.

Four 5.00 kg bowling balls are placed at the corners of an square whose sides are 1.50 m long. Calculate the magnitude and direction of the gravitational force on the lower left ball.
(1.42 X 10 -9 N, 45 o )

A geosynchronous satellite has a period of one day. The radius of the Earth is 6380 km and the mass of the Earth is 5.98 X 10 24 kg.
a) Convert the period to seconds (8.64 X 10 4 s) b) Calculate the height above the earth that a geosynchronous satellite must orbit. (Hint: use mv 2 /r , and subtract the radius of the earth) (3.59
X 7 m) c) Calculate the speed of the orbit. (3070 m/s)

A satellite orbits with a period of 5.00 hours. The radius of the Earth is 6380 km and the mass of the Earth is 5.98 X 10 24 kg.
a) Calculate the height of the satellite above the earth. (8.47 X 10 6 m) b) Calculate the speed of the orbit. (5.183X10
3 m/s)

1. The orbit of each planet is an ellipse, with the sun at one focus
2. Each planet sweeps out equal areas in equal time
3. T
1
2 = r
1
3
T
2
2 r
2
3

• Suppose the travel time in both cases is three days.
• Shaded areas are exactly the same area

Mars has a year that is about 1.88 earth years. What is the distance from Mars to the Sun, using the
Earth as a reference (r
ES
= 1.496 X 10 8 m)
T
1
2 = r
1
3
T
2
2 r
2
3
T
M
2 = r
M
3
T
E
2 r
E
3

r
M
3 = T
M
2 r
E
3
T
E
2 r
M
3 = (1.88y) 2 (1.496 X 10 8 m) 3
(1 y) 2 r
M
3 = 1.18 X 10 25 m 3 r
M
= 2.28 X 10 8 m

How long is a year on Jupiter if Jupiter is 5.2 times farther from the Sun than the earth?
T
J
2 = r
J
3
T
E
2 r
E
3
T
J
2 = r
J
3 T
E
2 r
E
3

T
J
2 = r
J
3 T
E
2 r
E
3
= (5.2) 3 (1 y) 2
(1) 3
T
J
2 = 141 y 2
T
J
= 11.9 y

How high should a geosynchronous satellite be placed above the earth? Assume the satellite’s period is 1 day, and compare it to the moon, whose period is 27 days. The average distance between the earth and the moon is 384,000 km.
T s
2 = r s
3
T
M
2 r
M
3 r s
3 = r
M
3 T s
2
T
M
2

r s
3 = r
M
3 T s
2
T
M
2 r
S
3 = r
M
3
729
= r
M
3 (1 day)
(27 day) 2
2
Take the 3 rd root of both sides r s
= r
M
9
The satellite must orbit 1/9 the distance to the moon. (4.27 X 10 7 m)

To derive Kepler’s Law, we will need two formulas.
F= Gm
1 m
J r 2 m
J m
1
F=m
1 v 2 r

Gm
1 m
J
= m
1 v 2 r 2 r
Gm
J
= v 2 r
Gm
J
= 4 p
2 r 2 r
T 2 = 4 p
2
T 2 r 3 Gm
J
Substitute v=2 p r
T

T 2 = 4 p
2 r 3 Gm
J
T
1
2 = 4 p
2 r
1
3 Gm
J
T
1
2 = T
2
2 r
1
3 r
2
3
We can do this for two
T
2
2 different moons
= 4 p
2 r
2
3 Gm
J

This form of the equation:
T 2 = 4 p
2
S could be the Sun,
Earth, or other body with satellites.
r 3 Gm
S
(solve it for m s
)
Useful for determining the mass of the central planet, using only the period and distance of one of the satellites.

What is the mass of the sun, knowing that the earth is 1.496 X 10 11 m from the sun.

T 2 = 4 p
2 r 3 Gm
S m
S
= 4 p
2 r 3
GT 2 m
S
= 4 p
2 (1.496 X 10 11 m) 3
(6.67 X 10 -11 N-m 2 /kg 2 ) (3.16 X 10 7 s) 2 m
S
= 2.0 X 10 30 kg

Calculate the mass of Neptune if you know that the period of its moon Galatea is 0.429 days, and the radius is 61,953 km from the center of Neptune.
(1.02 X 10 26 kg)

The Gemini 11 spacecraft sent two astronauts to a height of 1374 km above the earth’s surface.
The radius of the Earth is 6380 km and the mass of the Earth is 5.98 X 10 24 kg.
a) Calculate the speed of the orbit. (7173 m/s) b) Calculate the period of the satellite (1.89 hour)

The mass of Mars is 6.40 X 10 23 kg.
Calculate the period of its moon Phobos if
Phobos has an orbital radius of 9377 km.
(7.67 h)

Pluto has a radius of 1150 km and a mass of 1.20
X 10 22 kg.
a) Calculate the acceleration of gravity on Pluto
(0.605 m/s 2 ) b) Calculate the weight of a 70.0 kg person on
Pluto. (42.4 N) c) Calculate the acceleration of gravity on Pluto in terms of “g’s” (0.0618 g’s)

A student is given the following data and asked to calculate the mass of Saturn. The data describes the orbital periods and radii of several of
Saturn’s moons.
Orbital Period, T Orbital Radius, R
(seconds)
8.14 X 10 4
1.18 X 10 5
1.63 X 10 5
2.37 X 10 5
(m)
1.85 X 10 8
2.38 X 10 8
2.95 X 10 8
3.77 X 10 8

Let’s use this equation:
T 2 = 4 p
2 r 3 Gm
S
And rearrange it:
Gm
S
= 1
4 p
2 r 3 T 2
Once more:
1 = Gm
S
T 2 4 p
2 r 3

Calculate the following values and graph them.
1
T 2
G
4 p
2 r 3

1.60E-10
1.40E-10
1.20E-10
1.00E-10
8.00E-11
6.00E-11
4.00E-11
2.00E-11
0.00E+00
0.00E+00 5.00E-38 1.00E-37 1.50E-37 2.00E-37 2.50E-37 3.00E-37

Calculate the slope of the graph y = m x + b
1 = m
S
G
T 2 4 p
2 r 3 m s
= 5.9 X 10 26 kg

1. Gravity
2. Electromagnetic
3. Strong Nuclear Force
4. Weak Nuclear Force

2. a) 1.52 m/s 2 , center
4. 12 m/s
6. 14 m/s, no effect
8. a) 3.14 N b) 9.02 N b) 38.0 N, center
10. 27.6 m/s, 0.439 rev/s
12. 9.2 m/s
14. 11 rev/min
16.F
F
T1
T2
= 4 p
2 f 2 (m
1 r
1
= 4 p
2 f 2 m
2 r
2
+ m
2 r
2
)
18. 0.20

26. 1.62 m/s 2
28. 24.5 m/s 2
30. 0.91 g surface
32. 2.0 X 10 7 m
34. a) 9.8 m/s 2 b) 4.3 m/s 2
36. 9.6 X 10 17 N away from the Sun
38. 2.0 X 10 30 kg
40. 3.14 m/s 2 upward
42. 5.07 X 10 3 s (1.41 h), independent of mass
44.a) 58 kg b) 58 kg d) 39 kg e) 0 c) 77 kg

44.a) 58 kg (569N) b) 58 kg(569N) c) 77 kg (755 N) d) 39 kg (382N)
52. R
Icarus
= 1.62 X 10 11 m
54. 5.97 X 10 24 kg
56. 3.4 X 10 41 kg, 1.7 X 10 11 “Suns”
58.R
Europa
= 6.71 X 10 5 km
R
Ganymede
= 1.07 X 10 6 km
R
Callisto
= 1.88 X 10 6 km
68.1840 rev/day
70. a) 3000 m b) 5500 N c) 3900 N e) 0

A 1 kg yo-yo was swung in a circle at a constant speed. The force on the string was measured as the string was let out slowly.
0.5
0.6
0.7
0.8
0.9
Radius (m)
0.1
0.2
0.3
0.4
1.0
Force (N)
10.00
5.00
3.33
2.50
2.00
1.67
1.43
1.25
1.11
1.00

In a second experiment, the speed was changed while the length of the string (r) was kept constant
Speed (m/s)
1.0
2.0
3.0
4.0
5.0
6.0
Force (N)
1.0
4.0
9.0
16
25
36