Curves Forces act in straight lines An object will move in a straight line if there is NO force acting on it or if the net force acts in the direction of the motion BUT what if the net force acts at an angle to the motion ID: 539458
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Slide1
Chapter 5: Circular Motion; GravitationSlide2
Curves
Forces act in straight lines
An object will move in a straight line if there is NO force acting on it or if the net force acts in the direction of the motion.
BUT, what if the net force acts at an angle to the motion?Slide3
Come F
ull Circle
From the last example, what would happen if the magnitude of the force and its angle to the motion remained constant?Slide4
Uniform Circular Motion
An object that moves in a circle at constant speed is said to experience
uniform circular motion
. Slide5
Kinematics of UCM
a
R
= v
2
/ r
Remember:
frequency (f) is the number of times the object completes a circle per second.
period (T) is the amount of time it takes to complete a circle, or T = 1/fAlso remember v = d/t. In a circle, distance means circumference(2πr) and time means period. So v = Slide6
Example
A 150g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600m. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?Slide7
Answer
v = (2
π
r)/T = (2*3.14*0.600m)/(0.500s) = 7.54m/s
a
R
= v
2
/r = (7.54m/s)
2 / (0.600m) = 94.8m/s2 Slide8
Dynamics of UCM
From chapter 4 we learned that acceleration requires a force.
So what causes the centripetal acceleration that causes UCM?
That would be the centripetal force,
Σ
F
R
We start with
ΣF = ma and through the wonders of substitution we get ΣFR =
maR orΣFR = mv2 / r Slide9
Beware
Centripetal force is NOT a new kind of force.
The term “centripetal” simply tells you that the force is “center seeking”.
Example: Picture swinging a child around by the arms. In order to keep the child moving in a circle you have to pull in with your arms. You pulling with your arms is no different from you pulling a box. Slide10
Example
Estimate the force a person must exert on a string attached to a 0.150kg ball to make the ball revolve in a horizontal circle of radius 0.600m. Also note the ball makes 2.00 revolutions per second.Slide11
Solution
Σ
F
R
= mv
2
/ r = m(2
π
r)
2 / r = (0.150kg)(7.54m/s)2 / (0.600m) ≈ 14NSlide12
Example
A 0.150kg ball on the end of a 1.10m long cord(of negligible mass) is swung in a vertical circle.
A.) determine the minimum speed the ball must have at the top of its arc so that it continues moving in a circle.
B.) Calculate the tension in the cord at the bottom of the arc assuming the ball is moving at twice the speed
from part A.Slide13
You Going Into a Curve
When you are in a car that enters a curve you get pulled toward the door.
This is because of Newton’s First Law; you want to keep going straight.
So what makes you move toward the center of the curve?Slide14
The Car Going Into the Curve
The car also wants to go straight when it enters a curve.
What makes the car move in a curve?
What happens if this force is not enough?Slide15
Example
A 1000kg car rounds a curve on a flat road of radius 50m at a speed of 14m/s. Will the car make the turn or will it skid off if…
a. the pavement is dry and the coefficient of static friction is
μ
s
= 0.6
b. the pavement is icy and
μ
s
= 0.25Slide16
Solution
The only force acting in the horizontal direction is friction.
We need to see if this force provides enough centripetal force to maintain the curve.
F
R
=
ma
R
= mv
2/r = (1000kg * (14m/s)2)/50m = 3900N this is how much force is needed.Ffr
= μsFN = μsmg = (0.60)(1000kg)(9.8m/s
2
)=5900N this is the max friction by the car, so we are fine.Slide17
Solution
On the icy road the
F
fr
becomes,
F
fr
=
μ
sFN = μsmg = (0.25)(1000kg)(9.8m/s
2) = 2500NThis is too low. The car would skid off the road.Slide18
Slamming
on the Brakes
The situation gets even worse if your wheels lock up. (granted this doesn’t happen much anymore thanks to the wonders of anti-lock brakes)
When the wheels are rolling there is a spot on the tire that is stationary to the ground, so the friction is static.
When the tire locks and slides as a whole unit, the friction is kinetic, which is lower.Slide19
Handling the Curves
There are a few ways we can make curves safer.
Reduce v (slow down when entering a curve)
Increase r (make the curve more gradual)
Increase friction (new asphalt technology)
Make the curve an incline (called a banked curve)Slide20
Banked Curves
By banking the curve we create a normal force that is at an angle, which creates 2 components.
The x component of F
N
(F
N
sin
θ
)
adds to the FR. We can set the horizontal normal force, Fnsin
θ equal FR, mv2/r to find the correct angle.Slide21
Example
What banking angle is need for an express way off-ramp curve of radius 50m at a design speed of 50km/hr (or 14m/s)Slide22
Solution
F
N
sin
θ
= mv
2
/r
We need to find F
N We know the vertical forces are balanced (because the car is not floating)So FN cosθ
= mgor FN = mg/cosθWe can plug this into FN sinθ
= mv
2
/r Slide23
Solution
sin
θ
= mv
2
/r
mg tan
θ
= mv
2/rtanθ = v2 /rg
tanθ = (14m/s)2 / (50m)(9.8m/s2) = 0.40Using tan-1 we get
θ
= 22
o
Slide24
Falling Bodies
At this point Newton had already constructed his 3 laws when he was pondering falling objects.
He already concluded that falling objects must have a force exerted on them, but what was doing the exerting?Slide25
High Minded
The story goes that Newton watched an apple fall to the ground and was struck by an inspiration: if gravity can reach tree tops and even mountain tops can it reach the moon?
It was this simple question (along with a lot of math) that set in motion a chain of thinking that connects Isaac Newton to Neil Armstrong to Curiosity and beyond.Slide26
Haters gonna
hate
Newton ran into a great deal of resistance to his theory of gravity.
What do you think was/were the issue(s)?Slide27
Pro-active
Newton did more than just bring up good questions; he set out to find the answers.
We already know that the acceleration due to gravity on the surface of the Earth is 9.8m/s
2
.
Let’s do an example that will cast some light on the nature of gravity.Slide28
That’s no space station
The Moon is 3.84x10
8
m from the Earth and travels in a near perfect circle. The period of the moon is 27.3days = 2.36x10
6
s. Find the centripetal acceleration,
a
R
, of the Moon. Slide29
Analysis
The distance from the center of the Earth to the Moon is 60 times the distance from the center of the Earth to its surface.
60
2
= 3600
Divide 9.8m/s
2
by 3600 and compare it to your answer from the last problem.
What do you conclude?Slide30
Mass Effect
We already stated in chapter 4 that objects with more mass need more force to reach the same acceleration as a less massive object.
We also know from Newton’s 3
rd
that if the Earth is pulling on the Moon, then Moon is pulling on the Earth.
Therefore, Newton concluded that the force of gravity is proportional to both the masses.Slide31
Put it all together
So far we got the force is related to 1/r
2
and also related to m
1
x m
2
Together they make NEWTON’S LAW OF UNIVERSAL GRAVITATION:Slide32
Near Earth Gravity
When we use on the surface of the earth, m
1
becomes the mass of the Earth (5.98E24kg), m
2
becomes the mass of the object on the surface, and r becomes the radius of the Earth (6.38E6m).Slide33
Weighty Subject Matter
As we have already stated in chapter 4, we call the force of gravity the weight of the object, mg.
So we can rewrite as
The mass cancels out and we get Slide34
Example
Estimate g on the top of Mt. Everest, 8848m above the Earth’s surface.Slide35
Solution
We need to add the distance from the surface to the summit to the radius of the Earth.
6.38E6m + 8.8E3m = 6.389E6m
We plug this value in as “r” inSlide36
Geosynchronous Orbit
An object in geosynchronous orbit is an object that hovers over the same spot above the Earth. It revolves at the same rate that the Earth rotates.
Why is this important?
How is this possible?Slide37
Ready for Launch
What is the necessary height above the Earth a satellite must reach to obtain geosynchronous orbit?
What is that satellite’s speed? (That is how fast does the Earth rotate?)Slide38
Solution
Start at F
R
=
ma
R
,
Substituting (
Gm
satmE)/r2 in for FR
and v2/r for aR gives us:(GmsatmE
)/
r
2
= m
sat
v
2
/r
Substituting (2
π
r)/T for v gives us
(
Gm
E
)/r
2
= (2
π
r)
2
/rT
2
(note that T = 1day = 86,400s)
We need to solve this for r. Slide39
Solution Cont.
Algebra gives us
Which equals 7.54 x 10
22
m
3
Taking the cube root gives us r = 4.23 x 10
7
m.Finally we are ready to find vSlide40
As The World Turns
We can solve
(
Gmm
E
)/r
2
=
mv
2/r for v and get,Slide41
Meet Johannes Kepler
Over 50 years before Newton published his 3 laws of motion and the law of gravitation, German astronomer Johannes Kepler(1571-1630) published many works including his 3 laws of planetary motion. (Kepler btw based his work off the earlier work of Tycho Brahe(1546-1601)).Slide42
Kepler’s 3 Laws
1: The path of each planet about the Sun is an ellipse with the Sun at one focus.
2: Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal periods of time.
3: The ratio of the squares of the periods of any two planets revolving about the Sun is equal to the ratio of the cubes of their mean distances from the Sun.Slide43
Our Focus is on 3
The last law can be represented mathematically as follows:
We can rearrange them to getSlide44
Newton over Kepler
Where Newton really shined was in his ability to show that all 3 of Kepler’s Laws can be derived mathematically using the Law of Universal Gravitation.Slide45
Imperfections Reveal the True Beauty
Since Kepler’s time we have gotten somewhat better at measuring the orbits the planets and we have seen that they do not exactly follow Kepler’s perfect ellipses.
Newton fixes this as well. He stated that everything pulls on everything else. Other planets tug on a planet and cause disturbances in its orbit.Slide46
Something is Wrong with Uranus
These disturbances actually lead to the discovery of 2 planets!
Astronomers could not account for the weirdness in the orbit of Uranus.
The only thing that could fix the math would be if there was more mass beyond Uranus.
This lead us to look for and find Neptune and Pluto.