This Week

• Circular motion• Going round the bend• Riding in a ferris wheel, the vomit comet• Gravitation

Our solar system, satellites (Direct TV)• The tides, Dark matter, Space Elevator

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1/22/2018 Physics 214 Summer 2018 2

Circular MotionCircular motion is very common and very important in our everyday life. Satellites, the moon, the solar system and stars in galaxies all rotate in “circular” orbits. The term circular here is being used loosely since even repetitive closed motion is generally not a perfect circle.

At any given instant an object that is not moving in a straight line is moving along the arc of a circle.

So if we understand motionin a circle we can understandmore complicated trajectories.

Remember at any instant the velocity is along the path of motion but the acceleration can be in any direction.

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Circular motion

If the velocity of an object changes direction then the object experiences an acceleration and a force is required.

This is centripetal acceleration and force and is directed toward the center of the circle.

This is the effect you feel rounding a corner in a car

a = v2/rF = ma = mv2/r

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Balance of forces

We need to understand the forces that are acting horizontally and vertically.

In the case shown the tension or force exerted by the string has components which balance the weight in the vertical direction and provide the centripetal force horizontally.

Tv = W = mg Th = mv2/r

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1D-02 Conical Pendulum

NET FORCE IS TOWARD THE CENTER OF THE CIRCULAR PATH

T sin(θ) = mv2/RT cos(θ) = mgv = sqrt( gR tan(θ) )

Period of the pendulumτ= 2πR/v,where R = L / sin(θ)τ= 2πsqrt( Lcos(θ)/g )

Could you find the NET force?

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1D-03 Demonstrations of Central Force

THE SHAPES/SURFACES OF SEMI-RIGID OBJECTS BECOME MORE CURVED TO PROVIDE GREATER CENTRAL FORCES DURING ROTATION.

2T cos (θ)= mv2/R

What will happen when it is

subjected to forces during rotation ?

T

θθ

T

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Cars

When a car turns a corner it is friction between the tires and the road which provides the centripetal force.

If the road is banked then the normal force also provides a force.

For a banked track there is a velocity for which no friction is required.

Ff

Above

W = mgFf

Ff

Rear

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Vertical circles

If v = 0 then N = mg

As v increases N becomes smaller

When v2/r = g the car becomes weightless. Same as the “vomit comet”

mg – N = mv2/rFerris wheel

g

At the bottomN - mg = mv2/r

At the top Mg – N = mv2/r

+ is always toward the center of the circle

W = mg

N

v

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1D-05 Twirling Wine Glass

THE GLASS WANTS TO MOVE ALONG THE TANGENT TO THE CIRCLE AND THE REACTION FORCE OF THE PLATE AND GRAVITY PROVIDE THE CENTRIPETAL FORCE TO KEEP IT IN THE CIRCLE

g

Same as

mv

string

WHAT IS THE PHYSICS THAT KEEPS THE

WINE FROM SPILLING ?

N + mg = mv2/R N > 0

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1D-07 Paper Saw

THE RADIAL FORCES HOLDING THE PAPER TOGETHER MAKE THE PAPER RIGID.

Is paper more rigid

than wood ?

Free fall versus Weightless

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Near the earths surface there is a force F = mg which is called the weight of an object. Since the force of gravity is always present then this force always is present. There are situations where if we were standing on a scale the apparent weight, the reading on the scale, could be smaller or larger than the real weight.• In an accelerating elevator

In the case of free fall our apparent weight is zero• In an elevator and the cable snaps.

The case of being weightless corresponds to the case where one is moving in a circle and the force of gravity is exactly equal to the centripetal force required. Since the force of gravity is vertical the type of cases where weightlessness happens are

•At the top of a ferris wheel if mg = mv2/r•At the top of a hump back bridge if mg = mv2/r

Planes and satellites

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http://www.avweb.com/news/airman/184318-1.html

Vomit CometThis uses a plane moving in a circle so that mg = mv2/r Although this only occurs exactly at one point in the arc the The trajectory of the plane is designed so that the effect of gravity is very small for a much longer time

https://www.youtube.com/watch?v=DpIsbayP-xg

Space stationAs a satellite circles the earth in a stable orbit we have all the time that the gravitational force exactly equals the centripetal force since the center of the earth is the center of the circle and the direction of gravity. So everything and everybody is weightless all the time independent of mass. That is toothbrushes and people are both weightless.

GmM/r2 = mv2/r or GM/r2 = v2/r

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Gravitation and the planetsAstronomy began as soon as man was able to observe the sky and records exist going back several thousand years. In particular the yearly variation of the stars in the sky and the motion of observable objects such as planets.

People observed the “fixed” North Star and, for example, the rising of Sirius signaling the flooding of the Nile.

Copernicus was the first person to advocate a sun centered solar system. Followed by Galileo who used the first telescopesTycho Brahe was the most famous naked eye astronomer.Kepler, his assistant used the data to draw quantitative conclusions.

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Keplers Laws1) Orbits are ellipses

2) The radius vector sweeps outareas in equal times equal

3) T2 proportional to r3

T is the period which for the earthis one year and r is the average radius

For circular motion with constant velocity vThe circumference of a circle is 2πR and the

Period T = 2πr/v

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Newton and GravitationNewton developed the Law of Gravitation

force between two objects isF = GM1m2/r2 .

The constant ofproportionality was measuredby Cavendish after more than 100 years

G = 6.67 x 10-11 N.m2/kg2.

Since at the earths surface mg = GmMe/r2 the experiment measured the mass of the earth

http://www.physics.purdue.edu/class/applets/NewtonsCannon/newtmtn.html

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Planetary orbits

For a simple circular orbit

GmM/r2 = mv2/r Period T = 2πr/v

where M is the mass of the sun and m the mass of the earth or M is the mass of the earth and m the mass of a satellite.

For a geosynchronous orbitperiod is 24 hours and height above the earths surface is 22,000miles

T2/r3 = 4π2/GMs

Calculations

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GmMe/r2 = mv2/r where Me is the mass of the earthand m the mass of a satellite.

At the earths surface mg = GmMe/Re2

v2 = gRe2 / r for a stable orbit

So at the earths surfaceV = ~ 17500mphFor a synchronous satellite at a height of 22,000 milesV = ~6500mph

V decreases with distance.The moon is losing energy because of the tides so it is moving away from us.

rv

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1D-04 Radial Acceleration & Tangential Velocity

AT ANY INSTANT, THE VELOCITY VECTOR OF THE BALL IS DIRECTED ALONG THE TANGENT. AT THE INSTANT WHEN THE BLADE CUTS THE STRING, THE BALL’S VELOCITY IS HORIZONTAL SO IT ACTS LIKE A HORIZONTALLY LAUNCHED PROJECTILE AND LANDS IN THE CATCH BOX.

Once the string is cut, where is the

ball going?

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1D-08 Ball in Ring Is the ball leaving

in a straight line or continuing this circular path?

THE FORCE WHICH KEEPS THE BALL MOVING CIRCULAR IS PROVIDED BY THE RING. ONCE THE FORCE IS REMOVED, THE BALL CONTINUES IN A STRAIGHT LINE, ACCORDING TO NEWTON’S FIRST LAW.

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Summary of Chapter 5Circular motion and centripetal acceleration and force.

Fc = mv2/rFerris wheel, car around a corner or over a hill.

Gravitation and Planetary orbits

For a simple circular orbit

GmM/r2 = mv2/r M is the mass of the sun and m the mass of the earth.

v2 = GM/r T = 2πr/vT2 = 4π2r2/v2 = 4π2r3/GMs

T2/r3 = 4π2/GMs

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Examples of circular motion

W = mg

N

v

mg – N = mv2/r

N

mgN - mg = mv2/r

v

Looking down

N

N = mv2/r

Side

mg

Ff

mg = Ff

Vertical motion

mg + T = mv2/r top

T - mg = mv2/r bottom

v

Tmg

Our World

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Moon and tides

Tides are dominantly due to the gravitational force exerted by the moon. Since the earth and moon are rotating this effect also plays a role. The moon is locked to the earth so that we always see the same face. Because of the friction generated by tides the moon is losing energy and moving away from the earth.

anim0012.mov

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Dark MatterFor the orbit of a body of mass m about a much more massive body of mass M GmM/r2 = mv2/r and GM/r = v2. In fact M is the mass inside the orbit, that is the sun could be nearly as big as the orbit of the earth and it would not change anything. If we look at stars in motion in galaxies we find there is not enough normal matter to provide the necessary gravitational force. We believe this is caused by a new form of matter, which we call dark matter, and it comprises 25% of the energy in our Universe.(normal matter = 4.4%)

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Space Elevator

•The Space Elevator is a thin ribbon, with a cross-section area roughly half that of a pencil, extending from a ship-borne anchor to a counterweight well beyond geo-synchronous orbit. •The ribbon is kept taut due to the rotation of the earth (and that of the counterweight around the earth). At its bottom, it pulls up on the anchor with a force of about 20 tons. The ribbon is 62,000 miles long, about 3 feet wide, and is thinner than a sheet of paper. It is made out of a carbon nanotube composite material. •Electric vehicles, called climbers, ascend the ribbon using electricity generated by solar panels and a ground based booster light beam.

•This project does not exist although design work continues

100,000km

vstationcounterweight

http://www.youtube.com/watch?v=F2UZDHHDhog

http://www.pbs.org/wgbh/nova/sciencenow/3401/02.html

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Trajectories to other planetsTo launch a space craft from earth to sayMars or Jupiter is quite complicated since all the planets are moving including the earth. One needs to be able to calculate a trajectory that minimizes the amount of fuel required. That is why in nearly all launches there are specific time windows which are optimum.

Worked Questions and Problems

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Questions Chapter 5Q6 A ball on the end of a string is whirled with constant speed in a counterclockwise horizontal circle. At point A in the circle, the string breaks. Which of the curves sketched below most accurately represents the path that the ball will take after the string breaks (as seen from above)? Explain.

•

4 3

2

1A

Path number 3

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Q8 For a ball twirled in a horizontal circle at the end of a string, does the vertical component of the force exerted by the string produce the centripetal acceleration of the ball? Explain.

Q9 A car travels around a flat (unbanked) curve with constant speed.

A. Show all of the forces acting on the car.

B. What is the direction of the net force act.

Vertical component balances the weightHorizontal component provides the acceleration

Ff

RearN

mg

The force acts toward the center of the turn circle

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Q11 If a curve is banked, is it possible for a car to negotiate the curve even when the frictional force is zero due to very slick ice? Explain.

Q10 Is there a maximum speed at which the car in question 9 will be able to negotiate the curve? If so, what factors determine this maximum speed? Explain.

Yes. The friction between the tires and the road

Yes there is just one speed. If the car moves too slowly it will slide down. If it moves to fast it will slide up.

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Q12 If a ball is whirled in a vertical circle with constant speed, at what point in the circle, if any, is the tension in the string the greatest? Explain. (Hint: Compare this situation to the Ferris wheel described in section 5.2).

Q19 Does a planet moving in an elliptical orbit about the sun move fastest when it is farthest from the sun or when it is nearest to the sun? Explain by referring to one of Kepler’s laws.

The tension is the greatest at the bottom because the string has to support the weight and provide the force for the centripetal acceleration.

When it is nearest

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Q20 Does the sun exert a larger force on the earth than that exerted on the sun by the earth? Explain.

Q23 Two masses are separated by a distance r. If this distance is doubled, is the force of interaction between the two masses doubled, halved, or changed by some other amount? Explain.

The magnitude of the forces is the same they are a reaction/action pair

The force reduces by a factor of 4

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Wearth = m gearth = 180 lb

Wmoon = m gmoon

gmoon = 1/6 gearth

Wmoon = m 1/6 gearth = 1/6 m gearth = 1/6 (180 lb)

gmoo

n

The acceleration of gravity at the surface of the moon is about 1/6 that at the surface of the Earth (9.8 m/s2). What is the weight of an astronaut standing on the moon whose weight on earth is 180 lb?

Ch 5 E 14

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Time between high tides = 12 hrs 25 minutes.High tide occurs at 3:30 PM one afternoon.a) When is high tide the next afternoonb) When are low tides the next day?

a) 3:30 PM + 2 (12 hrs 25 min)

= 3:30 PM + 24 hrs + 50 min

= 4:20 PM

b) Low tide the next day = 4:20 PM - 6 hr 12 min 30 s = 10:07:30 AM2nd Low tide = 10:07:30 AM + 12 hrs 25 min = 10:32:30 PM

T= 12hrs 25min

T= 12hrs 25min

high tide

low tide

t

Ch 5 E 16

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A Ferris wheel with radius 12 m makes one complete rotation every 8 seconds.a) Rider travels distance 2r every rotation. What speed do riders

move at?b) What is the magnitude of their centripetal acceleration?c) For a 40 kg rider, what is magnitude of centripetal force to keep him

moving in a circle? Is his weight large enough to provide this centripetal force at the top of the cycle?

d) What is the magnitude of the normal force exerted by the seat on the rider at the top?

e) What would happen if the Ferris wheel is going so fast the weight of the rider is not sufficient to provide the centripetal force at the top?

Ch 5 CP 2

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a) S = d/t = 2r/t = 2(12m)/8s = 9.42 m/s

b) acent = v2/r = s2/r = (9.42m/s)2/12m = 7.40 m/s2

c) Fcent = m v2/r = m acent = (40 kg)(7.40 m/s2) = 296 NW = mg = (40 kg)(9.8 m/s2) = 392 NYes, his weight is larger than the centripetal force required.

d) W – Nf = 296 N = 96 newtonsN

W

e) rider is ejected

Ch 5 CP 2 (con’t)

Fcent

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A passenger in a rollover accident turns through a radius of 3.0m in the seat of the vehicle making a complete turn in 1 sec.a) Circumference = 2r, what is speed of passenger?b) What is centripetal acceleration? Compare it to gravity (9.8

m/s2)c) Passenger has mass = 60 kg, what is centripetal force

required to produce the acceleration? Compare it to passengers weight.

3 m

a) s = d/t = 2(3.0m)/1 = 19m/s

c) F = ma = (60 kg)(118 m/s2) = 7080 NF = ma = m (12 g) = 12 mg = 12 weight

b) a = v2/r = s2/r = (19 m/s)2/3m = 118 m/s2 = 12g

Ch 5 CP 4

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The period of the moons orbit about the earth is 27.3 days, but the average time between full moons is 29.3 days. The difference is due to the Earth’s rotation about the Sun.a) Through what fraction of its total orbital period does the

Earth move in one period of the moons orbit?b) Sketch the sun, earth & moon at full moon condition.

Sketch again 27.3 days later. Is this a full moon?c) How much farther does the moon have to move to be in

full moon condition? Show that it is approx. 2 days.

a) Earth orbital period = 365 days = 0E

Moon orbital period = 27.3 days = 0M

0M/0E = 27.3/364 0.075

Ch 5 CP 6

1/22/2018 Physics 214 Summer 2018 39

c) For moon to achieve full moon condition, it must sit along the line connecting sun & earth. In part (a) we found that the earth has moved thru 0.075 of its full orbit in 27.3 days (see diagram (ii)). To be inline w/ sun and earth, moon must move thru same fraction of orbit (see diagram (iii)).

0.075 (27.3 days) 2 days.

b) Day 0Full Moon

27.3 Days LaterThis is not a full moon.

This is the next full moon.

S

S

S

E

E

E

M

M

(i)

(ii)

(iii)

M

Ch 5 CP 6 (con’t)