PHYS214 - Physics - Purdue University

Week 4 Physics 214Spring 2010

1

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 gravitationally bound orbits can be elliptical (more or less “squashed circles”.) At any given instant a moving object, if it’s not in a straight line is moving along the arc of a (maybe changing) 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.


2

Circular motion

If the velocity of an object changes direction then the object experiences an acceleration. This requires a force.These are called “centripetal” acceleration and centripetal force and are directed toward the center of the circle.This is the force on you rounding a corner in a car

a = v2/rF = ma = mv2/r


3

Circular motion

Where does this come from? a = Δv/Δt = v.v/rLook at the Δv vector: it’s proportional to v. (and is directed toward the center of the circle.)Also, the rate that Δv is changing is proportional to the rate that the angle between v1 and v2 is changing (that is, the rate that vector v is swinging around.) This rate of change is just v/r (the angle is in RADIANS)

a = v2/rF = ma = mv2/r


4

Angular Measure in Radians

The “natural” measure of an angle is in RADIANS, which is just the arc length subtended by the angle, divided by the radius of the circle.A full circle, which is 360 degrees, has an arc length (circumference) of 2πR. Divide this arc by R and you get 2π radians is a full circle is 360 degrees. So 360 deg/2π rad = 57.296 [deg/rad] = 1.00000You can multiply or divide by this number and units to convert back and forth. Just use algebra on the [units] to know which to do.

RArc Length


5

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 right after the string breaks (as seen from above)? Explain.

•

4 3

2

1A

Path number 3

A=1, B=2, C=3 D=4.


6

Balance of forces

We need to understand the forces that are acting along both horizontal and vertical directions.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

r is the HORIZONTAL distance to the center of the circle.


7

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 some centripetal force.For each banked track there is a velocity for which no friction is required. The horizontal component of N exactly matches the needed FcNote Ff adjusts itself (up to break point)And it is the HORIZONTAL component of Ff which may help supply the needed Fcentrip.Note: Nv must balance W.Question: if v = 0, where does Ff point?

Ff

View from Above

W = mg

Ff

Ff

Rear view

A. Upslope B. Downslope


8

Vertical circles

If v = 0 then N = mgAs v increases N becomes smaller When v2/r = gThe car becomes weightless. Same as weightless-training airplane, for astronauts

mg – N = mv2/rFerris wheel

g

At the bottomN - mg = mv2/rAt the top mg – N = mv2/rWith seatbelt, at top if v2 is big enough, seatbelt would holdyou down, your “weight” would be upside down!

We choose+ always toward the center of the circle

W = mg

N

v

If car were running on “capture rails”that could also hold it down, big v would cause N to point downward


9

Response to a thoughtful questionMonday a student asked me how we can know if the rotating disks had moved 7o or maybe one or more full revolutions + 7o as the bullet went half a meter.

Logically, that could be the case, but quantitatively you can pretty easily rule it out by extra outside knowledge – namely, that the bullet is travelling pretty fast.

At 17 revolutions per second, if the disk made one full revolution, the bullet’s speed is about 0.5m x 17 “per s”[d = v t and v = d/t and t is 1/17 second for one rev.]

This is about 8.5 m/s, and if you fire a bullet at a can maybe 25.5 m away, it sure takes a lot less than 3 seconds for the can to jump, as you can observe.


10

Response to another thoughtful question

Last Friday, several students wanted a better explanation of the inertia ball (where you can break several strings while the single upper suspending string survives!)

I think the easiest way to think about it is to use the concept of IMPULSE. The hammer causes the rod in the lower string loops to speed up VERY briskly (very large acceleration). The heavy ball would need a fair amount of impulse to reach full speed.

Impulse is Ft, and if the interval t is very short, that would mean a LOT of force in the lower loops while the upper loop doesn’t exert a lot of extra force (it’s the NET force that generates motion and impulse.)


11

Gravitation and the planetsAstronomy began as soon as man was able to observe the sky. Records exist going back several thousand years. In particular, of 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 telescopes (the moons of Jupiter are like a small solar system.)Tycho Brahe was the most famous naked-eye astronomer.Kepler, his assistant used the data to draw quantitative conclusions.


12

Keplers Laws

1) Orbits are ellipses. For 0 “eccentricity” circle2) The radius vector sweeps outequal areas in equal times3) T2 proportional to r3

T is the period. For the earth, Tis one year and r is the average radius (distance to the sun). Earth’s eccentricity is small, but does affect climate cyclically.For circular motion with constant velocity vthe circumference of a circle is 2πr and thePeriod T = 2πr/v = distance (arc length) / speed


13

Keplers Laws

Period T = 2πr/v = distance (arc length) / speed

So v = 2πr/T = circumf./period = 2π 94M mi/1 yr

v ≈ 2π . 94 . 106 mi/ π . 107 s = 18.4 mi/s

Compare with escape velocity from earth, which I think is around 7 mi/s


14

Newton and Gravitation

Newton developed the Law of Universal Gravitation force between ANY two objects is proportional to M1m2/r2 . The (very small) constant ofproportionality was measuredby Cavendish more than 100 yearslater G = 6.67 x 10-11 N.m2/kg2.

Since at the earths surface mg = m(GMe/r2) the experiment measured the mass of the earth

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


15

Newton and Gravitation

The force between ANY two objects is ATTRACTIVE and = GM1m2/r2 .

N3 (third law) happens here too: each M pulls the other, equal and opposite.

For a spherical object, all its mass acts as if at the CENTER (provided the observer is outside the sphere!).


16

Quiz question

Two objects each of mass 1 Metric ton are 2 meters apart (between centers). What is the force between them? G = 6.67x10-11 and one Metric ton = 1,000kg

A. 1.7x10-11 N B. 1.7x10-17 N C. 1.7x10-5 N

D. 1.7x10-8 N E. 1.7x106 N

F=GMM/r2 = 6.67E-11Nm2/kg2 x 10E3kgx1E3kg/(2m)2

F = 6.67/4 x E(-11+6)N exponent is -5 answer is C.


17

Planetary orbitsFor a simple circular orbit GmM/r2 = mv2/r 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 an earth satellite.] Cancel m & solve

GM/r3=v2/r2 but v/r = 2π/T

GM/r3=(2π/T)2

T2/r3 = 4π2/GMs

Use average radius for elliptical orbits.

Works for other attractors: Earth, Jupiter.

Use right Mass


18

Satellite Orbits

For a geosynchronous orbitperiod is 24 hours and height above the earths surface is ~22,000 miles (add ~4000 miles to get the distance to the CENTER)

T2/r3 = 4π2 /GMe

R3 =T2GMe /4π2

T is 24 . 3600 s, G=6.6710-11, Me= 5.971024 kg, take cube root both sides to find R in meters, etc.

Geosynchronous orbit must be CIRCULAR, and will be quite off-scale on the above picture.

Also is equatorial.


19

Geosynchronous OrbitsThe notion of a geosynchronous satellite for communication purposes was first published in 1928 (but not widely so) by Herman Potočnik.[1] The idea of a geostationary orbit was first published on a wide scale in a paper entitled "Extra-Terrestrial Relays — Can Rocket Stations Give Worldwide Radio Coverage?" by Arthur C. Clarke, published in Wireless Worldmagazine in 1945.[2] In this paper, Clarke described it as a useful orbit for communications satellites. As a result this is sometimes referred to as the Clarke Orbit.[3] Similarly, the Clarke Belt is the part of space approximately 36,000 km (22,000 mi) above sea level, in the plane of the equator, where near-geostationary orbits may be implemented.

Geostationary orbits are useful because they cause a satellite to appear stationary with respect to a fixed point on the rotating Earth. As a result, an antenna can point in a fixed direction and maintain a link with the satellite. The satellite orbits in the direction of the Earth's rotation, at an altitude of 35,786 km (22,236 mi) above ground. This altitude is significant because it produces an orbital period equal to the Earth's period of rotation, known as the sidereal day.


20

Summary of Chapter 5Circular motion and centripetal

acceleration and force.Fc = mv2/r

Ferris wheel, car around a corner or over a hill. “Circular” pendulum.

Gravitation and Planetary orbitsFor a simple circular orbit GmM/r2 = mv2/r where M is the mass of the sun and m the mass of the earth. v2 = GM/r T = 2πr/v

T2 = 4π2r2/v2 = 4π2r3/GMs

T2/r3 = 4π2/GMs

At any given instant, FG is given by the distance BETWEEN CENTERS OF MASS

Wrong r !! use AVG.


21

Examples of circular motion

W = mg

N

v

mg – N = mv2/r

N

mgN - mg = mv2/r

v

Looking down on stunt driver’s cylindrical cage

N

N = mv2/r

Side view

mg

Ff

mg = Ff

Vertical motion

mg + T = mv2/r top

T - mg = mv2/r bottom

v

Tmg

Swing object on a string in a vertical circle

Red force arrows should be equal, if enough static friction is N available. If μs is too small, not enough Ff and car will slide downwards.


22

On a rotating platform, spinning at a certain constant rate, you move three times farther from the center. How many times more (or less) friction does it take to keep you from slipping off the turntable?

Ch 5 QUIZ Question

A. 1/9 B. 1/3 C. same D. 3 E. 9Note that both v and r change when you move. Think about how the distance around the circle changes when r increases, and remember that you go around once in the same amount of time wherever you are on the turntable.

New v (call it v’) is three times faster. New r (call it r’) is three times bigger

New mv2/r = m(3v)2/3r = (9/3)mv2/r : Three times as much Fc is needed.


23

Moon and tidesEarth tides are dominantly due to the gravitational force exerted by the moon. Since the Earth turns faster than the Moon orbits, the tidal bulges sweep around the Earth, with high tides happening approx. twice a day. Because of the friction generated by tides the Moon is gaining energy from the Earth’s spin, and moving away from the earth.

http://www.sfgate.com/getoutside/1996/jun/tides.html

whytides.gif

anim0012.mov

X

E+M Centerof Mass is 80 times closer to Earth since Earth is 80 times heavier than Moon

Earth & Moon both orbit around their mutual CofM


24

Moon, Sun, and tidesThe tidal bulge away from the moon is where “centrifugal force” is bigger than gravitational force. The bulge towards the moon is where the gravity outweighs the centrifugal force. This works because the entire earth moves with the same angular velocitiy around the CofM. The outward bulge is where FG is less and Fc (because of v2) is more, than required for the orbit of the Earth’s center. And vice versa for the Moon-ward bulge. The Sun’s effect is equally important!!

http://www.sfgate.com/getoutside/1996/jun/tides.htmlwhytides.gif

anim0012.mov

X

E+M Centerof Mass is 76 times closer to Earth’s center, since Earth is 76 times heavier than Moon.

Earth & Moon both orbit around their mutual CofM, ~3000 mi from Earth’s center

~240,000 mi

~384,000 km


25

Moon and tidesThe Moon itself has a tidal bulge, due to ITS orbit. The Earth pulls on this bulge and keeps the Moon “locked” with the bulge facing the Earth. [This required the Moon to lose some of its spin energy early-on in the life of the solar system.]

The Earth spins faster than the Moon orbits, and in the same direction, so the Earth’s tidal bulges “torque” the Moon’s orbit. The Earth gives up “angular momentum” to the Moon’s orbit.This actually moves the Moon’s orbit further away and slows down the Earth’s rotation. This ongoing effect is measurable (3.8 cm/yr).Also geologically measurable (Devonian seashells with growth layers showing ~400 days per year!!) Because the Earth spun faster then, days were shorter, year has not changed.

Someday far in the future, the Earth’s spin may get locked to the Moon’s orbit!

http://www.sfgate.com/getoutside/1996/jun/tides.htmlwhytides.gif

anim0012.mov

X

E+M Centerof Mass is 76 times closer to Earth since Earth is 76 times heavier than Moon

Earth & Moon both orbit around their mutual CofM, ~3000mi from Earth’s center


26

The period of the moon’s orbit about the earth is 27.3 days, but the average time between full moons is 29.5 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 (I’d have called it TE)

Moon orbital period = 27.3 days = 0M ( “ TM)0M/0E = 27.3/365.25 ≈ 0.0747

Ch 5 CP 6


27

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 hasmoved thru 0.075 of its full orbit in 27.3 days (see diagram (ii)). To be in-line w/ sun and earth, moon must move thru same fraction of orbit (see diagram (iii)). This answer is only approx. (but not too bad) since in 2 days the earth-sun line has advanced even further!

0.0747 (27.3 days) ≈ 2.04 days. Apparent lunar month ≈ 29.34 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)


28

c) Exact solution: θE = 2π(t/365.25) = θM = 2π(t/27.3 - 1) with t in daysHere, earth makes a fraction of a revolution, and the moon makes one full revolution plus the SAME fraction of the next revolution, to line up with the Sun-Earth direction. We have to subtract off the full revolution to compare angles.

Solve: 1 + t/365.25 = t/27.3; 1 = -0.002738 t + 0.036630 t = 0.033892 tt = 29.505 days compare with approx. 29.34 days previous slide

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)

Fraction of a revolution -------------

Fraction of 2 pi radians angle


29

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. M is the mass inside the orbit. If we look at stars in motion in galaxies we find that the orbital speeds do NOT fall off as quickly as you’d expect from the VISIBLE (shining) stars. There’s extra invisible mass boosting the effective M inside each orbit. We now know that 25% of our Universe is Dark Matter, only about 5% is “baryonic” matter (made of nuclei and atoms) and only about 1% – 2% of the Universe is made of shining stars!!

v~1/√r

Total M within r from center must grow linearly with r Green curve

v (and v2) constant


30

Space Elevator (courtesy of Arthur C. Clarke)

•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. – and over ¼ of the way from Earth to Moon•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. •Electric vehicles, called climbers, ascend the ribbon using electricity generated by solar panels (i.e. photovoltaic) facing a ground based booster light beam (laser.) •In addition to lifting payloads from earth to orbit, the elevator can also release them directly into lunar-injection or earth-escape trajectories. •The baseline system weighs about 1500 tons (including counterweight) and can carry up to 15 ton payloads, easily one per day. •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. •The climbers travel at a steady 200 kilometers per hour (120 MPH), do not undergo accelerations and vibrations, can carry large and fragile payloads, and have no propellant stored onboard. •The climbers are driven by earth based lasers.•Orbital debris are avoided by moving the floating anchor ship on Earth, and the ribbon itself is made resilient to local space debris damage. •The elevator can increase its own payload capacity by adding ribbon layers to itself. There is no limit on how large a Space Elevator can be!

100,000km

vstationcounterweight


31


•The Space Station can be beyond geosynchronous height (as shown) and can launch objects with greater than escape velocity. This is because at its location, it is LESS gravitationally bound to the Earth AND it has MORE than the Velocity for a circular orbit AT THAT DISTANCE.

•Basically, the “free orbit” from the blue Space Stationis not actually an orbit, it won’t even close into an ellipse, the path is too fast and straight for that.

If the ribbon breaks, the station and the counterweight will go flying off away from the earth and not return.

100,000km



32


•Engineering note: The strongest steel cable is much too heavy, for its strength, to do this job. A 10 mile long steel cable suspended (magically) near the Earth’s surface, would break under its own weight. We need something WAY lighter (per unit of strength) to make the space elevator.

•Carbon nanotubes are incredibly strong. They are like rolled-up tubes of single-layer graphite (a hexagonal network of Carbon atoms, like chicken wire.) The Carbon-Carbon bond achieves this (somewhat the way diamond, which is a crystal of pure carbon, is so hard.)

100,000km



33

1C-05 Velocity of Rifle BulletMEASURING SPEEDS OF OBJECTS MOVING VERY FAST MAY NOT BE DIFFICULT.THIS TECHNIQUE IS THE SAME ONE USED TO MEASURE SPEEDS OF MOLECULES.

We know that the distance between two disks is L. If the second disk rotates an angle Δθ before the bullet arrives, the time taken by this rotation is

t = Δθ / 2πn , where 2πn is the angular frequency of the shaft, in radians/s.

Therefore we come up with v = L / t = 2πn L / Δθ .

Note: θ is in radians, and n by itself is the revolution rate, in revols./sor Hz.

How can we measure the

speed of a bullet ?


34

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) )‘cuz tan() = v2/gR

Note that T’s cancel, and m’s cancel.Which of the m’s is inertial?Which of the m’s is grav.?

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

Could you find the NET force?


35

1D-03 Demonstrations of Central Force

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

Fc= mv2/R

The force comesfrom the stripswanting to un-flex

What will happen when it is subjected

to forces during rotation ?

T

θθ

T


36

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?


37

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 (and the wine in the glass).

g

Same as

mv

string

WHAT IS THE PHYSICS THAT KEEPS THE

WINE FROM SPILLING ?

N + mg = mv2/R N > 0


38

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


39

Q11 If a curve is banked, is it possible for a car to negotiate thecurve 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.


40

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


41

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


42

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) = 30 lb

gmoon

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


43

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


44

1D-07 Paper Saw

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

Is paper more rigid

than wood ?


45

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.


46

A Ferris wheel with radius 12 m makes one complete rotation every 8 seconds.a) Rider travels distance 2πr 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


47

a) S = d/t = 2πr/t = 2π(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)

r = 12mFcent


48

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 = 2πr, 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 = 2π(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

PHYS214 - Physics - Purdue University

Documents

Transcript of PHYS214 - Physics - Purdue University