Circular Motion and Gravitation - Killeen, TX · Circular Motion & Gravitation Rene’ McCormick, AP Strategies, Inc. 1 CIRCULAR MOTION AND GRAVITATION An object moves in a …

Circular Motion & GravitationRene’ McCormick, AP Strategies, Inc. 1

CIRCULAR MOTION AND GRAVITATION

An object moves in a straight line if the net force on it acts in the direction of motion, or is zero.If the net force acts at an angle to the direction of motion at any moment, then the object moves in acurved path.

KINEMATICS OF UNIFORM CIRCULAR MOTION

! An object that moves in a circle at constant speed, v, is said to experienceuniform circular motion." The magnitude of the velocity remains constant, but the direction

of the velocity is continuously changing ." Since acceleration is defined as the rate of change in velocity, a

change in direction of v constitutes an acceleration just as does achange in magnitude.

" THUS, an object revolving in a circle is continuouslyaccelerating, even when the speed remains constant.

! Acceleration is defined as:

a = v2 - v1 = ) v )t ) t

" where )v is the change in velocity during the short time interval ) t" Consider a nonzero time interval--during )t, the particle moves

from point A to point B, covering a distance )l along the arc whichsubtends an angel )2.

" Look at figure 5-2 b). Let )t be VERY small. Then )R and )2are also very small and v2 will be almost parallel to v1 and )v willbe essentially perpendicular to them.

" Thus, the )v vector points inward toward the center of the circleand a, by definition above, is in the same direction as )v, it too mustpoint toward the center of the circle and centripetal acceleration(center-seeking) is born! a.k.a. radial acceleration, aR since it isdirected along a radius toward the center of the circle.

! To determine the magnitude of aR consider 5-2 a) again. Because CA is z to v1, and CB is z tov2, it follows that the angle )2, defined as the angle between CA and CB, is also the anglebetween v1 & v2. " Hence, vectors v1, v2, & ) v form a triangle that is geometrically similar to triangle ABC." Taking )2 small (letting )t be VERY small), we can write

)v . )R where v = v1 = v2 since constant velocity v r

" This is an exact equality when )t approaches zero, for then the arc length )R equals thecord length AB. Since we want to find the instantaneous acceleration, for which )tapproaches zero, we write the above expression as an equality and solve for )v:


)v = v )R r

To get the centripetal acceleration, aR, we divide )v by )t:

aR = ) v = v )R = v2 )t r )t r

and since )R/)t is the linear speed, v of the object,

CENTRIPETAL ACCELERATION: aR = v2 r

SUMMARY: An object moving in a circle of radius r with constant speed v has an acceleration whosedirection is toward the center of the circle and whose magnitude is aR = v2/r.

" This acceleration depends on v & r. The greater v, the fasterthe velocity changes direction; and the larger the radius, the lessrapidly the velocity changes direction

" The acceleration vector points toward the center of the circle.BUT the velocity vector always points in the direction of themotion.

" THUS, the v and a vectors are always z at every point in thepath for uniform circular motion.

" Really smashes the notion that a and v are always in the samedirection! In projectile motion a = g and is always acting down,now, in circular motion a is z to v!

" frequency--f, the number of revolutions per second." period--T, time required for one complete revolution. Equal to 1/f

For an object revolving in a circle @ constant speed: v = 2BBBB r since in one revolution, the object travels one circumference T

Example 5.1A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m. Theball makes 2.00 revolutions in a second. What is its centripetal acceleration?


Example 5.2The Moon’s nearly circular orbit around Earth has a radius of about 384,000 km and a period T of 27.3days. Determine the acceleration of the Moon toward the Earth.

DYNAMICS OF UNIFORM CIRCULAR MOTION

According to Newton’s Second Law (GF = ma), an object that is accelerating must have a net forceacting upon it. Yep, we call it centripetal force.

! GFR = maR and aR = v2/r so....

CENTRIPETAL FORCE : GGGGFR = maR = m v2 and the force is directed toward the center! r

!!!! If this net force were not applied, it would obey Newton’s first law and fly off in a straight line!! Ever heard of centrifugal (center fleeing) force? Doesn’t exist. There is no outward force!

" Ever swung an object on a string above your head? The misconception comes from“feeling” a pull on your hand from the string. This is simply Newton’s 3rd law inreaction to the inward force you are putting on the string to keep the object moving in acircle.

"""" If you let go AND there was a centrifugal force acting, then the object would flyOOUTward when you released the string. Doesn’t happen--it flies off tangentially to thec ircular path.


Example 5.3Estimate the force a person must exert on a string attached to a 0.150-kg ball to make the ball revolve ina horizontal circle of radius 0.600m, as in example 5.1. The ballmakes 2.00 revolutions persecond.

Example 5.4The game of tetherball is played with a ball tied to a pole with a string. Whenthe ball is struck, it whirls around the pole as shown in figure 5-8. In whatdirection is the acceleration of the ball, and what causes its acceleration?


Example 5.5A 0.150-kg ball on the end of a 1.10 m-long cord (negligible mass) is swungin a vertical circle. Determine the minimum speed the ball must have at thetop of its arc so that it continues moving in a circle. b) Calculate the tensionin the cord at the bottom of the arc assuming the ball is moving at twice thespeed of part a).


Example 5.6A rider on a Ferris wheel moves in a vertical circle of radius r at a constant speed v. Is the normal forcethat the seat exerts on the rider at the top of the circle less than, more than, or the same as, the force theseat exerts at the bottom of the circle?

A CAR ROUNDING A CURVE

Why do you feel thrust outward as a car rounds a curve? You tend to travel in a straight line while thecar is traveling in a curved path. The car itself must have an inward force exerted on it if it is to move ina curve. On a flat road, this force is supplied by friction between the tires and the pavement. [It’s staticfriction as long as the tires are not slipping.] If there is not enough friction, the car skids out of a circularpath and into a more nearly straight one.

Example 5.7A 1,000 kg car rounds a curve on a flat road of radius 50m at a speed of 50 km/h.Will the car make the turn, or will it skid, if:a) the pavement is dry and :s = 0.60?

b) the pavement is icy and :s = 0.25?


The situation is worse if the wheels lock--stop rotating--when the brakes are applied too hard. As longas the tires are rolling, the bottom of the tire is at rest against the road at each instant, so static frictionexists. BUT, if the wheels lock, the tires slide and the friction force, which is now kinetic friction, isless. Moreover, when the road is wet or icy, locking of the wheels occurs with less fore on the brakepedal since there is less road friction to keep the wheels turning rather than sliding. ABS brakes aredesigned to limit brake pressure just before the point where sliding would occur, by means of delicatesensors and a fast computer--resist the urge to “pump” the brakes--it causes the sensors and computer tohave to start all over each time you pump.! banking--the banking of curves can reduce the chance of skidding because the normal force the

road will have a component toward the center of the circle, thus reducing the reliance on friction. ! For a given banking angle, 2, there will be a speed for which NO FRICTION at all is required.

" This is when the horizontal component of the normal force toward the center of the curve,FN sin 2, is just equal to the force required to give the vehicle its centripetal acceleration.

" That is when

FN sin 2 = m v2 r

Example 5.8a) For a car traveling with a speed, v, around a curve of radius r, determine aformula for the angle at which a road should be banked so that no friction isrequired.

b) What is this angle for an expressway off-ramp curve of radius 50m at adesign speed of 50 km/h?


NONUNIFORM CIRCULAR MOTION

Circular motion at constant speed occurs when the force is directed toward thecenter. What if it is directed at an angle like in fig. 5-15 a?

The force has 2 components:! The component directed toward the center of the circle gives us FR and

gives rise to aR AND keeps the object moving in a circle.! The component tangent to the circle, Ftan, acts to increase or decrease the

speed, and thus gives rise to a component of the acceleration tangent tothe circle atan.

! When the speed of the object is changing, a tangential component of force is acting." When you first start revolving a ball around your head, you must

give it a tangential acceleration. You do this by pulling on thestring with your hand displaced from the center of the circle.

" The tangential component of the acceleration is equal to the rateof change of the magnitude of the velocity of the object:

atan = )v )t

" the radial (centripetal) acceleration arises from the change in the direction of the velocity

aR = v2 r

" the tangential acceleration always points in a direction tangent to the circle AND is in thedirection of the motion IF the speed is increasing and antiparallel IF the speed isdecreasing. In any case, atan & aR are always perpendicular and the total vectoracceleration, a is the sum of these two:

Example 5.9A racing car starts from rest in the pit area and accelerates at a uniform rate to a speed of 35 m/s in 11 s,moving on a circular track of radius 500m. Assuming constant tangential acceleration, find a) the tangential acceleration

b) the centripetal acceleration when the speed is 30 m/s.


NEWTON’S LAW OF UNIVERSAL GRAVITATION--noone calls it the 4th!

Newton was wondering about the force keeping the Moon in its near circular orbit when an apple felland bopped him on his noggin. [poetic licence taken] Since falling bodies accelerate, they must have aforce acting upon them.

! this was met with great resistance since most forces are contact forces--gravitation “acts at adistance”.

! A calculation was in order!" At the Earth’s surface, an object is accelerated 9.8 m/s2

" But, what is the moon’s aR? In example 5.2, we found it to be .00272 m/s2 which is1/3600 g.- That means that the accel. of the Moon toward the Earth is about 1/3600 as great

as the acceleration of objects at the Earth’s surface" The moon is 384,000 km from the Earth which is about 60 times the Earth’s radius of

6380 km. - That means the Moon is 60 times farther from the Earth’s center than are objects

at the Earth’s surface." BUT 60 x 60 = 3600. Again, that number 3600!

! Newton concluded that the gravitational force exerted by the Earth on any object decreases withthe square of its distance, r from the Earth’s center." The Moon, being 60 Earth’s radii away, feels the pull from the center as 1/602 = 1/3600

times as strong as if would if it were at the Earth’s surface.! Newton also realized that mass mattered! His 3rd law dictates that it be proportional to both

masses. Thus, F % mem. r2

Drum roll please.......

Newton’s Law of Universal Gravitation: Every particle in the universe attracts every otherparticle with a force that is proportional to the product of their masses and inversely proportionalto the square of the distance between them. This force acts along the line joining the two particles.

F = G m1m2 r2

! Henry Cavendish measured G in 1798, about 100 years post Newton.! G = 6.67 x 10-11 NCM2/kg2


Example 5.10A 50 kg person and a 75 kg person are sitting on a bench so that their centers are about 50 cm apart. Estimate the magnitude of the gravitational force each exerts on the other.

Example 5.11What is the force of gravity acting on a 2,000 kg spacecraft when it orbits two Earth radii from theEarth’s center (that is, a distance rE = 6380 km above the Earth’s surface)? The mass of the Earth is 5.98x 1024 kg.

Example 5.12Find the net force on the Moon ( mass = 7.35 x 1022 kg) due to the gravitational attraction of both theEarth and the Sun (mass = 1.99 x 1030 kg), assuming they are at right angles to each other?


GRAVITY NEAR THE EARTH’S SURFACE; GEOPHYSICAL APPLICATIONS

Apply Newton’s Gravitational Law to an object at the Earth’s surface:

m1 becomes mEm2 becomes mr becomes the distance of the object from the Earth’s center or rE = 6.38 x 106 mthis force is the weight of the object, so.....

mg = G m mE rE

2

Hence

g = G mE rE

2

Until G was measured, the mass of the Earth was not known. Let’s calculate:

Example 5.13Estimate the effective value of g on the top of Mt. Everest, 8848 m above the Earth’s surface. That is,what is the acceleration due to gravity of objects allowed to fall freely at this altitude?

SATELLITES AND “WEIGHTLESSNESS”

How do you get a satellite up? Launch it really fast with a rocket.What keeps a satellite up? Its high tangential speed.


If it stops moving, it will crash to Earth. If it moves too fast, it flies off in a straight line path in the samedirection as its tangential velocity.

For satellites that move in a circle, theiracceleration is approx. v2/r

The force that gives a satellite this accelerationis the force of gravity.

Apply Newton’s second law:

E FR = maR

G mmE = m v2 r2 r

where m is the mass of the satellite.r is the distance of the satellite from the Earth’s center and is equal to rE + hh is the height above the Earth’s surface

Example 5.14A geosynchronous satellite is one that stays above the same point on the equator of the Earth. Suchsatellites are used for such purposes as cable TV transmission, for weather forecasting, and ascommunication relays. Determinea) the height above the Earth’s surface such a satellite must orbit

b) such a satellite’s speed


WEIGHTLESSNESS

Let’s look at a falling elevator. We see an elevator at REST with a bag hanging from a spring scale. The scale reading indicates the downward force exerted on it by the bag which is equal and opposite tothe force exerted by the scale upward on the bag. Call this force W.

Since the mass, m, is at rest and NOT accelerating,

3 F = ma ANDw-mg = 0 thusw = mg as expected!

Let’s move that elevator! Now 3F = ma and is NOT equal to zero so.....

3 F = maw-mg = maw = mg + ma

Choosing 8 as +IF acceleration is 8, w is greater than mg.IF acceleration is 9, w is less than mg.

These are called apparent weights not to be confused with actual weight, mg.

Further, if the a = ½ g 8, then we findw = mg + m ½ g = 1 ½ mgAND we say the bag is experiencing 1 ½ g’s.

Next, if the a = - ½ g 9, then we findw = mg - m ½ g = ½ gAND the scale reads ½ the weight.


Finally, if the cable breaks and the elevator is in free fall, then a = -g, then we find w = mg + m(-g) = 0AND voila! APPARENT WEIGHTLESSNESS

Satellites are in free fall– they are falling toward Earth or they wouldn’t be in orbit, they’d be travellingin a straight line path according to Newton’s first law!

Don’t confuse with real weightlessness. When a spacecraft is NOT in orbit it is free from the Earth orMoon’s gravity and indeed weightless!

Circular Motion and Gravitation - Killeen, TX · Circular Motion & Gravitation Rene’ McCormick, AP Strategies, Inc. 1 CIRCULAR MOTION AND GRAVITATION An object moves in a …

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