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Circular Motion & Gravitation Rene’ McCormick, NMSI. 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 a curved path. KINEMATICS OF UNIFORM CIRCULAR MOTION ! An object that moves in a circle at constant speed, v, is said to experience uniform 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 a change in magnitude. " THUS, an object revolving in a circle is continuously accelerating, even when the speed remains constant. !Acceleration is defined as: a = v 2 - v 1 = Δ v Δt Δ t " where Δv is the change in velocity during the short time interval Δ t " Consider a nonzero time interval--during Δt, the particle moves from point A to point B, covering a distance Δl along the arc which subtends an angel Δθ. " Look at figure 5-2 b). Let Δt be VERY small. Then Δℓ and Δθ are also very small and v 2 will be almost parallel to v 1 and Δv will be essentially perpendicular to them. " Thus, the Δv vector points inward toward the center of the circle and a, by definition above, is in the same direction as Δv, it too must point toward the center of the circle and centripetal acceleration (center-seeking) is born! a.k.a. radial acceleration, a R since it is directed along a radius toward the center of the circle. ! To determine the magnitude of a R consider 5-2 a) again. Because CA is ٣ to v 1 , and CB is ٣ to v 2 , it follows that the angle Δθ, defined as the angle between CA and CB, is also the angle between v 1 & v 2. " Hence, vectors v 1 , v 2 , & Δ v form a triangle that is geometrically similar to triangle ABC. " Taking Δθ small (letting Δt be VERY small), we can write Δv Δℓ where v = v 1 = v 2 since constant velocity v r " This is an exact equality when Δt approaches zero, for then the arc length Δℓ equals the cord length AB. Since we want to find the instantaneous acceleration, for which Δt approaches zero, we write the above expression as an equality and solve for Δv:
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Circular Motion and Gravitation 5 5 circular motion and...Circular Motion & Gravitation Rene’ McCormick, NMSI. 3 Example 5.2 The Moon’s nearly circular orbit around Earth has a

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Page 1: Circular Motion and Gravitation 5 5 circular motion and...Circular Motion & Gravitation Rene’ McCormick, NMSI. 3 Example 5.2 The Moon’s nearly circular orbit around Earth has a

Circular Motion & Gravitation Rene’ McCormick, NMSI. 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 a curved path. KINEMATICS OF UNIFORM CIRCULAR MOTION ! An object that moves in a circle at constant speed, v, is said to

experience uniform 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 a change in magnitude.

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

!Acceleration is defined as:

a = v2 - v1 = Δ v Δt Δ t

" where Δv is the change in velocity during the short time interval Δ t " Consider a nonzero time interval--during Δt, the particle moves

from point A to point B, covering a distance Δl along the arc which subtends an angel Δθ.

" Look at figure 5-2 b). Let Δt be VERY small. Then Δℓ and Δθ are also very small and v2 will be almost parallel to v1 and Δv will be essentially perpendicular to them.

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

directed along a radius toward the center of the circle. ! To determine the magnitude of aR consider 5-2 a) again. Because CA is to v1, and CB is to

v2, it follows that the angle Δθ, defined as the angle between CA and CB, is also the angle between v1 & v2. " Hence, vectors v1, v2, & Δ v form a triangle that is geometrically similar to triangle ABC. " Taking Δθ small (letting Δt be VERY small), we can write

Δv ≈ Δℓ where v = v1 = v2 since constant velocity v r

" This is an exact equality when Δt approaches zero, for then the arc length Δℓ equals the cord length AB. Since we want to find the instantaneous acceleration, for which Δt approaches zero, we write the above expression as an equality and solve for Δv:

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Δv = v Δℓ r To get the centripetal acceleration, aR, we divide Δv by Δt: aR = Δ v = v Δℓ = v2 Δt r Δt r and since Δℓ/Δt 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 whose direction is toward the center of the circle and whose magnitude is aR = v2/r.

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

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

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

" Really smashes the notion that a and v are always in the same direction! In projectile motion a = g and is always acting down, now, in circular motion a is 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 = 2π r since in one revolution, the object travels one circumference T Example 5.1 A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

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Example 5.2 The Moon’s nearly circular orbit around Earth has a radius of about 384,000 km and a period T of 27.3 days. Determine the acceleration of the Moon toward the Earth. DYNAMICS OF UNIFORM CIRCULAR MOTION According to Newton’s Second Law (ΣF = ma), an object that is accelerating must have a net force acting upon it. Yep, we call it centripetal force. ! ΣFR = maR and aR = v2/r so.... CENTRIPETAL FORCE : ΣFR = 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 in reaction to the inward force you are putting on the string to keep the object moving in a circle.

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

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Example 5.3 Estimate the force a person must exert on a string attached to a 0.150-kg ball to make the ball revolve in a horizontal circle of radius 0.600 m, as in example 5.1. The ball makes 2.00 revolutions per second. Example 5.4 The game of tetherball is played with a ball tied to a pole with a string. When the ball is struck, it whirls around the pole as shown in figure 5-8. In what direction is the acceleration of the ball, and what causes its acceleration?

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Example 5.5 A 0.150-kg ball on the end of a 1.10 m-long cord (negligible mass) is swung in a vertical circle. 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 of part a).

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Example 5.6 A rider on a Ferris wheel moves in a vertical circle of radius r at a constant speed v. Is the normal force that the seat exerts on the rider at the top of the circle less than, more than, or the same as, the force the seat 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 the car is traveling in a curved path. The car itself must have an inward force exerted on it if it is to move in a curve. On a flat road, this force is supplied by friction between the tires and the pavement. [It’s static friction as long as the tires are not slipping.] If there is not enough friction, the car skids out of a circular path and into a more nearly straight one. Example 5.7 A 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 μs = 0.60? b) the pavement is icy and μs = 0.25?

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The situation is worse if the wheels lock--stop rotating--when the brakes are applied too hard. As long as the tires are rolling, the bottom of the tire is at rest against the road at each instant, so static friction exists. BUT, if the wheels lock, the tires slide and the friction force, which is now kinetic friction, is less. Moreover, when the road is wet or icy, locking of the wheels occurs with less fore on the brake pedal since there is less road friction to keep the wheels turning rather than sliding. ABS brakes are designed to limit brake pressure just before the point where sliding would occur, by means of delicate sensors and a fast computer--resist the urge to “pump” the brakes--it causes the sensors and computer to have 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, θ, 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 θ, is just equal to the force required to give the vehicle its centripetal acceleration.

" That is when FN sin θ = m v2 r Example 5.8 a) For a car traveling with a speed, v, around a curve of radius r, determine a formula for the angle at which a road should be banked so that no friction is required. b) What is this angle for an expressway off-ramp curve of radius 50m at a design speed of 50 km/h?

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NONUNIFORM CIRCULAR MOTION Circular motion at constant speed occurs when the force is directed toward the center. 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 to the 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 the string with your hand displaced from the center of the circle.

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

atan = Δv Δt

" 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 the direction of the motion IF the speed is increasing and antiparallel IF the speed is decreasing. In any case, atan & aR are always perpendicular and the total vector acceleration, a is the sum of these two:

a ' a 2tan % a 2

R

Example 5.9 A 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.

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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 fell and bopped him on his noggin. [poetic licence taken] Since falling bodies accelerate, they must have a force acting upon them. ! this was met with great resistance since most forces are contact forces--gravitation “acts at a

distance”. ! 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 is

1/3600 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 with the 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 other particle with a force that is proportional to the product of their masses and inversely proportional to 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 N·M2/kg2

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Example 5.10 A 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 5.11 What is the force of gravity acting on a 2,000 kg spacecraft when it orbits two Earth radii from the Earth’s center (that is, a distance rE = 6380 km above the Earth’s surface)? The mass of the Earth is 5.98 x 1024 kg.

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

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GRAVITY NEAR THE EARTH’S SURFACE; GEOPHYSICAL APPLICATIONS Apply Newton’s Gravitational Law to an object at the Earth’s surface: m1 becomes mE m2 becomes m r becomes the distance of the object from the Earth’s center or rE = 6.38 x 106 m this 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 5.13 Estimate 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.

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If it stops moving, it will crash to Earth. If it moves too fast, it flies off in a straight line path in the same direction as its tangential velocity. For satellites that move in a circle, their acceleration is approx. v2/r The force that gives a satellite this acceleration is the force of gravity. Apply Newton’s second law: Σ 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 + h h is the height above the Earth’s surface Example 5.14 A geosynchronous satellite is one that stays above the same point on the equator of the Earth. Such satellites are used for such purposes as cable TV transmission, for weather forecasting, and as communication relays. Determine a) the height above the Earth’s surface such a satellite must orbit b) such a satellite’s speed

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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 to the force exerted by the scale upward on the bag. Call this force W. Since the mass, m, is at rest and NOT accelerating, ∑ F = ma AND w-mg = 0 thus w = mg as expected!

Let’s move that elevator! Now ∑F = ma and is NOT equal to zero so..... ∑ F = ma w-mg = ma w = mg + ma Choosing ↑ as + IF acceleration is ↑, w is greater than mg. IF acceleration is ↓, w is less than mg. These are called apparent weights not to be confused with actual weight, mg. Further, if the a = ½ g ↑, then we find w = mg + m ½ g = 1 ½ mg AND we say the bag is experiencing 1 ½ g’s. Next, if the a = - ½ g ↓, then we find w = mg - m ½ g = ½ g AND the scale reads ½ the weight.

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Finally, if the cable breaks and the elevator is in free fall, then a = -g, then we find w = mg + m(-g) = 0 AND voila! APPARENT WEIGHTLESSNESS Satellites are in free fall– they are falling toward Earth or they wouldn’t be in orbit, they’d be travelling in 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 or Moon’s gravity and indeed weightless!