Uniform Circular Motion

Horizontal CircleVertical CircleBanked Curve

Unbanked CurveNewton’s Law of Universal Gravitation

Kepler’s Laws

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Uniform Circular Motion

Motion of an object traveling at a constant speed in a circular path.

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Uniform Circular Motion

r

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Horizontal Circle

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Velocity

Velocity is constant in magnitude only.

The direction is changing with time.

V

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Centripetal Acceleration

ac = V2

R

Acceleration is occurring because the direction of the velocity is changing.

Acceleration is always toward the center of the circular path.

ac

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Centripetal Force

Fc = mac Net force (Fc) and

acceleration always act in the same direction.

The centripetal force keeps the object moving in a circle

Fc

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Centrifugal Force

Fictitious force. Doesn’t exist

Force and acceleration acting outward.

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Derived Equations for ac and Fc

a = V2

R

a = (2πR/T)² R

a = 4π²R T²

F = ma

F = m V2

R

F = m 4π²R T²

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Example 1. Period

How long does it take a plane traveling at a speed of 110 m/s, to fly once around a circle whose radius is 2850 m?

Given:V = 110 m/s V = 2πrR = 2850 m TT = ? T = 2πr V

= 2π(2850 m) 110 m/s

T = 160 s

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Example 2. Frequency

It takes 2 min to twirl a ball in a circle 50 times. What is the frequency?

f = 50 2 min. X 60 s min

f = 0.4/s

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Example 3. VelocityA stopper is twirled in a horizontal circle whose

radius is 0.75 m. The stopper travels once around the circle in 0.30 s.

What is its velocity?Given:R = 0.75 m V = 2πrT = 0.30 s T

= 2π(0.75 m) 0.30 s

V = 16 m/s

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Example 4. Acceleration A runner moving at a speed of 7.5 m/s rounds

a bend with a radius of 25 m. Determine the centripetal acceleration of the runner.

Givens: a = V2

v = 7.5 m/s R

r = 25 m = (7.5 m/s)²

a = ? 25 m

a = 2.3 m/s²

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Example 5. Acceleration

A race car travels at a constant speed around a circular track whose radius is 2.6 km. If the car goes around the track in

360 s, what is the magnitude of the centripetal acceleration of the car?

Given: ac = 4π² rR = 2.6 x 10³ m T²T = 360 s = 4 π² (2.6 x 10³ m ) (360s)²

ac = 0.79 m/s²

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Example 6. Centripetal Force

A 0.015 kg ball is shot from the plunger of a pinball machine. Because of a centripetal force, the ball follows a circular arc whose radius is 0.25 m at a speed of 0.68 m/s. Calculate the centripetal force.

Given:M = 0.015 kg Fc = m V²R = 0.25 m rV = 0.68 m/s = (0.015 kg)(0.68 m/s)² Fc = ? 0.25 m Fc = 0.028 N

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Example 7. Centripetal Force

A child is twirling a 0.0120 kg ball on a string in a horizontal circle whose radius is 0.l00 m. The ball travels once around the circle in 0.500 s. Determine the centripetal force on the ball.

Given:M = 0.0120 kg Fc = m 4π² rR = 0.100 m T²T = 0.500 s = (0.0120 kg)4 π² (0.100 m) (0.500 s)²

Fc = 0.189 N

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Example 8. Centripetal Force

A 0.350 kg ball is twirled in a horizontal circle. The centripetal acceleration is 0.75 m/s². Calculate the centripetal force.

Given: Fc = mac

m = 0.350 kg = (0.350 kg)(0.75 m/s²)a = 0.75 m/s²

Fc = 0.26 N

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Vertical Circle

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Direction of forces when object is at top and bottom of vertical circle

TopFnet = T + W Bottom

Fnet = T - W

Fc = Fnet

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F c equations for vertical circle

Object at Top

T + W = mV2

R

Object at Bottom

T – W = mV2

R

Fc = Fnet

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Example 9. Top of Vertical Circle A 2.0 kg object is attached to a 1.5 m long string and

swung in a vertical circle at a constant speed of 12 m/s. What is the tension in the string when the object is at the top of its path?

Given: T + W = mV²m = 2.0 kg Rr = 1.5 m T = mV² - WV = 12 m/s RT = ? = (2.0 kg)(12 m/s)² - (2.0 kg)(9.80 m/s²)

1.5 m

T = 1700 N

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Example 10. Vertical Circle

What is the tension in the string when the object is at the bottom of its path?

Given: T – W = m V²/rm= 2.0 kg T = m V² + WV = 12 m/s Rr = 1.5 m = (2.0 kg)(12m/s)²+ 19.6 N 1.5 m T = 2100 N

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Unbanked Curve

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F c is the Static Friction, (fs)

The static friction acts toward the center of circular motion.

fs = Fc

µsFN = m V²/Rµsmg = m V²/R

µsg = V²/R

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Unbanked Curve

Racing on a flat track, a car going 32 m/s rounds a curve 56 m in radius. What would be the minimum coefficient of static friction between tires and road that would be needed for the car to round the curve without skidding?

Given: µs g = V²/R

V = 32 m/s µs =(32 m/s)² /(56 m)(9.80 m/s²)

r = 56 m µs = 1.87

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Banked Curve

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Banked Curve Fc = Wx

The x-component of the weight is acting towards the center of the circular path

Wx = Fc

mg sin Ө = Fc

mg sin Ө = m V²/R g sin Ө = V²/R

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Example 11. Banked Curve

A 60.0 kg speed skater comes into a curve of 20.0 m radius and banked at 12º. Calculate the speed needed to negotiate the curve?

Given: g sin Ө = V²/Rm = 60.0 kg V² = Rg sin Ө R = 20.0m =(20.0 m)(9.80 m/s²)(sin

12º)V = ? V = 6.38 m/s

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Motion and Forces on a Banked Curve These problems deal with an object (a car, for example)

navigating a circular curve at constant speed.  The radius of the curve is fixed at 10 m, but the speed and mass of the object, the angle at which the curve is banked, the coefficient of static friction, and the gravitational field strength can all be varied.

Two views of the object are shown:  i)  overhead showing the circular path with acceleration and velocity vectors, ii) ground level showing the forces acting on the object.  As the values of input parameters are changed, the vectors change in response.  A "fric-o-meter" gauges the ratio of the actual static friction to the maximum available static friction.  This allows the user to determine whether the object will skid off the curve or slide down to the center

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Motion and Forces on a Banked Curve

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Conical Circular Path

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Conical Circular Path Fc = Tx

The x-component of T provides the Fc to keep the object movingin a circular path.

T sinӨ = Fc

T sinӨ = m V²/R

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Using two equations to find V

T cos Ө = mg T sin Ө = mV2/r

T = mg T = mV2/r cos Ө sin Ө

T sin Ө = mg T cos Ө mV2/r

V2 = rg tan Ө

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Example 12. Conical PathThe Swing Ride at an amusement park travels at 15.0 m/s

and the 12.0 m swings make an angle of 25.0º with the vertical as they move in a circle. What is the tension in the rope of a swing carrying a 50.0 kg person?

Given: T sinӨ = m V²/RV = 15.0 m/s T = m V²Ө = 25.0º r sinӨ M = 50.0 kg = (50.0 kg)(15.0 m/s)²R = 12.0 m (12.0 m)(sin 25.0º)

T = 2218 N

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Newton’s Law of Universal Gravitation

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Newton’s Law of Universal Gravitation

If the path of a planet were an ellipse, then the net force on the planet varies inversely with the square of the distance between the planet and the sun.

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Symbols represent

Fg = gravitational force of attraction

G = gravitational constant,

6.67 x 10-11 Nm²/kg²(Henry Cavendish)

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Fg is directly proportional to m1m2 indirectly proportional to

R²

Mass of one object doubled, Fg is doubled.

Mass of one of the objects is halved, Fg is halved.

Distance between the centers of masses is doubled, Fg is decreased to ¼.

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Kepler’s Laws

Three Laws of Planetary Motion

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Law 1

Path of the planets are ellipses with the sun at one focus.

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Law 2

An imaginary line extending from the sun to a planet will sweep out equal areas in equal amounts of time.

Planets move fastest when closest to the sun.

Planets move slowest when farthest from the sun.

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Shaded segments = 30 days

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Law 3

The ratio of the squares of the periods of any two planets in orbit around the Sun is equal to the ratio of the cubes of their distances from the sun.

Ta² = ra³

Tb² rb³Finkster 2007

Motion of Planets and Satellites

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Satellite in Orbit Uniform circular motion if always

at the same height above Earth Path has both vertical and

horizontal components Velocity of horizontal component

great enough… path will follow a curve that matches the curve of Earth. Projectile is in orbit.

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Weightlessness

Occurs when the downward force of gravity is unbalanced and there is no upward force acting on a satellite in orbit.

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Deriving Equations

F = F

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Example 1. F = F

F = ma F = Gm1m2

r²

ma = Gm1m2

r²

a = Gm r²

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Example 2. F = F

W = mg Fg = Gm1m2

r²

mg = Gm1m2

r²

g = Gm

r²

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Example 3. F = F

Fc= mV² Fg = Gm1m2

r r²

mV² = Gm1m2

r r² V² = Gm r

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Example 4. F = F

Fc = m4π²r Fg = Gm1m2

T² r²

m4π²r = Gm1m2

T² r²

m = 4π²r³ GT²

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