Uniform Circular Motion the motion of an object traveling in a circular path an object will not travel in a circular path naturally an object traveling in a circular path must have a net force acting upon it!

The object wants to travel in a straight line! It cannot because a force (friction, tension, normal, etc.) constantly pulls/pushes toward the center every split second That net force has a very specific name

The net force that produces circular motion is called Centripetal Force (F c ). v v v v a a a a The acceleration vector is considered to be in a constant direction because it is always towards the center of the path The velocity is tangent to the path- the object wants to travel in a straight line

Centripetal Force (F c ) depends upon the mass, speed squared and radius of the object: F c = mv 2 r This is a net force, therefore: a c = v 2 r

m =.013 kg r =.85 m T =.65 s F c = ? F c = mv 2 r v = ∆d ∆t = 2πr T = 2(3.14)(.85 m).65 s = 8.2 m/s = (.013 kg)(8.2 m/s) 2.85 m = 1.0 N A rubber stopper of mass 13.0 g is swung at the end of a cord.85 m long with a period of.65 s. What is the tension in the cord?

An object of mass 15.0 g is spun from a.750 m string so that the centripetal force acting on the object is 4.50 N. If the object is spun in a horizontal circle at a height above the floor of 2.00 m and then released, how far away will it land? m =.0150 kg r =.750 m F c = 4.50 N ∆y = 2.00 m ∆x = ? Horizontally: ∆x = ? ∆x = v x ∆t v x = ? ∆t = ? Vertically: v o = 0 a = m/s 2 ∆y = m ∆t = ?

∆t = 2∆y a 2(-2.00 m) m/s 2 =.639 s v = F c r m (4.50 N)(.750 m).0150 kg = 15.0 m/s ∆x = v x ∆t = (15.0 m/s)(.639 s) = 9.58 m

Imagine firing a projectile with a very high speed from a very high mountaintop: With an increased speed:

Faster and faster, farther and farther until: At this point, the weight force is the centripetal force: mg = mv 2 r The speed needed to “orbit” would depend only upon the acceleration due to gravity and the radius of the path.

Critical Velocity-- The minimum velocity needed to maintain a circular path when gravity is a factor.  the critical velocity only depends upon the radius of the path and the acceleration due to gravity! v = rg this equation is ONLY for finding the minimum velocity needed to maintain a circular path!

1) A car with a mass of 1250 kg rounds a curve where the coefficient of friction is measured to be.185. If the radius of the curve is 195 m, what speed must the car be traveling? 2) A student spins a 15.0 g rubber stopper above his head from a.750 m string. The tension in the string is measured to be 8.00 N. He lets the stopper go and it lands 12.8 m away. How high is he spinning the stopper above the floor before he releases it? 3) A student spins a 13.5 g stopper from a 56.0 cm string and times 12 revolutions as taking 6.56 s. What is the tension in the string?

How fast would you have to throw a baseball in order to get it to come all the way around back to you? Use 6.37 X 10 6 m for the radius of the earth, and, of course, ignore air resistance. A roller coaster car of mass 725 kg enters a loop with a circumference of 511 m. What minimum speed must the car maintain in order to successfully navigate the loop?

Newton’s Law of Universal Gravitation Newton, in studying gravity, discovered that gravitation is not limited to planets, but applies to all bodies in the universe. The force of attraction between any 2 bodies in the universe depends directly upon the masses of the bodies. This force also depends inversely upon the square of the distances between the center of the masses.

This means if the mass of an object changes, so does its gravity. This also means that if the distance between the centers of the objects changes, the gravity will change. Therefore, if you move closer to the center of the earth, you will experience a greater gravity and you will weigh more! Conversely, if you move out into space, you will weigh less!

Newton’s Law of Universal Gravitation can be written in equation form: F = Gm 1 m 2 r 2 This is a force due to gravity- another way of finding F w ! G  Newton’s Universal Gravitational Constant G = 6.67 X N  m 2 /kg 2

What is the weight of a 1250 kg object that is in the payload bay of the shuttle orbiting at 7.25 X 10 6 m above the surface of the earth? m = 1250 kg d = 7.25 X 10 6 m F w = ? F = Gm 1 m 2 r 2 r = (7.25 X X 10 6 )m = X 10 7 m F = (6.67X Nm 2 /kg 2 )(1250kg)(5.96X10 24 kg) (1.362 X 10 7 m) 2 = 2680 N

1) What is the gravitational force of attraction between a N man on earth and the moon, which has a mass of 7.27 X kg. The center of the moon is 3.90 X 10 9 m away from the surface of the earth. 2) A satellite on the surface of the earth has a weight of 12, 800 N. When it is in orbit, its weight is 3200 N. How far above the surface of the earth is the satellite orbiting? 3) What is the force of attraction between a 90.0 kg boy and his 65.0 kg girlfriend sitting 1.23 m away?

Rotary Motion Angular Kinematics Motion about an internal axis Describing the motion of a spinning/rotating object Defining movement in terms of rotations!

All quantities are linear unless specifically stated as angular/rotational: All motion begins with understanding displacement: Displacement (∆d): distance in a direction Angular Displacement (∆Ø): rotations in a particular direction QuantityLinear Angular Displacementmeter(m) revolution(rev) radian (rad) radian = radius length 1 rev = 2π rad= 6.28 rad

Angular Velocityrev/s or rad/s Angular Accelerationrev/s 2 or rad/s 2 In the rotary system, all motion is based upon how much an object has turned! (rev or rad--> ø ) Linear MotionAngular Motion v = v o + a∆t ω = ω o + α∆t ∆d = v o ∆t a∆t 2 ∆ø = ω o ∆t α∆t 2 v = √v o 2 + 2a∆d ω = √ω o 2 + 2α∆θ

Find the angular displacement (in radians) after 15.0 s for a wheel that accelerates at a constant rate from rest to 725 rev/min in 10.0 min. ∆ø = ?(rad) ∆t 1 = 15.0 s ω 0 = 0 ω = 725 rev/min ∆t 2 = 10.0 min = 600 s 725 rev/min(6.28 rad/rev)(1 min/60 s) = 75.9 rad/s ∆ø = ω i ∆t +.5α∆t 1 2 α = ω – ω 0 ∆t 2 = 75.9 rad/s s =.127 rad/s 2 =.5(.127 rad/s 2 )(15.0 s) 2 = 14.3 rad

A car tire is decelerated from 25.0 rev/s to rest in 20.0 s. If the radius of the tire is 30.0 cm, how far does the car move down the road in this time? ω 0 = 25.0 rev/s ω = 0 ∆t = 20.0 s r = 30.0 cm =.300 m ∆ø = ? ∆ø = ω i ∆t +.5α∆t 2 a = ω – ω 0 ∆t = rev/s 20.0 s = rev/s 2 =(25.0 rev/s)(20.0 s) +.5(-1.25rev/s 2 )(20.0 s) 2

= 250 rev This is how many times the wheel turned! How linear distance covered relates to rotational displacement: ∆d = (# rev)(Circumference) ∆d = (# rad)(radius length) ∆d = (250 rev)(2)(3.14)(.300 m) C = 2π r = 471 m

3) A bicycle slows down from 8.40 m/s to rest over a distance of 115 m. If the diameter of the wheels are 68.0 cm, how many times must they have turned over that 115 m? 1) A car tire makes 65.0 revolutions in slowing from 35.0 rad/s to 12.0 rad/s. How long did it take for the car to slow? 2) A car tire of radius 40.0 cm slows to a stop from an angular speed of 40.0 rad/s in 25.0 s. How far did the car move down the road during this time?