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## Circular Motion

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**Rotating**Turning about an internal axis Revolving Turning about an external axis**Linear speed, v**How far you go in a certain amount of time Miles per hour, meters per second Rotational (angular) speed, How many times you go around in a certain amount of time Revolutions per minute, rotations per hour, radians per second**Which horse has a larger linear speed on a merry go round,**one on the outside or one on the inside? Outside. Which horse has a greater rotational speed? Neither, all the horses complete the circle in the same amount of time.**Uniform Circular Motion, UCM: moving in a circle with a**constant speed. Question: Is there a constant velocity when an object moves in a circle with a constant speed? No, the direction changes, therefore the velocity changes. If the velocity changed, the object is actually ACCELERATING even while moving at the same speed.**Suppose an object was moving in a straight line with some**velocity, v. According to Newton’s 1st Law of Motion, “An object in motion continues that motion unless a net external force acts on it”. If you want the object to move in a circle, some force must push or pull it towards the center of the circle. A force that pushes or pulls an object towards the center of a circle is called a centripetal force Centripetal means “center seeking”**The centripetal acceleration is along the same line as the**radius of the circle, so it is a “radial” acceleration. It is given by acentripetal = v2 / r Where r is the radius of the circle and v is the velocity of the object.**Centripetal force**Since SF= ma, and acentripetal = v2/r, the net centripetal force is given by**Lots of forces can help in pushing or pulling an object**towards (or away from) the center of a circle. Sometimes it takes more than one force to get an object to move in uniform circular motion. Centripetal force is NOT a new kind of force. If an object moves in a circle (or an arc), there must be at least one force that is acting toward the center of the circle.**When can these forces be centripetal forces?**Gravity? Moon revolving around the Earth Tension? Twirling a pail at the end of a string Friction? Cars rounding a curve. Air Resistance (“Lift”)? Airplane or birds flying in a circle. Normal? Riders in a carnival ride**What happens if the string breaks? Which way will the ball**move? The ball will continue to move in a straight line path that is “tangent” to the circle.**Tension in a string as a centripetal force**A student twirls a rock around and around in a horizontal circle at the end of the string. The only force that contributes to a NET centripetal force is the tension in the string.**Example**A boy twirls a ½ kg rock in a horizontal circle on the end of a 1.6 meter long string. If the velocity of the rock was 4 m/s, what is the Tension in the string? m = ½ kg r = 1.6 m v = 4 m/s The only centripetal force is Tension. T = m (v2 / r) T = ½ (42 / 1.6) T = 5 N**Example**How fast was the ½ kg rock moving if the Tension was 10 N and the string was 1.6 m long? m = ½ kg r = 1.6 m T = 10 N T = m(v2 / r) Tr/m = v2 10 x 1.6 / .5 = v2 v = 5.7 m/s**Friction along a surface as a centripetal force**A 1500 kg race car goes around a curve at 45 m/s. If the radius of the curve is 100 m, how much friction is required to keep the car on the track? What is m, the coefficient of friction? m = 1500 kg v = 45 m/s r = 100 m The centripetal force is friction. f = m(v2/r) f = 1500 x (452 / 100) f = 30375 N f = mN m= f / N but what is N? N = mg = 15000 N m = 30375 N / 15000 N m = 2.02**How do you find the velocity if it is not directly provided?**Velocity = distance / time In circular motion, the distance traveled is all around the circle… the circumference. The circumference = 2pr So… v = 2pr / T**Vertical loops**Twirling a rock at the end of a string in a vertical loop. At the top of the loop, both the Tension and the weight point towards the center of the circle! SF = T + mg = m(v2/r) At the bottom of the loop, the Tension points toward the center, the weight away from the center: SF = T – mg = m(v2/r)**What about an object on a vertical track?**At the top of the track, both the Normal force (the track pushing against the ball) and the weight point down toward the center of the circle, therefore, they are both positive: SF = N + mg = m(v2/r) At the bottom of the track, the Normal force points toward the center and the weight points away from the center: SF = N – mg = m(v2/r)**Loop the Loop**What is the minimum speed that a rider must be moving at in order to complete a loop the loop of radius 12 m? At the top of the loop, both the Normal force and weight point towards the center of the circle, so SFcentripetal = N + mg = mv2 / r However, at the minimum required speed, contact is lost for a moment at the top of the loop, so that… The Normal force goes all the way to ZERO. The weight is the only centripetal force when the rider is moving at the minimum required speed. mg = m(v2/r) g = v2/r v2 = rg v2 = 12 x 9.8 v = 10.84 m/s**Which COMPONENT of the Tension provides a centripetal force**on the object?**Free body diagram for an object that is “topping” a**hill. Which force is the positive centripetal force??**The Normal force**In some cases the normal force can contribute to the net centripetal force. For example, on the carnival ride where the riders stand against the walls of the circular room and the floor drops out! And yet, the rider does not slide down! Draw the free-body diagram! What keeps them from sliding down? The wall pushes against the rider toward the center of the circle. N = m v2 / r Also, if he is not sliding down, we know f = mg and therefore…..mN = mg We can combine those two equations (divide one by the other!) f N mg**Banked Tracks**What angle is necessary for a car to complete a turn without sliding even if the road is frictionless?**“Artificial Gravity”**Occupants of a space station feel weightless because they lack a support (Normal) force pushing up against their feet. By spinning the station as just the right speed, they will experience a “simulated gravity” when the Normal force of the floor pushing up on their feet becomes a centripetal force. The closer their centripetal acceleration, v2/r is to g, the acceleration due to Earth’s gravity, the more they feel the sensation of normal weight.**Artificial Gravity Example…**A circular rotating spacestation has a radius of 40 m. What linear velocity, v, must be maintained along the outer edge, to maintain a sense of “normal” gravity? We want the centripetal acceleration, v2/r (due to the rotation) to be the same as the acceleration due to gravity on Earth- 9.8 m/s2 v2/r = 9.8 m.s2 v2 = 9.8 x 40 v = 19.8 m/s**The Gravitational Force**Newton’s Universal Law of Gravitation states that every particle in the universe exerts an attractive force on every other particle. Where “G” is the “universal gravitational constant” G = 6.67 x 10-11 Nm2/kg2**What happens to the Force if one of the masses is doubled?**1. It will now be F x ? What happens to the Force if both of the masses were doubled? 2. It will now be F x ? What happens to the Force if one of the masses is doubled and then other one is halved? 3. It will now be F x ?**This is an “inverse square” law, since the**Force is proportional to the inverse of the distance squared. Example: At twice the distance, the gravitational force between two objects would be less. How much less?**Two objects are separated by some distance, d. How would**the gravitational force differ if the distance was tripled? 1/9 the original force What if the distance was 4d? 1/16 the original force 4. the distance was 5d? 5. The distance was 10d? ½ d? 6. At ½ d, the force would by F x ??**Example: Two masses of 5 kg and 9 kg are separated by 1.5 m.**What is the gravitational force they exert on each other? How do you enter all those numbers in your calculator? Use your exponent button (EE) for “G”!! Do NOT type in “ x 10^ ” 6.67E-11*5*9÷1.52 = F = 1.3 x 10-9 N • Enter 1.3 (one decimal place) • Enter the exponent, -9 G = 6.67 x 10-11**What is the gravitational force between a 600 kg mass and a**850 kg mass if they are 0.4 meters apart? • Enter the number with 1 decimal place • Enter the exponent G = 6.67 x 10-11**Example: Two masses of 3 x 103 kg and 1.8 x 1015 kg are**separated by d = 1.4 x 1021m. What is the gravitational force they exert on each other? How do you enter all those numbers in your calculator? Use your exponent button!! 6.67E-11*3E3*1.8E15÷1.4E212 = • Enter the number with 1 decimal place. • Enter the exponent G = 6.67 x 10-11**If the gravitational force between a 95 kg mass and a 120 kg**mass is 4 x 10-4N, how far apart are they? What’s the shortcut to get d2 out of the denominator? Trade places with F!! And don’t forget to take the square root! 13. Enter the value of “d” (two decimal places) G = 6.67 x 10-11**NET Gravitational Force**Two masses pull on the central mass. How would you get the NET gravitational force? Subtract the two forces.**NET Gravitational Force**Two masses pull on the left mass. How would you get the NET gravitational force? Add the two forces. (Be careful about your distances!)**NET Gravitational Force**Two masses pull on the mass at the origin. How would you get the NET gravitational force? Pythagorize the two forces. arc tan for angle.**Cavendish and “G”, the gravitational constant**Henry Cavendish, a British scientist, first devised an experiment to determine “G” in 1797. He suspended two small known masses from a “torsion wire” of which he knew the strength. These two small masses were gravitationally attracted to two large known masses, which caused the wire to twist until the torsion force was balanced by the gravitational force. Because he knew the strength of the torsion force, he also knew the strength of the gravitational force. With known masses, known Force, and known distance, the only “unknown” left was G! * You need to know who first determined “G”,**Finding “g”**Weight is the gravitational force a planet exerts. Weight = Gravitational Force mg = G “g”, the acceleration due to gravity can be found by canceling an “m”. The distance, d, is measured from the center of the planet to the location of interest. (often, the radius) The acceleration due to gravity, “g”, is also called the “gravitational field strength”. planet**How large is “g” on the planet Venus, which has a mass**of 4.87 x 1024 kg and has a radius of 6,050,000 meters? 6.67E -11 x 4.87 E24 ÷ 6,050,0002 = g = 8.87 m/s2**Example: An asteroid of radius 500 m has a mass of 6.5 x**1013 kg. What is the gravitational field strength at its surface? 6.67E -11 x 6.5 E13 ÷ 5002 = g = 0.0173 m/s2 How much would a 60 kg astronaut WEIGH on this asteroid? W = mg W = 60 kg x 0.0173 m/s2 W = 1.04 N**Aristotle**Geocentric universe 384 BC “geocentric” – Earth centered universe…… WRONG!**Ptolemy, 83 AD**Ptolemy (also geocentric universe) presented his astronomical models in convenient tables, which could be used to compute the future or past position of the planets, the Sun, and Moon, the rising and setting of the stars, and eclipses of the Sun and Moon. His model showed the planets turning in small circles as they orbited the Earth! The tables actually produced fairly good predictions, but his model and his geocentric universe was….. WRONG! Ptolemy was also the first to use latitude and longitude lines.**Copernicus 1473heliocentric universe**“sun-centered” universe Although others before him had proposed that the planets orbit the sun rather than the Earth, Copernicus was the first to publish mathematical evidence**Tycho Brahe**• 1546 • Built “The Castle of the Stars” • Had an accident in a duel • Died an unusual death…

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