Dynamics of Uni Cicular Motion勻速圓周運動的動力學(xué)

1、dynamics of uniform circular motion uniform circular motion is the motion of an object traveling at a constant (uniform) speed on a circular path.the period (t) is the time it takes an object to make complete circle or revolution.because velocity = distance / timethe velocity of an object around a c

2、ircle is v = 2r (the distance around a circle) t (time it takes for one revolution)this fan has a radius of .7m and is rotating at 90 rev/min. how fast is the end tip of the fan blade going?convert 90 rev / min to the time it takes for one revolution.v = 2r = 2(3.14)(.7m) = m/s t 1.67 secthis fan ha

3、s a radius of .7m and is rotating at 90 rev/min. how fast is the end tip of the fan blade going?r = .7mcentripetal acceleration uniform circular motion emphasizes that velocity vector is constant but the direction of this constant velocity is not constant. by definition, acceleration is the change i

4、n velocity over time. the “change in velocity” for an object in ucm is the change in the velocity vector. the acceleration that occurs is toward the center of the circle. it is the force that is pulling the object “in” centripetal acceleration the centripetal acceleration (the acceleration toward th

5、e center of a circle) of an object depends on the velocity of the object and the radius of the circle and is expressed as:ac = v2 rd = 100mhow fast would the outer most part of the space station have to spin in order to have a centripetal acceleration of 9.8 m/s2 how fast would the outer most part o

6、f the space station have to spin in order to have a centripetal acceleration of 9.8 m/s2 ?a = v2 / r a = 9.8 m/s r = 100m/2 = 50msolving for v:v = /(a r) = /(9.8m/s2)(50m) = 22.1 m/sd = 100ma bobsled goes down a track, moving at a constant velocity throughout all three turns. at which turn will ther

7、e be the greatest centripetal acceleration?ab cd all the sameif the bobsled moves at 35 m/s throughout the course and turn b has a radius of 20m and turn c is 50m how many gs of acceleration is the sledder feeling through turns b & c ?centripetal force the net force required to keep an object of

8、 mass m, moving at a constant velocity v, on a circular path. f = mv2 rnote that v2/r = accelerationa 50 kg person and a 100 kg person ride the “enterprise” at valleyfair. the ride has a radius of 7 meters and takes 4 seconds per revolution. how much centripetal force does each person feel at point

9、a & c?“gondolas start out hanging from the ride, then go to a horizontal position by centrifugal force from the spinning ride, which then lifts to a vertical position turning riders upside down.” a cdbat what point will a person feel the most centripetal force?abcd e= all the samecentripetal for

10、ce lab reviewa person on a centrifuge with a radius of 6m spins around once in 2 seconds. the person is in contact with a scale that is reading 1,100 lbs. what is the mass of the person being spun?how many gs is thisperson experiencing?1n = .22 lb 2.2lbs = 1kga person on a centrifuge with a radius o

11、f 6m spins around once in 2 seconds. the person is in contact with a scale that is reading 1,100 lbs. what is the mass of the person being spun? how many gs is this person experiencing?start by finding the velocity of the centrifuge: v = 2r / t t=2sec r= 6mv= 2(3.14)(6) / 2 = 18.8m/sthen, using fc =

12、 mv2/r, rearrange the equation solving for mm = r fc / v2 convert 1,100 lbs to newtons1,100 lbs x 1 n = 5000n 1 .22 lbs m = 6 m(5000 n) / 18,82 = 84.9 kg = massac = v2 / r = 18.82 / 6 = 58.9 m/s258.9m/s2 / 9.8m/s2 = 6.0 gs of accelerationis it harder for the man to hold his partner when the partner

13、is hanging straight down and is stationary or when the partner is swinging through the straight down position?it is harder for the man to hold his partner when the partner is swinging through.when stationary, the man only needs to hold the force of the partner against gravity . ex. 50kg x 9.8m/s2 =

14、490 nwhen the partner swing, the man needs to hold with the force from gravity, plus the centripetal force of the partner because they are accelerating (changing v) a soapbox racer is making a turn on asphalt at 20 m/s. the coefficient of frictionstatic is .9. the radius of the turn is 40 m. will th

15、e soapbox crash into the hay? why or why not? when a car turns a corner, there is a centripetal force toward the center of the circle. the force that is being applied to the center of the circle is coming from static friction between the car and the ground. if the rubber on the car cannot handle the

16、 centripetal force, the car will slide. static friction is keeping the car from slidingfc = fn = mv2/r mg = mv2/rv2 = mgr / m = grv = sqrt .9(9.8)(40) = 18.8 m/s max speed around the turn banked turnswhen a car turns a flat corner, static friction provides the centripetal force. the need of static f

17、riction can be eliminated by making a bank in the turn.the centripetal force becomes a vector component of the normal force. in the case above, it is = fn(cos ).if there is a set amount of bank, the can be a perfect speed to take the turn at without needing static friction. but, if you go too slow,

18、you may slip, too fast and you may skidbanked turns*this equation indicates that for a given speed, the centripetal force needed for a turn of radius r can be obtained by banking the turn at an angle, independent of the mass of the vehicle.tan = v 2 r gmr. shaf hota racetrack is designed with a 40o

19、bank angle and a radius of 240 m at the ends. at what speed is the track designed for cars to make an ideal turn? v = sqrt (tan )rg = sqrt (tan 40)(240m)(9.8m/s2)= 44.4 m/sa racecar enters a banked turn that has a 40o angle. the car is going 34 m/s into the turn. at what distance should the driver l

20、ocate his car if he wishes to take the turn without depending on friction?40odmg140.5m d = 140.5 sin 50 = 183.4 musing the equation tan = v2 / rg we want to find the radius of the turn. once we have the radius, well us trig to find d r = v2 / tan (g) = 342 / tan 40 (9.8) = 140.5 m 50oa racecar enter

21、s a banked turn that has a 40o angle. the car is going 34 m/s into the turn. at what distance should the driver locate his car if he wishes to take the turn without depending on friction?40od fc = fn cos = mv2 / r if we can find the radius of the turn, we can find d using trig. solving for r, r = mv

22、2 since fn = mg/sin50, well substitute this in fn cos r = mv2 = v2 (sin50) = 342 (sin50) = 885 = 140.5m mgcos g cos 9.8(cos 500) 6.3 (sin50)fnfc =fn(cos ) = mv2 / r50omg fn sin 50 mg = 0 (no vertical acceleration)50o140.5md = 140.5 sin 50 =183.4 msatellites in orbitthere is only one speed that a sat

23、ellite can have if the satellite is to remain in an orbit with a fixed radius.the centripetal force acting on satellites is gravity:fc = g m1m2 = mv2 r2 rv = gme t = 2r3/2 r gmea rocket is launched to a height of 50 miles above the earths surface. how fast does the rocket need to be going in order to stay in orbit? (1 mile = 1.6 km)(earths radius = 6,380,000 m)(earths mass = 5.98 x 1024)how many miles per hour is the earth orbiting the sun? (93,000,000 miles radius)(mass of sun = 2 x 1030)synchronous satellites are satellites