2 Uniform Circular Motion Motion in a circular path at constant speed Speed constant, velocity changing continually Velocity changing direction, so there is acceleration Called centripetal acceleration, since it is toward the center of the circle, along the radius Value can be calculated by many formulas, first isac = v2/r
3 Example ac = v2 =__(6.0 m/s)2 = 36 (m/s)2 = 2.9 m/s2 r 12.5 m 12.5 m A bicycle racer rides with constant speed around a circular track 25 m in diameter. What is the acceleration of the bicycle toward the center of the track if its speed is 6.0 m/s?ac = v2 =__(6.0 m/s)2 = 36 (m/s)2 = 2.9 m/s2r m m
4 Rotation and Revolution Rotation-Around an Internal Axis-Earth rotates 24 hours for a complete turn Linear (tangential) versus rotational speed Linear is greater on outside of disk or merry-go-round, more distance per rotation Linear is smaller in middle of disk, less distance per rotation. Rotational speed is equal for both Rotations per minute (RPM) Linear speed is proportional to both rotational speed and distance from the center
5 Rotation and Revolution Revolution-Around an External Axis-Earth revolves days per trip around sun Same relationship between linear and revolutional speeds as with rotational Planets do not revolve at the same revolutional speeds around the sun
6 PeriodAnother important measure in UCM is period, the time for 1 rotation or revolution Since x=v0t , this implies that vT = 2r and thus T= 2pr/ v Rearranging differently, v= 2pr/ T and then inserting it into the acceleration equation ac = v2/r = 4p2r/T2
7 ExampleDetermine the centripetal acceleration of the moon as it circles the earth, and compare that acceleration with the acceleration of bodies falling on the earth. The period of the moon's orbit is 27.3 days. According to Newton's first law, the moon would move with constant velocity in a straight line unless it were acted on by a force. We can infer the presence of a force from the fact that the moon moves with approximately uniform circular motion about the earth. The mean center-to-center earth-moon distance is 3.84 x 108 m.
8 Exampleac= 4p2r = 4p2(3.84 x 108) T= 27.3 da (24 hr/da)(3600 s/hr) = 2.36 x 106 s T2 (2.36 x 106)2 ac = 2.72 x 10-3 m/s2 The ratio of the moon's acceleration to that of an object falling near the earth is ac = 2.72 x l0-3 m/s2 = 1 g 9.8 m/s2 3600
9 FrequencyThe number of revolutions per time unit Value is the inverse of the period, 1/T Units are sec-1 or Hertz (Hz) Inserting frequency into the ac equation ac=4p2f 2r
10 ExampleAn industrial grinding wheel with a 25.4-cm diameter spins at a rate of 1910 rotations per minute. What is the linear speed of a point on the rim? The speed of a point on the rim is the distance traveled, 2pr, divided by T, the time for one revolution. However, the period is the reciprocal of the frequency, so the speed of a point on the rim, a distance r from the axis of rotation, isv = 2prfv= (2)(25.4cm/2)(1910/1 min)(1min/60s)v = 2540cm/s = 25.4 m/s.
11 Angular VelocityVelocity can be defined in terms of multiples of the radius, called radians There are 2p radians in a circle, and so the angular velocity w = v/r In terms of period w= 2p/T In terms of frequency w=2pf
12 ExampleAt the Six Flags amusement park near Atlanta. The Wheelie carries passengers in a circular path with a radius of 7.7 m. The ride makes a complete rotation every 4.0 s. (a) What is a passenger's angular velocity due to the circular motion? (b) What acceleration does a passenger experience?a) The ride has a period T = 4.0 s. We can use it to compute the angular velocity as2= 2 rad = rad/s = 1.6 rad/sT 4.0 s(b) Because the riders travel in a circle, they undergo a centripetal acceleration given byac= 2r = (/2 rad/s)2(7.7m) = 19m/s2.Notice that this is almost twice the acceleration of a body in free fall.
13 Angular Velocity and Acceleration Any real object that has a definite shape can be made to rotate – solid, unchanging shapeAngular displacement -- q -- Radians around circular pathAngular velocity -- w --radians per second, angle between fixed axis and point on wheel changes with timeAngular acceleration -- a -- increase of w , when angular velocity of the rigid body changes, radians per seconds squared
14 Rotational Kinematics Rotational velocity, displacement, and acceleration all follow the linear forms, just substituting the rotational values into the equations:q= wot + 1/2 at2 wf2 = wo2 + 2aqwf=w0 +at q=(w0 +wf) t/ 2q= x/r a= a/r w= v/r
15 ExampleThe wheel on a moving car slows uniformly from 70 rads/s to 42 rads/s in 4.2 s. If its radius is 0.32 m:a. Find a b. Find q c. How far does the car go?a. a= Dw = (42-70) rads/s = -6.7 rads/s2Dt sb. q= wot + 1/2 at2 = (70)(4.2) + 1/2(-6.7)(4.2)2 = 235 radsc. q = x / r in rads so x = q r = (0.32)(235) = 75 m
16 ExampleA bicycle wheel turning at 0.21 rads/s is brought to rest by the brakes in exactly 2 revolutions. What is its angular acceleration?q = 2 revs = 2(2p) radians = 4p rads wf=0 rads/s wo= 0.21 rads/sUse angular equivalent of vf2 = vo2 + 2ax which iswf2 = wo2 + 2aq(0)2 = (0.21)2 + 2a(4p)a = -(0.21)2 = x10-3 rad/s22(4p)
17 Forces in Circular Motion Centripetal ForceForce toward the center from an object, holding it in circular motionAt right angle to the path of motion, not along its distance, therefore does NO work on objectExamplesGravitation between earth and moonElectromagnetic force between protons and electrons in an atomFriction on the tires of a car rounding a curveEquation is Fc=mac = mv2/r
18 ExampleApproximately how much force does the earth exert on the moon? Moon’s period is 27.3 days Assume the moon's orbit to be circular about a stationary earth. The force can be found from F = ma. The mass of the moon is 7.35 x 1022 kg. Fc= mac = m 42r T2 Fc = (7.35 x 1022 kg)42(3.84 x 108m) ((27.3 days)(24 hr/day)(3600 s/hr))2 Fc=2.005 x 1020 N.
19 Forces in Circular Motion Centrifugal “Force”Not a true force, but really the result of inertia“Centrifugal force effect” makes a rotating object fly off in straight line if centripetal force fails
20 ExampleImagine a giant donut-shaped space station located so far from all heavenly bodies that the force of gravity may be neglected. To enable the occupants to live a “normal” life, the donut rotates and the inhabitants live on the part of the donut farthest from the center. If the outside diameter of the space station is 1.5km, what must be its period of rotation so that the passengers at the periphery will perceive an artificial gravity equal to the normal gravity at the earth's surface?The weight of a person of mass m on the earth is a force F = mg.The centripetal force required to carry the person around a circle of radius r isF =mac = m 42rT2We may equate these two force expressions and solve for the period T:mg =m42rT2 T=2 =2p =55s = 0.92 min.
21 Banked Curves“Banking” road curves makes turns without skidding possible For angle q, there is a component of the normal force toward the center of the curve, thus supplying the centripetal force. The other component balances the weight force. FN sin q = mv2/r FN cos q = mg tan q= v2/gr thusly q = tan-1 (v2/gr) This equation can give the proper angle for banking a curve of any radius at any linear speed
22 Banked Curve ExampleA race track designed for average speeds of 240 km/h (66.7 m/s) is to have a turn with a radius of 975 m. To what angle must the track be banked so that cars traveling 240km/h have no tendency to slip sideways? Determine q fromq = tan-1 (v2/ g r)= tan-1 (66.72/9.81(975))= 24.9o
23 Law of Universal Gravitation Newton’s first initiative for the Principia was investigating gravity From his 3rd law, he proposed that each object would pull on any other object He likewise noted differences due to distance His final relationship was that Force was proportional to masses and inversely proportional to distance squared Using a constant Fg = Gm1m2 r2
24 Center of GravityNewton found that his law would only work when measuring from the center of both objectsThis idea is called the center of gravitySometimes it is at the exact center of the objectSometimes it may not be in the object at allAll forces must be from the CG of one object to the CG of the other object
25 Universal Gravitation Constant G was elusive to find since gravity is a weak force if masses are small Cavendish developed a device which made measurement of G possible The value of G is 6.67 x N m2 This puts Fg in Newtons kg2 G can be used then to find values of many astronomical properties
26 ExampleConsider a mass m falling near the earth's surface. Find its acceleration in terms of the universal gravitational constant G. The gravitational force on the body is F = GmME r2 ME = mass of the earth r = the distance of the mass from the center of the earth, essentially the earth's radius. The gravitational force on a body at the earth's surface is F = mg. mg= GmME or g =GME r2 r2 Both G and ME are constant, and r does not change significantly for small variations in height near the surface of the earth. The right-hand side of this equation does not change appreciably with position on the earth’s surface, so replace r with the average radius of the earth RE g = GME RE2
27 ExampleShow that Kepler’s third law follows from the law of universal gravitation. Kepler’s third law states that for all planets the ratio (period)2/ (distance from sun)3 is the same.Make the approximation that the orbits of the planets are circles and that the orbital speed is constant.The sun's gravitational force on any planet of mass m isF= GmMr2M =the mass of the sun. Because the mass of the sun is so much larger than the mass of the planet, we can assume, as Kepler did, that the sun lies at the center of the planetary orbit. The circular orbit implies a centripetal force. This net force for circular motion is provided by the gravitational force. Equating these two forces, we getFc=GmM = 4p2mr Rearranging gives T2 = 4p2r T r GM
28 ExampleUse the law of universal gravitation and the measured value of the acceleration of gravity g to determine the average density of the earth. The density, r of an object is defined as its mass per unit volume: r = m/V where m is the mass of the object whose volume is V From a previous example g = GME RE2 Substitute for M an expression involving r, r = ME/V. If we take the earth to be a sphere of radius RE. Then r = ME and ME= 4/3pRE3 r 4/3pRE3 The equation for g can then be rewritten in terms of the density as g = G(4/3pRE3 r) = 4/3GpRE r
29 Density Example Upon rearranging, we find the density to be r = 3g 4pREGInserting the numerical values, we getr = (9.81 m/s2)4p(6.38 x 106 m)(6.67 x N m2/ kg2r = 5.50 x 103 kg/m3
30 Escape VelocityEarth spacecraft must get entirely away from the earth to go on to other planets This requires giving a spacecraft enough energy to overcome the gravitational potential energy of earth This gives an equation such that Where M and R vary according to the celestial object involved
31 Black HolesIf the escape velocity is equal to the speed of light, gravity will keep even light from escaping--the idea behind the black hole Conjecture due to observations from space Theory is a supergiant star collapses in on itself creating super strong gravity at a small point
32 Black HolesGravity is great due to small distance with huge mass Gravity only great near the object, at distance gravity is no different
33 Keplers LawsFirst Law: Each planet travels ina an elliptical path around the sun, and the sun is at one of the focal points
34 Keplers LawsSecond Law: An imaginary line is drawn from the sun to any planet sweeps out equal areas in equal time intervals.
35 Keplers LawsThird Law: The square of a planet’s orbital period (T2) is proportional to the cube of the average distance (r3) between the planet and the sun or T2 ∝ r3
37 General Definition of Torque Let F be a force acting on an object, and let r be a position vector from a rotational center to the point of application of the force. The magnitude of the torque is given by = 0° or = 180 °:torque are equal to zero = 90° or = 270 °: magnitude of torque attain to the maximum
38 Torque Units and Direction The SI units of torque are N.mTorque is a vector quantityTorque magnitude is given byTorque will have directionIf the turning tendency of the force is counterclockwise, the torque will be positiveIf the turning tendency is clockwise, the torque will be negativeThe work done by the torque is given by
39 Moment of inertiaIn rotation problems, its not only the mass of an object that is important but also its locationThe spacial distribution of the mass of an object is called theMoment of inertia (I)I =1/2 MR 2
40 Newton’s Second Law for a Rotating Object When a rigid object is subject to a net torque (≠0), it undergoes an angular accelerationThe angular acceleration is directly proportional to the net torqueThe angular acceleration is inversely proportional to the moment of inertia of the objectThe relationship is analogous to
42 Extended Work-Energy Theorem The work-energy theorem tells usWhen Wnc = 0,The total mechanical energy is conserved and remains the same at all timesRemember, this is for conservative forces, no dissipative forces such as friction can be present
43 Total Energy of a System A ball is rolling down a rampDescribed by three types of energyGravitational potential energyTranslational kinetic energyRotational kinetic energyTotal energy of a system
44 Conservation of Mechanical Energy Remember, this is for conservative forces, no dissipative forces such as friction can be present
45 Work-Energy in a Rotating System The work done on the body by the external torque equals the change in the rotational kinetic energyThe work equals the negative of the change in potential energyConservation of Energy in Rotational Motion
46 Problem Solving Hints Choose two points of interest One where all the necessary information is givenThe other where information is desiredIdentify the conservative and non-conservative forcesWrite the general equation for the Work-Energy theorem if there are non-conservative forcesUse Conservation of Energy if there are no non- conservative forcesUse v = w to combine termsSolve for the unknown
47 BuoyancyThe “upthrust” or buoyancy is equal to the weight of the displaced fluidV = volume of liquid displacedG = the acceleration of gravityObjects that are less dense than water will floatTo find where an object sits in the water, find the ratio of its density to the density of water
48 Bernoulli’s principle Bernoulli's Principle: slower moving air below the wing creates greater pressure and pushes up.
49 Moving fluids Q = volumetric flow rate A = cross sectional area v = fluid velocity= the angle between the direction of the fluid flow and a vector normal to AMass flow rate (m)When a pipe is constricted, the mass flow rate is conserved
50 Bernoulli’s principle for an incompressible fluid constantWhere v = fluid velocityg = acceleration due to gravityh = heightp = pressure= fluid density
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