Physics 101 Lecture 10 Rotation

Transcription

1 Physics 101 Lecture 10 Rotation Assist. Pro. Dr. Ali ÖVGÜN EMU Physics Department

2 Rotational Motion Angular Position and Radians Angular Velocity Angular Acceleration Rigid Object under Constant Angular Acceleration Angular and Translational Quantities Rotational Kinetic Energy Moments o Inertia

3 Angle and Radian What is the circumerence S? s = ( p )r can be deined as the arc length s along a circle divided by the radius r: = is a pure number, but commonly is given the artiicial unit, radian ( rad ) s r p = Whenever using rotational euations, you must use angles expressed in radians s r r s

4 Conversions Comparing degrees and radians p ( rad) =! 360 p ( rad) =! 180 Converting rom degrees to radians p ( rad ) = ( degrees) 180 Converting rom radians to degrees 180 (deg rees) = p! ( rad) 360 1rad = = 57.3 p

5 Rigid Object A rigid object is one that is nondeormable n The relative locations o all particles making up the object remain constant n All real objects are deormable to some extent, but the rigid object model is very useul in many situations where the deormation is negligible This simpliication allows analysis o the motion o an extended object

6 Angular Position Axis o rotation is the center o the disc Choose a ixed reerence line Point P is at a ixed distance r rom the origin As the particle moves, the only coordinate that changes is As the particle moves through, it moves though an arc length s. The angle, measured in radians, is called the angular position.

7 Angular Displacement The angular displacement is deined as the angle the object rotates through during some time interval D = - SI unit: radian (rad) This is the angle that the reerence line o length r sweeps out i

8 Average and Instantaneous Angular Speed The average angular speed, ω avg, o a rotating rigid object is the ratio o the angular displacement to the time interval -i D w = = avg t -t Dt The instantaneous angular speed is deined as the limit o the average speed as the time interval approaches zero lim w D t 0 SI unit: radian per second (rad/s) D º = Dt Angular speed positive i rotating in counterclockwise Angular speed will be negative i rotating in clockwise i d dt

9 Average Angular Acceleration The average angular acceleration, a, o an object is deined as the ratio o the change in the angular speed to the time it takes or the object to undergo the change: a avg w - w Dw t -t Dt i = = i t = t i : w i t = t : w

10 Instantaneous Angular Acceleration The instantaneous angular acceleration is deined as the limit o the average angular acceleration as the time goes to 0 SI Units o angular acceleration: rad/s² Positive angular acceleration is counterclockwise (RH rule curl your ingers in the direction o motion). n n i an object rotating counterclockwise is speeding up i an object rotating clockwise is slowing down Negative angular acceleration is clockwise. n n Dw º = Dt lim a D t 0 dw dt i an object rotating counterclockwise is slowing down i an object rotating clockwise is speeding up

11 Rotational Kinematics A number o parallels exist between the euations or rotational motion and those or linear motion. v avg = x t - x - t i i = Dx Dt i = = Under constant angular acceleration, we can describe the motion o the rigid object using a set o kinematic euations n These are similar to the kinematic euations or linear motion n The rotational euations have the same mathematical orm as the linear euations w avg - D t -t Dt i

12 Analogy with Linear Kinematics Start with angular acceleration: Integrate once: Integrate again: w a w at = ò dt = i + dw a = dt v = v + at 1 ( t) dt t t = ò wi + a = i + wi + a 1 x = x + vt+ at i i Just substitute symbols, and all o the old euations apply: x Þ v Þ w a Þ a i

13 Comparison Between Rotational and Linear Euations

14 Ex:1 A Rotating Wheel A wheel rotates with a constant angular acceleration o 3.5 rad/s. I the angular speed o the wheel is.0 rad/s at t = 0 (a) through what angle does the wheel rotate between t = 0 and t =.0 s? Given your answer in radians and in revolutions. (b) What is the angular speed o the wheel at t =.0 s? w = t i a = = s rad rad / / s s w - = = i??

15 Relationship Between Angular and Linear Quantities Every point on the rotating object has the same angular motion Every point on the rotating object does not have the same linear motion Displacement Speeds s v Accelerations =r = wr a=ar

16 Speed Comparison The linear velocity is always tangent to the circular path n Called the tangential velocity The magnitude is deined by the tangential speed Ds D = r D Ds = = 1 Dt rdt r Ds Dt v w = or v = rw r

17 Acceleration Comparison The tangential acceleration is the derivative o the tangential velocity Dv Dv = Dt = rdw Dw r = ra Dt a t = ra

18 Speed and Acceleration Note All points on the rigid object will have the same angular speed, but not the same tangential speed All points on the rigid object will have the same angular acceleration, but not the same tangential acceleration The tangential uantities depend on r, and r is not the same or all points on the object v w = or v = rw a = ra r t

19 Centripetal Acceleration An object traveling in a circle, even though it moves with a constant speed, will have an acceleration n Thereore, each point on a rotating rigid object will experience a centripetal acceleration v ( rw) a r = = = r r rw

20 Resultant Acceleration The tangential component o the acceleration is due to changing speed The centripetal component o the acceleration is due to changing direction Total acceleration can be ound rom these components a= a + a = r a + r w = r a + w 4 4 t r

21 Rotational Kinetic Energy An object rotating about z axis with an angular speed, ω, has rotational kinetic energy Each particle has a kinetic energy o K i = ½ m i v i Since the tangential velocity depends on the distance, r, rom the axis o rotation, we can substitute v i = wr i => K i = ½ m i w r i

22 Rotational Kinetic Energy There is an analogy between the kinetic energies associated with linear motion (K = ½ mv ) and the kinetic energy associated with rotational motion (K R = ½ Iw ) Rotational kinetic energy is not a new type o energy, the orm is dierent because it is applied to a rotating object Units o rotational kinetic energy are Joules (J)

23 Work-Energy Theorem or pure Translational motion The work-energy theorem tells us 1 Wnet = DKE = KE - KEi = mv - Kinetic energy is or point mass only, ignoring rotation. Work Wnet = dw = F d s ò ò 1 mv i Power dw d s P = = F = F v dt dt

24 Mechanical Energy Conservation Energy conservation When W nc = 0, Wnc = D K + DU K + U = Ui + Ki The total mechanical energy is conserved and remains the same at all times 1 1 mv i + mgyi = mv + mgy Remember, this is or conservative orces, no dissipative orces such as riction can be present

25 Total Energy o a System A ball is rolling down a ramp Described by three types o energy n Gravitational potential energy U = Mgh n Translational kinetic energy n Rotational kinetic energy 1 Mv 1 = Iw 1 1 E = MvCM + Mgh + Iw Total energy o a system K t Kr = CM

26 Work done by a pure rotation Apply orce F to mass at point r, causing rotation-only about axis Find the work done by F applied to the object at P as it rotates through an ininitesimal distance ds! dw = F d s = F cos(90 -j) ds = F sinjds = Fr sinjd Only transverse component o F does work the same component that contributes to torue dw =td

27 Work-Kinetic Theorem pure rotation As object rotates rom i to, work done by the torue I is constant or rigid object Power tw =t = = dt d dt dw P ò ò ò ò ò = = = = = i i i i i d I d dt d I d I d dw W w w w a t 1 1 i I I d I d I W i i w w w w w w - = = = ò ò

28 An motor attached to a grindstone exerts a constant torue o 10 N-m. The moment o inertia o the grindstone is I = kg-m. The system starts rom rest. n n n Find the kinetic energy ater 8 s 1 t K = Iw = 1600J Ü w = w + at = 40 rad/s Ü a = = 5 rad/s I i Find the work done by the motor during this time W = òtd = t ( - ) = = 1600J i i 1 ( - ) = wt+ 160 rad at = 1 1 i i W = K - Ki = Iw - Iwi = 1600 J Find the average power delivered by the motor n P avg dw 1600 = = = 00 watts dt 8 Find the instantaneous power at t = 8 s P = tw = = 400 watts

29 Work-Energy Theorem For pure translation 1 1 W = D K = K - K = mv - mv net cm cm, cm, i i For pure rotation 1 1 Wnet = D Krot = Krot, - Krot, i = Iw - Iwi Rolling: pure rotation + pure translation W = D K = ( K + K )- ( K + K ) net total rot, cm, rot, i cm, i æ1 1 ö æ1 1 ö = ç Iw + mv - ç Iwi + mvi è ø è ø

30 Energy Conservation Energy conservation When W nc = 0, W = D K + DU nc K + K + U = K + K + U rot, cm, rot, i cm, i i The total mechanical energy is conserved and remains the same at all times I wi + mvi + mgyi = Iw + mv + mgy Remember, this is or conservative orces, no dissipative orces such as riction can be present total

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38 Total Energy o a Rolling System A ball is rolling down a ramp Described by three types o energy n Gravitational potential energy U = Mgh n Translational kinetic energy 1 Kt = Mv n Rotational kinetic energy 1 Kr = Iw 1 E = Mv + Mgh + 1 Iw Total energy o a system

39 Problem Solving Hints Choose two points o interest n One where all the necessary inormation is given n The other where inormation is desired Identiy the conservative and non-conservative orces Write the general euation or the Work-Energy theorem i there are non-conservative orces n Use Conservation o Energy i there are no nonconservative orces Use v = rw to combine terms Solve or the unknown

40 A Ball Rolling Down an Incline A ball o mass M and radius R starts rom rest at a height o h and rolls down a 30 slope, what is the linear speed o the ball when it leaves the incline? Assume that the ball rolls without slipping. 1 mv i + mgy i Iw i = mv + mgy Mgh + 0 = Mv Iw 1 Iw I = MR Mgh = Mv + MR = Mv + w = 5 v R v R 1 5 Mv 10 v = ( gh) 7 1/

41 Rotational Work and Energy A ball rolls without slipping down incline A, starting rom rest. At the same time, a box starts rom rest and slides down incline B, which is identical to incline A except that it is rictionless. Which arrives at the bottom irst? Ball rolling: 1 1 mvi + mgyi + I 1 1 mgh = mv + Iw Box sliding 1 1 w i = mv + mgy + Iw 1 1æ ö 7 = mv + mr ( v / R) mv ç = è5 ø mv i + mgyi = mv + mgy 1 sliding: mgh = mv 7 rolling: mgh = mv 10

42 Blocks and Pulley Two blocks having dierent masses m 1 and m are connected by a string passing over a pulley. The pulley has a radius R and moment o inertia I about its axis o rotation. The string does not slip on the pulley, and the system is released rom rest. Find the translational speeds o the blocks ater block descends through a distance h. Find the angular speed o the pulley at that time.

43 Find the translational speeds o the blocks ater block descends through a distance h. K + K + U = K + K + U rot, cm, rot, i cm, i i m1 v + mv + Iw ) + ( m1gh - m gh) = ( 1 I ( m1 + m + ) v = mgh - m1gh R v = é ( m - m1 ) gh ê ëm1 + m + I / R ù ú û 1/ Find the angular speed o the pulley at that time. w = v R = 1 R é ( m - m1 ) gh ê ëm1 + m + I / R ù ú û 1/

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45 Example: A roll o toilet paper is held by the irst piece and allowed to unurl as shown in the diagram to the right. The roll has an outer radius R = 6.0 cm, an inner radius r = 1.8 cm, a mass m = 00 g, and alls a distance s = 3.0 m. Assuming the outer diameter o the roll does not change signiicantly during the all, determine the inal angular speed o the roll the inal translational speed o roll the angular acceleration o the roll the translational acceleration o the roll the tension in the sheets

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50 Example: The top shown below consists o a cylindrical spindle o negligible mass attached to a conical base o mass m = 0.50 kg. The radius o the spindle is r = 1. cm and the radius o the cone is R = 10 cm. A string is wound around the spindle. The top is thrown orward with an initial speed o v 0 = 10 m/s while at the same time the string is yanked backward. The top moves orward a distance s =.5 m, then stops and spins in place. Using energy considerations determine the tension T in the string

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52 Example: Calculating Helicopter Energies A typical small rescue helicopter has our blades: Each is 4.00 m long and has a mass o 50.0 kg. The blades can be approximated as thin rods that rotate about one end o an axis perpendicular to their length. The helicopter has a total loaded mass o 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy o the helicopter when it lies at 0.0 m/s, and compare it with the rotational energy in the blades. (a) Sketch o a our-blade helicopter. (b) A water rescue operation eaturing a helicopter rom the Auckland Westpac Rescue Helicopter Service.

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54 Example: Energy in a Boomerang A person hurls a boomerang into the air with a velocity o 30.0 m/s at an angle o 40.0 with respect to the horizontal It has a mass o 1.0 kg and is rotating at 10.0 rev/s. The moment o inertia o the boomerang is given as I = 1/1 ml where L = 0.7 m. (a) What is the total energy o the boomerang when it leaves the hand? (b) How high does the boomerang go rom the elevation o the hand, neglecting air resistance?

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57 Moment o Inertia o Point Mass For a single particle, the deinition o moment o inertia is I = mr n m is the mass o the single particle n r is the rotational radius SI units o moment o inertia are kg. m Moment o inertia and mass o an object are dierent uantities It depends on both the uantity o matter and its distribution (through the r term)

58 Moment o Inertia o Point Mass For a composite particle, the deinition o moment o inertia is I = åm r = m r + m r + m r + m r +... n n i i m i is the mass o the ith single particle r i is the rotational radius o ith particle SI units o moment o inertia are kg. m 1 1 Consider an unusual baton made up o our sphere astened to the ends o very light rods Find I about an axis perpendicular to the page and passing through the point O where the rods cross I = åm r i i = mb + Ma + mb + Ma = Ma + mb

59 The Baton Twirler Consider an unusual baton made up o our sphere astened to the ends o very light rods. Each rod is 1.0m long (a = b = 1.0 m). M = 0.3 kg and m = 0. kg. (a) Find I about an axis perpendicular to the page and passing through the point where the rods cross. Find K R i angular speed is (b) The majorette tries spinning her strange baton about the axis y, calculate I o the baton about this axis and K R i angular speed is

60 Moment o Inertia o Extended Objects Divided the extended objects into many small volume elements, each o mass Dm i We can rewrite the expression or I in terms o Dm I = lim åri D mi = r dm D m 0 i i Consider a small volume such that dm = r dv. Then ò I = ò r r dv I r is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known

61 Densities You know the density (volume density) as mass/unit volume n r = M/V = dm/dv => dm = rdv We can deine other densities such as surace density (mass/unit area) n s = M/A = dm/da => dm = sdv Or linear density (mass/unit length) n l = M/L = dm/dx => dm = ldv

62 Moment o Inertia o a Uniorm Rigid Rod The shaded area has a mass n dm = l dx Then the moment o inertia is L/ M Iy = òr dm= ò x dx -L/ L 1 I = ML 1

63 Moment o Inertia or some other common shapes

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