Rotational Motion. Every quantity that we have studied with translational motion has a rotational counterpart

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1 Rotational Motion & Angula Momentum

2 Rotational Motion Evey quantity that we have studied with tanslational motion has a otational countepat TRANSLATIONAL ROTATIONAL Displacement x Angula Position Velocity v Angula velocity acceleation a Angula acceleation Mass m Inetia I Momentum p Angula Momentum L Foce F Toque

3 Angula Position s Aclength Radius (fom axis of otation) Measued in adians all angula quantities will be measued in adians (NOT degees)

4 Angula Position In tanslational motion, position is epesented by a point, such as x. x = 3 x 0 5 p/2 linea In otational motion, position is epesented by an angle, such as, and a adius,. p 0 3p/2 angula

5 Angula Displacement Linea displacement is epesented by the vecto Dx. Dx = x p/2 linea Angula displacement is epesented by D, which is not a vecto, but behaves like one fo small values. p D 60 3p/2 0 angula

6 Angula Displacement D 0 Compae to Dx = x x o Which diection is positive (by convention)? Positive Counteclockwise Negative - Clockwise

7 Tangential vs. angula displacement A paticle that otates though an angle D also tanslates though a distance s, which is the ac length defining its path. This distance s is elated to the angula displacement D by the equation s = D, as we saw in a pevious slide. s D

8 EXAMPLE: (a) What is the angula position,, if we go aound a cicle two times? Ans. = s = 2(2p) = 4π (b) Say you go a quate tun moe, what is D? Ans. D = o = 9π 2 4π = π 2 (c) What is the aclength coveed in total between Pobls. (a) and (b) if the adius of the cicle is 3 m? Ans. s = = (3 m)(4p + π 2 ) =3m(9π 2 ) = 27π 2 m

9 P.O.D. 1: Two synchonous communications satellites ae put into an obit whose adius is = 4.23 x 10 7 m. The two adjacent satellites have an angula sepaation of 2. Find the ac length, s, that sepaates the satellites.

10 Speed and velocity s v T The instantaneous velocity of a paticle in a cicula path has magnitude v T = ds and is tangent to dt the cicle. The same paticle otates with an avg. angula velocity = Dθ = θ f θ i Dt Dt o instantaneous angula velocity = dθ dt Tangential and angula speeds ae elated by the equation v T =. v T D

11 Angula Velocity avg inst D Dt d dt Units ad/s, ev/s Compae to v avg = x t and v inst = dx dt Diection Same as displacement (positive is counteclockwise, negative is clockwise) Magnitude Angula Speed

12 Deivation of tangential velocity fomula s ds dt d dt Take the deivative of each side of s = with espect to t. v Substituting = dθ dt and v T = ds dt

13 OBSERVATIONS: All points on object have the same angula speed, Those points fathe fom the cente have a geate linea (tangential) speed, v T These points have to tavel fathe in the same time Deivation of tangential velocity fomula avg D Dt v T

14 P.O.D. 2: A gymnast on a high ba swings though two evolutions in a time of 1.90 s. (a) Find the aveage angula velocity (in ads/s) of the gymnast. (b) What is he angula displacement (in ad) afte t = 1 s. (c) Find he tangential velocity if the gymnast is 1.8 m tall.

15 Acceleation Tangential acceleation is given by s a T a T = dv T dt Instantaneous angula acceleation of this paticle is given by = d dt Aveage Angula acceleation of this paticle is given by = D Dt Tangential and angula acceleations ae elated by the equation a = a T D

16 avg Angula Acceleation D Dt Compae to a avg = v t and a inst = dv dt Angula Acceleation is how the angula velocity changes with time d Units ad/s inst 2 dt This is not the centipetal acceleation that tells about a tanslational acceleation

17 Sample Poblem A DVD spins at an angula velocity of 3 ad/s and 5 seconds late is spinning at a velocity of 8 ad/s. Its adius is 0.02 m. Find (a) Angula acceleation, = ω t =ω f ω i t = (a) Tangential acceleation, a T (a) Centipetal acceleation, a c, at you final angula velocity

18 P.O.D. 3: A jet awaiting cleaance fo takeoff is momentaily stopped on the unway. The fan blades ae otating with an angula velocity of 110 ad/s, whee the negative sign indicates a clockwise otation. As the plane takes off, the angula velocity of the blades eaches 330 ad/s in a time of 14 s. (a) Find the angula acceleation, assuming it to be constant. (b) Find the tangential acceleation, if the fan blade has a adius of 1.5 m.

19 Centipetal Acceleation F c F g A pendulum is swinging back and foth. At the bottom of the swing the foce of gavity is pulling it downwads but it doesn t fall down. This means thee must be a foce pulling upwads to balance it out. This is the centipetal foce. Since F = ma, the cente-seeking acceleation is called centipetal acceleation and is given by: v 2 a 2 c

20 Tangential Acceleation Deivation of Acceleation v T dv dt T d dt Take the deivative of each side of v = with espect to t. dv dt T Substituting = d dt a T Substituting a = dv dt

21 Tangential Acceleation vs. Angula Acceleation vs. Centipetal Acceleation The net acceleation is the sum of the tangential and centipetal acceleations. a a c v 2 2 a T = 2 2 t a a a

22 Sample Poblem A compact disk otates about an axis though its cente accoding to (t) 1 3 t 3 6t (a) Detemine its angula velocity and angula acceleation at time t = 5 seconds. Ans. To find the angula velocity, take the fist deivative. To find the angula acceleation, take the second deivative '(t) ' '(t) d dt d dt 2t t 2 6 (5) 2(5) ads 2 s ads s

23 Sample Poblem (Cont.) A compact disk otates about an axis though its cente accoding to (t) 1 3 t 3 6t (a) What is the linea speed of a point 20 cm fom the cente at t = 5 s? (b) What is the linea acceleation at 0.5 m at t = 5 s? Ans. To find the linea speed, use the elationship, whee (5) we obtained fom the pevious slide: v = = (19 ad/s)(0.20 m) = 3.8 m/s. To find the linea acceleation, use the elationship a c = 2 = (19 ad/s) 2 (0.20 m) = 72.2 m/s 2

24 P.O.D. 4: The gyoscope of a plane is spinning accoding to θ t = 3t 3 + 6t 2 ½t. (a) Find the aveage angula velocity of the gyoscope between t = 2 s and t = 5 s. (b) Find the instantaneous angula acceleation at t = 2 s. (c) Find the linea speed, v T, at t = 2 s of a point on the gyoscope if it has a adius of 0.05 m. (d) Find the linea (tangential) acceleation, a T, at t = 2 s of the gyoscope. (e) Find the centipetal acceleation, a c, of a point on the edge of the gyoscope.

25 Constant Angula Acceleation Ou kinematics equations have angula equivalents Just as with thei linea countepats, these only wok fo constant acceleation

26 Fist Kinematic Equation v = v o + at (linea fom) Substitute angula velocity fo linea velocity. Substitute angula acceleation fo linea acceleation. = o + t (angula fom)

27 Second Kinematic Equation x = x o + v o t + ½ at 2 (linea fom) Substitute angle fo position. Substitute angula velocity fo linea velocity. Substitute angula acceleation fo linea acceleation. = o + o t + ½ t 2 (angula fom)

28 Thid Kinematic Equation v 2 = v o2 + 2a(x x o ) Substitute angle fo position. Substitute angula velocity fo linea velocity. Substitute angula acceleation fo linea acceleation. 2 = o2 + 2( o )

29 Sample Poblem An automobile stats fom est and fo 20.0 s has a constant linea acceleation of 0.8 m/s 2 to the ight. Duing this peiod, the ties do not slip. The adius of the ties is m. At the end of the 20.0 s inteval what is the angle though which each wheel has otated? Ans. The angula acceleation can be found fom the fomula a = α = a α = 0.8 m s 2 = 2.42 ad/s m The angula acceleation should be negative because the tie spins clockwise. To find the angula displacement: (t) = o t + ½t 2 = 0(20 s) + ½(2.42 ad/s 2 )(0.20 s) 2 = 484 ad

30 P.O.D. 5: The blades of an electic blende ae whiling with an angula velocity of +375 ad/s while the puee button is pushed in. When the blend button is pessed, the blades acceleate and each a geate angula velocity afte the blades have otated though an angula displacement of ad (seven evolutions). The angula acceleation has a constant value of ad/s 2. (a) Find the final angula velocity of the blades. (b) Find the angula displacement of the blades afte 10 seconds. (c) Find the change in tangential velocity of the blades fom the puee to the blend position if they have a adius of 0.02 m.

31 ANGULAR MOMENTUM!, INERTIA, ETC.

32 Angula Momentum Angula momentum depends on linea momentum and the distance fom a paticula point. It is a vecto quantity with symbol L. If and v ae then the magnitude of angula momentum w/ esp. to point Q is given by L = p = mv. In this case L points out of the page. If the mass wee moving in the opposite diection, L would point into the page. The SI unit fo angula momentum is the kg m 2 / s. (It has no special name.) Angula momentum is a conseved Q v m quantity. A toque is needed to change L, just a foce is needed to change p. Anything spinning has angula has angula momentum. The moe it has, the hade it is to stop it fom spinning.

33 Angula Momentum: Geneal Definition If and v ae not then the angle between these two vectos must be taken into account. The geneal definition of angula momentum is given by a vecto coss poduct: L = p This fomula woks egadless of the angle. Fom coss poducts, the magnitude of the angula momentum of m elative to point Q is: L = p sin = m v. In this case, by the ight-hand ule, L points out of the page. If the mass wee moving in the opposite diection, L would point into the page. v m Q

34 Moment of Inetia vs. Angula Momentum Any moving body has inetia. (It wants to keep moving at constant v.) The moe inetia a body has, the hade it is to change its linea motion. Rotating bodies possess a otational inetia called the moment of inetia, I. The moe otational inetia a body has, the hade it is change its otation. Fo a single point-like mass w/ espect to a given point Q, I = m 2. Fo a system, I = the sum of each mass m times its espective distance fom the point of inteest. I = m 2 Q m 1 1 Q 2 m 2 I = m 2 = m 2 + m 2

35 Moment of Inetia of vaious shapes

36 Moment of Inetia Example Two mey-go-ounds have the same mass and ae spinning with the same angula velocity. One is solid wood (a disc), and the othe is a hollow metal ing. Which has a bigge moment of inetia elative to its cente of mass? m m Ans. I is independent of the angula speed. Since thei masses and adii ae the same, the ing has a geate moment of inetia. This is because moe of its mass is fathe fom the axis of otation. Since I is bigge fo the ing, it would moe difficult to incease o decease its angula speed.

37 Toque & Angula Acceleation Newton s 2 nd Law, as you know, is F net = ma The 2 nd Law has a otational analog: net = I A foce is equied fo a body to undego acceleation. A tuning foce (a toque) is equied fo a body to undego angula acceleation. The bigge a body s mass, the moe foce is equied to acceleate it. Similaly, the bigge a body s otational inetia, the moe toque is equied to acceleate it angulaly. Both m and I ae measues of a body s inetia (esistance to change in motion).

38 Example: The toque of an Electic Saw Moto The moto in an electic saw bings the cicula blade up to the ated angula speed of 80.0 ev/s in ev. One type of blade has a moment of inetia of 1.41 x 10-3 kgm 2. What net toque (assumed constant) must the moto apply to the blade? Ans. Fist we need to convet ou values into ad/s fo calculation puposes. o t 240 ev x 2p = 1508 ads? 80 evs/s x 2p = 503 ad/s 0 ad/s? We can find the angula acceleation fom 2 = o2 + 2 Solving fo : = 2 2 o = (503 ad/s)2 0 2 = ad/s 2 2(1508 ad/s) 2 = I = 1.41 x 10-3 kgm ad/s 2 = Nm

39 P.O.D. 6: A Chinese sta of mass kg and adius 0.03 m is thown by Dwight fom The Office at his advesay, Jim Halpet. The Chinese sta is thown fom est. If its final angula velocity is 15 evs/s afte 3 sec, (a) find the angula acceleation of the Chinese sta (in ad/s 2 ). (b) Find the toque of the Chinese sta (Assume the Chinese sta is hoop-shaped).

40 Linea Momentum vs. Angula Momentum If a net foce acts on an object, it must acceleate, which means its momentum must change. Similaly, if a net toque acts on a body, it undegoes angula acceleation, which means its angula momentum changes. Recall, angula momentum s magnitude is given by L = mv (if v and ae pependicula) v m So, if a net toque is applied, angula velocity must change, which changes angula momentum. Poof: net = F net = m a = m Dv = DL t t So net toque is the ate of change of angula momentum, just as net foce is the ate of change of linea momentum. continued on next slide

41 Linea & Angula Momentum (cont.) Hee is yet anothe pai of simila equations, one linea, one otational. Fom the fomula v =, we get L = mv = m ( ) = m 2 = I This is vey much like p = mv, and this is one eason I is defined the way it is. In tems of magnitudes, linea momentum is inetia times speed, and angula momentum is otational inetia times angula speed. L = I p = m v

42 Compaison: Linea & Angula Momentum Linea Momentum, p Tendency fo a mass to continue moving in a staight line. Paallel to v. A conseved, vecto quantity. Magnitude is inetia (mass) times speed. Net foce equied to change it. The geate the mass, the geate the foce needed to change momentum. Angula Momentum, L Tendency fo a mass to continue otating. Pependicula to both v and. A conseved, vecto quantity. Magnitude is otational inetia times angula speed. Net toque equied to change it. The geate the moment of inetia, the geate the toque needed to change angula momentum.

43 Example: Spinning Ice Skate Suppose M. Stickman is sitting on a stool that swivels holding a pai of dumbbells. His axis of otation is vetical. With the weights fa fom that axis, his moment of inetia is 600 kgm 2 and he is spinning at an angula velocity of 20 ad/s. When he pulls his ams in as he s spinning, the weights ae close to the axis, so his moment of inetia gets to 400 kgm 2. What will be his angula velocity at this moment? Ans. I = L = I 600 kgm 2 (20 ad/s) = 400 kgm 2 12,000 = ad/s =

44 P.O.D. 7: An atificial satellite (m = 1500 kg) is placed into an elliptical obit about the eath. Telemety data indicate that its point of closes appoach (called the peigee) is p = 8.37 x 10 6 m fom the cente of the eath, while its point of geatest distance (called the apogee) is A = 25.1 x 10 6 m fom the cente of the eath. The speed of the satellite at the peigee is v p = 8450 m/s. (a) Find its speed v A at the apogee. (b)find its angula momentum at any point in its obit.

45 Rotational Kinetic Enegy A paticle in a otating object has otational kinetic enegy: K i = ½ m i v i 2, whee v i = i (tangential velocity) The whole otating object has a otational kinetic enegy given by: 1 K K m 2 2 R i i i i i KR m I 2 i i i

46 Rotational Kinetic Enegy example: A thin walled hollow cylinde (mass = m h, adius = h ) and a solid cylinde ( mass = m s, adius = s ) stat fom est at the top of an incline. Both cylindes stat at the same vetical height h o. All heights ae measued elative to an abitaily chosen zeo level that passes though the cente of mass of a cylinde when it is at the bottom of the incline. Ignoing enegy losses due to etading foces, detemine which cylinde has the geatest tanslational speed upon eaching the bottom.

47 Rotational Kinetic Enegy example (cont.): Ans. At the top of the incline the cylinde have only gavitational potential enegy. At the bottom of the incline this enegy has conveted into tanslational kinetic and otational kinetic enegy. E in = E out GPE in = TKE out + RKE out mgh = ½mv f2 + ½I f 2 The angula velocity can be elated to the linea velocity v f by f = v f Substituting the given values and fo the angula velocity: m h gh o = ½m h v f2 + ½I( v f h ) 2 Fo the hollow cylinde, the moment of inetia is given by: I = m 2 Substituting: m h gh o = ½m h v f2 + ½(m h h2 )( v f h ) 2 Simplifying: gh o = ½v f2 + ½ h2 ( v f h ) 2 gh o = ½v f2 + ½ h2 v f2 gh o = ½v f2 + ½v f 2 gh o = v f 2 gh o = v f h 2

48 Rotational Kinetic Enegy example (cont.): Ans. At the top of the incline the cylinde have only gavitational potential enegy. At the bottom of the incline this enegy has conveted into tanslational kinetic and otational kinetic enegy. E in = E out GPE in = TKE out + RKE out mgh = ½mv f2 + ½I f 2 The angula velocity can be elated to the linea velocity v f by f = v f Substituting the given values and fo the angula velocity: m s gh o = ½m s v f2 + ½I( v f s ) 2 Fo the solid cylinde, the moment of inetia is given by: I = ½m 2 Substituting: m s gh o = ½m s v f2 + ½(½m s s2 )( v f s ) 2 Simplifying: 4 gh o = ½v f2 + ½ ½ s2 ( v f s ) 2 gh o = ½v f2 + ¼ s2 v f2 gh o = ½v f2 + ¼v f 2 gh o = ¾v f 2 3 gh o= v f s 2

49 PROBLEM 8:

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