Linea and angula analogs Linea Rotation x position x displacement v velocity a T tangential acceleation Vectos in otational motion Use the ight hand ule to detemine diection of the vecto! Don t foget centipetal acceleation! a R = a c = v / Kinematic equations fo angula and linea motion. Kinematic Equations v = v o + at = o + t Kinematic Equations x = x o + v ot + / at = o + ot + / t Kinematic Equations 3 v = v o + a(x x o) = o + ( o) Rotational Inetia Rotational analog of mass Fo point masses I = m I: otational inetia (kg m ) m: mass (kg) : adius of otation (m) Fo solid objects I = dm Paallel Axis Theoem I = I cm + M h I: otational inetia about cente of mass M: mass h: distance between axis in question and axis though cente of mass Kinetic Enegy K tans = ½ M v cm K ot = ½ I K combined = ½ M v cm + ½ I Rolling without slipping uses both kinds K = ½ M v cm + ½ I v = K = ½ M v cm + ½ I cm v cm /R o K = ½ M R + ½ I cm Toque Toque is the otational analog of foce. A twist (wheeas foce is a push o pull). Toque is a vecto) = F = F sin R: moment am length F: foce : angle between moment am and point of application of foce. = I (think F = ma) : toque I: otational inetia : angula acceleation Wok in otating systems W ot = (think W = F d) W ot : wok done in otation : toque : angula displacement Powe in otating systems P ot = (think P = F v) P ot : powe expended : toque : angula velocity Static Equilibium = 0 F = 0 Angula momentum Fo a paticle L = p Fo a system of paticles L = Li Fo a igid body L = I (think P = mv) Consevation of Angula Momentum Angula momentum of a system will not change unless an extenal toque is applied to the system. L B = L A I B = I A (one body) l b = l a (system of paticles) Angula momentum and toque = dl/dt (think F = dp/dt) : toque L: angula momentum t: time Toque inceases angula momentum when paallel. Toque deceases angula momentum when antipaallel. Toque changes the diection of the angula momentum vecto in all othe situations. Pecession The otating motion made by a spinning top o gyoscope. Pecession is caused by the inteaction of toque and angula momentum vectos. = dl / dt = F
MULTIPLE CHOICE PRACTICE PROBLEMS. D. A wheel spinning at 3 m/s unifomly acceleates to 6 m/s in 4 s. Its adius is 0 cm. How fa aound the wheel will a speck of dust tavel duing that inteval? A) 6 m D) 8 m B) 9 m E) 30 m C) m Ans. a = v v o t = 6m s 3m s 4 s = ¾ m/s v = v o + a x (6 m/s) = (3 m/s) + (¾ m/s )x 36 = 9 + 3/x 7 = 3/x 8 m = x The adius is not elevant.. B. If an object of adius 3 m that expeiences a constant angula acceleation stating fom est, otates 0 ads in s, what is its angula acceleation? A).5 ad/s D) 0 ad/s B) 5 ad/s E) 5 ad/s C) 7.5 ad/s Ans. = o + ot + / t 0 ad = 0 ad + (0 ad/s)( s) + ½ ( s) 0 = ½ ( s) 0 = 5 ad/s = 3. B. A bicycle moves at constant speed ove a hill along a smoothly cuved suface as shown above. Which of the following best descibes the diections of the velocity and the acceleation at the instant it is at the highest position? A) The velocity is towads the ight of the page and the acceleation is towads the top of the page. B) The velocity is towads the ight of the page and the acceleation is towads the bottom of the page. C) The velocity is towads the ight of the page and the acceleation is towads the bottom ight of the page. D) The velocity is towads the ight of the page and the acceleation is towads the top ight of the page. E) The velocity is towads the top ight of the page and the acceleation is towads the bottom ight of the page. Ans. Since the bike is moving at constant speed, we don t have to woy about tangential acceleation (a T). The only acceleation is a c, the centipetal acceleation. QUICK REVIEW. The net foce on the bike will be the diffeence of the downwads and upwad foces acting on the bike. The upwad foce is FN. The downwads foce is Fw. FNET = m a = m a c = Fw FN (assuming down is +) Since ac = v and Fw = m g, substituting: m v = m g FN What is the maximum velocity the bike could go so as to not lose contact with the hill? Assuming a cicula hill, we make the contact foce (F N) between the hill and the bike minimum at this maximum speed. So m v m v m v v = m g FN = m g (0) = m g = g Fw FN v = g If the bike was not moving at constant speed aound the cicle, then the answe would have been C. since you would have not only the centipetal acceleation, a c, but also the tangential acceleation, a T. a c a T a
Base you answes to questions 4 and 5 on the following situation. An object weighing 0 N swings at the end of a ope that is 0.7 m long as a simple pendulum. At the bottom of the swing, the tension in the sting is N. 4. A. What is the magnitude of the centipetal acceleation at the bottom of the swing)? A) m/s D) m/s B) 4 m/s E) m/s C) 0 m/s FT Ans. At the bottom of the swing the foce diagam of the pendulum looks as shown in the Figue at ight. Fw The net foce on the pendulum bob will be the diffeence of the downwads and upwad foces acting on the bob. The upwad foce is FT. The downwads foce is Fw. FNET = m a = m a c = Fw FT (assuming down is +) Solving fo a c and plugging in ou given values: m a c = F w F T a c = F w F T m Fw = m g 0 N N a c = 0 N = m (0 m/s ) kg a c = m s The magnitude is just the absolute value of the answe, a c without the diection ( ). 5. A. What is the speed of the object at the bottom of the swing? A) 0.6 m/s D).4 m/s B). m/s E) 7. m/s C).0 m/s Ans. The velocity can be obtained fom the fomula fo the centipetal acceleation: ac = v v = a c = ( m s) (0.7 m) v = 0.36 = 0.6 m/s 6. C. How fa fom the left end of the od should the foce be placed to maintain equilibium? A) 0 cm D) 40 cm B) 0 cm E) 50 cm C) 5 cm Ans. Since the od is unifom, we can assume that its cente of mass is at its geometic cente. Since the ba is.0 m long, the x cm is at 0.5 m. So we have a downwad foce of F = m g = 6 kg(0 m/s ) = 60 N at 0.5 m away fom the pivot point. To balance this out we need c = cc (clockwise and counte-clockwise) F = F (40 N) = (0.5 m) (60 N) 40 = 30 = 0.75 m Fom the left end, this is 0.5 m. 7. B. What foce is applied to the od by the pivot? A) 0 N D) 60 N B) 0 N E) 00 N C) 40 N Ans. We need to find the net foce fo equilibium to exist (fo the ba to be still) fogetting about toque fo the moment. F NET = F F F NET = 60 N 40 N = 0 N Base you answes to questions 6 and 7 on the pictue below, which epesents a igid unifom od with a mass of 6 kg and a length of.0 m is pivoted on the ight end. It is held in equilibium by an upwad foce of 40 N.
8. D. A unifom wooden boad of mass 0 M is held up by a nail hammeed into a wall. A block of mass M ests L/ away fom the pivot. Anothe block of a cetain mass is hung a distance L/3. The system is in static equilibium. Ans. E in = E out GPE in = TKE out + RKE out mgh = ½mv + ½I mgh = ½mv + ½ ( 5 m )( v ) gh = ½v + ½ ( 5 ) v g h = ½v + 5 v g h = 7 0 v 0 7 gh = v 0 (0 m 7 s)(7 m) = v 00 m = v 0 m/s = v What is the measue of the mass labeled "?"? A) M D) 3M B) M E) M 3 C) M Ans. To balance this out we need c = cc (clockwise and counte-clockwise) F = F (L/) (M g) = (L/3) (M g) M g L/ = M g L/3 3MgL gl = M 3M = M 9. B. The angula velocity of a otating disk with a adius of m deceases fom 6 ads pe second to 3 ads pe second in seconds. What is the linea acceleation of a point on the edge of the disk duing this time inteval? A) Zeo D) 3/ m/s B) 3 m/s E) 3 m/s C) 3/ m/s Ans. We will use the two elationships: a T = and = o t = o = 3 =.5 ad/s t s a T = = (.5 ad/s )( m) = 3 m/s = 3ad s 6ad s 0. E. A solid sphee of adius 0. m and mass kg is at est at a height 7 m at the top of an inclined plane making an angle 60 with the hoizontal. Assuming no slipping, what is the speed of the cylinde at the bottom of the incline? A) Zeo D) 6 m/s B) m/s E) 0 m/s C) 4 m/s. B. A spinning object with moment of inetia I inceases in angula speed fom = 0 to a in t seconds. What is the aveage powe deliveed to the object duing this inteval t? A) I a/t B) I a /t C) I a /t D) I a /t E) I a /t Ans. We will have to use the fomula fo powe in otational motion P ot = and the fomula fo toque = I P ot = I ( o ) ( a 0 ) t t a a t.? What is the moment of inetia of a spinning object of adius 0.5 m and mass 6 kg moving at 5 m/s, if it has a kinetic enegy of 00 J? A) kg m D) 8 k gm B) kg m E) 0 kg m C) 4 kg m Ans. TKEcombined = TKE + RKE TKEcombined = ½mv + ½I TKEcombined = ½mv + ½I( v ) 00 J = ½(6 kg)(5 m/s) + ½I( 5m s 0.5 m ) 00 = 75 + 50I 5 = 50I ½ = I
3. B. Which of the following objects has the least kinetic enegy at the bottom of the incline if they all have the same mass and adius? A) cylinde D) all have the same B) sphee E) not enough infomation C) hoop Ans. the sphee has the most mass concentated the closest to its axis of otation (I = /5m ). This means it has the geatest inetia and thus the geatest otational kinetic enegy (RKE = ½I ) all othe things being equal. 4. C. Which of the following objects has the geatest otational kinetic enegy at the bottom of the incline if they all have the same mass & adius? A) cylinde D) all have the same B) sphee E) not enough infomation C) hoop Ans. the hoop has the most mass concentated the futhest away fom the axis of otation (I = m ). This means it has the geatest inetia and thus the geatest otational kinetic enegy (RKE = ½I ) all othe things being equal. 5. C. A solid cylinde of adius. m and mass kg is at est at a height. m at the top of an inclined plane making an angle 60 with the hoizontal. Assuming no slipping, what is the speed of the cylinde at the bottom of the incline? A) Zeo D) 6 m/s B) m/s E) 0 m/s C) 4 m/s Ans. E in = E out GPE in = TKE out + RKE out mgh = ½mv + ½I mgh = ½mv + ½ ( m )( v ) gh = ½v + ½ ( ) v g h = ½v + 4 v g h = 3 4 v 4 3 gh = v 4 (0 m 3 s)(. m) = v 4 3 (0)( 6 5 ) = v 6 = v 4 m/s = v 6. B. What is the atio of the moment of inetia of a cylinde of mass m and adius to the moment of inetia of a hoop of the same mass and same adius? A) : D) :4 B) : E) 4: C) : Ans. I cylinde I hoop m m = 7. E. A 4 kg object moves in a cicle of adius 8 m at a constant speed of m/s. What is the angula momentum of the object with espect to an axis pependicula to the cicle and though its cente? A) N s D) 4 m /s B) 6 N m/kg E) 64 kg m /s C) kg m/s Ans. L = p = m v = 8 m 4 kg m/s = 64 kg m /s 8. E. A solid cylinde with diamete 0 cm has an angula velocity of 0 ad/s and angula momentum of kg m /s. What is its mass? A) 0. kg D) 5 kg B) kg E) 0 kg C) kg Ans. L = I L = m kg m /s = m (0. 0 m) (0 ad/s) = 0.m 0 kg = m
FREE RESPONSE Ff FN Fw (b) Ans. The foces along the plane ae the component of the weight along the plane F x and the foce of fiction, F f. F net = ma = F x F f = mg sin F f Befoe otational motion, we could define the foce of fiction F f stictly in tems of linea vaiables. But it is the fiction that is making the sphee oll. Let s take a look at fiction F f and its ole in making the ball oll down the incline. toque is given by = I
Toque is also F whee the foce F that is making the ball spin as it is in contact with the incline is the fictional foce, F f. Substituting: F f = I Since the ball is spheical, we can eplace I with 5 m, the fomula fo the moment of inetia. We can also eplace since a =, whee a is the tangential acceleation of the ball and is the adius of the sphee. F f =( 5 m ) a Solving fo F f: F f =( 5 m ) a = 5 ma Ff = 5 m a If we solve fo m a, we get 5 Ff = m a We now have two expessions fo m a: 5 Ff = m a and m a = m g sin Ff Setting them equal to each othe: 5 Ff = m g sin Ff Solving fo F f: 7 Ff = m g sin Ff = 7 m g sin Ff = 7 (6 kg) (0 m/s ) sin (30 ) = 8.6 N (c) Ans. To find the speed of the sphee we have to use ou consevation of enegy elationship. E in = E out GPE in = TKE out + RKE out mgh = ½mv + ½I m g d sin = ½mv + ½ ( 5 m )( v ) g d sin = ½v + ½ ( 5 ) v g d sin = ½v + 5 v g d sin = 7 0 v 0 7 g d sin = v 0 (0 m 7 s)(4 m) sin (30 ) = v 5.34 m/s = v (d) Ans. The hoizontal velocity ight as the ball comes olling off the oof would be vx = v cos 30 (see the Figue at ight) The ball will maintain this hoizontal velocity in the absence of a hoizontal acceleation. This is the velocity that will make the wagon move fowad. The momentum befoe and afte the collision must be conseved. p in = p fi m v in = m v fi m v x = m v fi m v cos 30 = m v fi m v cos 30 = v fi m 6 kg (5.34 m cos 30 )= v fi 8 kg s.54 m/s = v fi vy 30 v vx
FREE RESPONSE (a) Ans. The angula momentum of this igid unifom body is given by L = I = ( 3 M d ) since the given otational inetia is I = 3 M d. (b) The momentum will be conseved afte the impact. Lin = Lfi ( 3 M d ) = M v d
3 M d M d = v (c) If the collision is elastic then the kinetic enegy is also conseved. This means we have a second elationship between M and M : two equations and two unknowns. Ein = Eout RKEin = TKEout ½I = ½mv ( M 3 d ) = M( 3 M d = M 3 M d ) M d 9 M d 4 M d d M 3 d = M 9M M = M d 3M d M = M 3M M M = M 3 M M = 3 so M M = 3 (d) The momentum will be conseved afte the impact. Lin = Lfi I = p ( M 3 d ) = M v x Solving fo v as we did in Pobl. (a): = v 3 M d M x The masses cancel out. = v 3 d x Since the collision is elastic, we can call upon the kinetic enegy elationship as well as in Pobl. (c): Ein = Eout RKEin = TKEout ½I = ½mv ( M 3 d ) = M( 3 d ) x Simplifying and solving fo x: 3 M d = M 3 d = 9 d4 x 3 x 9 = d4 d 3 x = 9 d x = 3 d x = 3 3 d 9 d4 x
FREE RESPONSE 3
(a) Ans. The otational inetia is given by I = m. I = m L + m L = m L (b) The acceleation of the downwad block can be obtained fom F NET = m a = 4m a = F w F T 4m a = 4m g F T Solving fo F T: F T = 4m g 4m a The tension is elated to the toque that the od expeiences. toque is given by = I Toque is also F whee the foce F that is making the od spin is equal to F T, the tension in the ope. Substituting: F T = I In Pobl. (a), we found that the moment of inetia, I = m L fo the od-and-block appaatus. We can also eplace since a =, whee a is the tangential acceleation of the od-and-block appaatus and is equal to L. F T =(ml ) a Solving fo F T: F T =(ml ) a F T = ml a Setting the tensions equal to each othe and solving fo a: ml a = 4m g 4m a ml a + 4m a = 4m g Canceling out the mass m: L a + 4a = 4g a( L + 4) = 4g a( L +4 ) = 4g 4g a = L +4 a = g L + (c) The total kinetic enegy, by consevation of enegy, must be equal to the loss of potential enegy of the falling block of mass 4m: GPEin = 4mgD = TKEfi (d) The total kinetic enegy, by consevation of enegy, must be less than the loss of potential enegy of the falling block of mass 4m because some of the enegy has gone into aising the two spinning masses on the od-and-block appaatus. They now have some GPE. GPE in = 4mgD = TKE fi + GPEfi 4m FT Fw