PY1052 Problem Set 8 Autumn 2004 Solutions


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1 PY052 Poblem Set 8 Autumn 2004 Solutions H h () A solid ball stats fom est at the uppe end of the tack shown and olls without slipping until it olls off the ighthand end. If H 6.0 m and h 2.0 m, what is its speed when it olls off the end? The total mechanical enegy of the ball will be conseved we must just make sue to include both its otational and tanslational kinetic enegies, and also the elationship between the speed of the cente of mass and the angula speed of the otation fo olling without slipping, v Rω: KE i + PE i KE i + PE i 0 + mgh 2 mv2 f + 2 Iω2 f + mgh mgh 2 mv2 f + 2 [2 5 mr2 ][ v f R ]2 + mgh g(h h) 2 v2 f + 5 v2 f 7 0 v2 f 0 v f g(h h) (9.8 m/s2 )(4.0 m) 7.48 m/s (2) A hollow sphee of adius 0 cm and mass 20 kg is mounted so that it can otate feely about a hoizontal axis that is paallel to and 5.0 cm fom the cental axis of the sphee. (a) What is the moment of inetia of the sphee about this axis? (b) If the sphee is eleased fom est with its cental axis at the same height as the otation axis, what is the angula speed of the sphee as it passes though its lowest position? (a) We can use the paallelaxis theoem, which elates the moment of inetia we ae looking fo, I, with the moment of inetia fo otation about an axis paallel to ous and going though the cente of mass, I com (h is the distance between the two axes): I I com + Mh MR2 + M( 2 R)2 2 3 MR2 + 4 MR2 2 MR2 (20 kg)(0.0 m)2 2 I 0.83 kg m 2 (b) Now use consevation of enegy; the height of the cente of mass when the sphee passes though its lowest position is 5 cm lowe than its initial height:
2 PE i + KE i PE f + KE f Mgh i + 0 Mgh f + 2 Iω2 2 Iω2 Mg(h i h f ) 2 [ 2 MR2 ]ω 2 Mg(h i h f ) ω 2 24g(h i h f ) R 2 24g(hi h f ) ω R 2 24(9.8 m/s2 )(0.050 m) (0.0 m) 2 ω 0.3 ad/s (3) A hoop olls down a amp. (a) What must be the inclination of the amp in ode fo the tanslational acceleation of the cente of the hoop to have a magnitude of 0.0g? (b) If a fictionless block wee to slide down the incline tilted at that same angle, would its acceleation be moe than, less than o equal to 0.0g? Explain why without actually calculating the acceleation of such a block. (a) We deived in class an expession fo the tanslational acceleation of an object olling down an incline (it is also given on p. 250 of Halliday, Resnick & Walke): a com g sin θ + I com /MR 2 Hee a com is the acceleation of the cente of mass down the amp and I com is the moment of inetia of the olling object fo otation about an axis passing though its cente of mass. In the case of a hoop, this is I com MR 2. We can now find the needed inclination of the amp: a com a com 2 g sin θ g sin θ + MR 2 /MR 2 sin θ 2a com g sin θ 2(0.0g) g θ (b) The acceleation of the block would be geate than that of the hoop. The eason is that the acceleation comes about due to a tansfomation of the initial gavitational potential enegy of the object going down the amp into kinetic enegy. In the case of both the block and the hoop, the same amount of gavitational potential enegy is tansfomed into kinetic enegy in the case of the block, all of the kinetic enegy that is gained is tanslational, but in the case of the hoop, some of the kinetic enegy that is gained is otational, which means that the hoop has less tanslational kinetic enegy afte going the same distance down the amp.
3 (4) A 3.0 kg object with velocity v (5.0 m/s)î (6.0 m/s)ĵ is located at x 3.0 m, y 8.0 m and expeiences a foce F 7.0 N acting in the negative x diection. (a) What is the angula momentum of the object about the oigin? (b) What toque about the oigin acts on the paticle due to F? (c) At what ate is the angula momentum of the paticle changing with time? y y +8 F v α atan(8/3) 69 deg F α β x +3 x v β atan(6/5) 50 deg (a) The diagam shows the object, its velocity and the foce acting on it. The object s momentum is in the same diection as its velocity, and is equal to p m v. The angula momentum of the object about the oigin will be (φ is the angle between and p) L p L p sin φ p sin(α + β) (3.0 m) 2 + (8.0 m) m v (5.0 m/s) 2 + ( 6.0 m/s) m/s L (8.54 m)[(3.0 kg)(7.8 m/s)] sin(9) L 75 kg m 2 /s The diection of the angula momentum is given by the ighthand ule, and is into the page. (b) The toque exeted by the foce F is τ F τ F sin φ F sin(80 α) τ (8.54 m)(7.0 N)(sin() τ 55.8 N m The diection of the toque is given by the ighthand ule, and is out of the page. (c) The ate at which the angula momentum of the paticle is changing with time is given by dl dt τ 55.8 N m (5) A sanding disk with moment of inetia I kg m 2 is attached to an electic dill whose moto exets a constant toque of τ 6 N m. What ae (a) the angula momentum of the disk about its cental axis 33 ms afte the moto is tuned on and (b) the angula speed of the disk at this same time? (a) The angula momentum of the disk will be given by the expessions τ L/ t o L Iω; the fist of these is easie to use based on the infomation given: L τ t L 0 (6 N m)( s) L 0.53 N m s 0.53 J s
4 (b) Now that we have L, we can use the second of the elations given above to find ω the angula speed of the disk at the moto has been going 33 ms: ω L I 0.53 J s kg m ad/s Anothe way to get this is fom the elation τ Iα: α tau I 6 N m kg m ad/s 2 ω ω α t 0 + ( ad/s 2 )( s) ad/s m M (6) A small puck of mass m kg slides in a cicle on a fictionless table while attached to a hanging mass M kg by a cod though a hole in the table, as shown. (a) With what speed must the puck move to keep the hanging mass stationay if the adius of the cicle in which it moves is m? (b) What is the puck s angula momentum in this case? Anothe mass M 2 is gently hung below M, pulling on the cod and causing the adius of the cicula motion of the puck to decease to m. (c) What is the new velocity of the puck? (d) What is M 2? (a) The upwad tension in the cod must balance the downwad foce of gavity. This tension will also be equal to the centipetal foce: T M g 0 mv 2 M g 0 v M g m v 3.50 m/s (0.500 kg)(0.500 m)(9.8 m/s 2 ) kg (b) Teating the puck as a point mass moving in a cicle, its angula momentum will be L mv L (0.200 kg)(3.50 m/s)(0.500 m) kg m 2 /s (c) The angula momentum of the system is conseved. Theefoe,
5 L i L f mv i i mv f f v f v i i f v f (3.50 m/s) m/s (d) We can now find the mass M 2 using the same appoach as in pat (a) T (M + M 2 )g 0 mv 2 (M + M 2 )g 0 M 2 mv2 g M M 2 (0.200 kg)(4.38 m/s)2 (0.400 m)(9.8 m/s 2 ) M kg kg (7) A wheel with a moment of inetia I is otating feely at an angula speed of 800 ev/min on an axle with negligible moment of inetia. A second wheel with moment of inetia 2I that is initially at est is then suddenly coupled to the same axle, so that it also otates. (a) What is the angula speed of the combination of the two wheels otating about the axle? (b) What faction of the oiginal otational kinetic enegy is lost? (a) The key is that the angula momentum of the two wheels is conseved when they ae coupled: L befoe L afte I i ω i + I 2i ω 2i I f ω f + I 2f ω 2f Iω i + 2I(0) (I + 2I)ω f ω f ω i 3 (b) The kinetic enegy that is lost is given by KE KE befoe KE afte 267 ev/min ( 2 I ω 2 i + 2 I 2(0) 2 ) 2 I fω 2 f 2 Iω2 i 2 (3I)ω2 f and the faction that is lost is given by KE KE i 2 Iω2 i 2 (3I)ω2 f 2 Iω2 i ω2 i 3ω 2 f ω 2 i (800 ev/min)2 3(267 ev/min) 2 (800 ev/min) % of the initial kinetic enegy is lost when the wheels ae coupled.
6 m.5 kg m 2.0 kg 2 M 4.0 kg (8) A boad is balanced as is shown above. m and m 2 ae x 2.5 m and x 2.2 m to the left of the suppot point, which is just unde the cente of the boad. How fa is M fom the suppot point? We have thee conditions fo the equilibium: Fx 0 Fy 0 τ 0 In this case, thee ae no hoizontal foces, so we only have the last two of these equations. (i) F y 0 F suppot + m g + m 2 g + Mg + m boad g 0 whee F suppot is the upwad foce of the suppot and m boad is the mass of the boad (which we do not know). (ii) τ 0 We can choose any point about which to take the toques. It is convenient in this case to choose the cente of the boad as the point about which we take ou toques. The eason is that if we choose any othe point, we will get a toque equation that has thee unknowns: F suppot, m boad and X, whee X is the distance fom the suppot to the mass M. This means that we ll have two equations with thee unknowns, since we ae not given the mass of the boad. If we take toques about the cente of the boad, neithe the weight of the boad no F suppot exet toques, and the only unknown in ou toque equation will be X: τ m gx sin(90) + m 2 gx 2 sin(90) MgX sin(90) 0 X m x +m 2 x 2 0 M (.5 kg)(2.5 m)+(2.0 kg)(.2 m) X.54 m 4.0 kg (b) A bowle thows a bowling ball of adius R cm along a lane. The ball slides on the lane with an initial speed v m/s and initial angula speed ω 0 0. The coefficient of kinetic fiction between the ball and lane is µ k 0.2. The kinetic fiction foce acting on the ball causes a linea acceleation of the ball while also poducing a toque that causes an angula acceleation of the ball. When the speed v has deceased enough and the angula speed ω has inceased enough, the ball stops sliding and then olls smoothly. (a) What then is v in tems of ω? (b) Duing the sliding, what ae the ball s tanslational and angula acceleation? (c) How fa does the ball slide? (a) When the ball has slowed down enough to oll smoothly, the linea and angula velocities must be elated by the usual fomula fo smooth olling: v Rω. (b) The ball s tanslational acceleation is given by F ma µ k N µ k mg ma µ k g a
7 whee N is the nomal foce and m is the mass of the bowling ball. acceleation is given by The ball s angula τ Iα µ k NR µ k mgr [ 2 5 mr2 ]α 5µ k g 2R α Hee, τ RF k sin φ, whee F k is the foce of kinetic fiction and φ is the angle between the adius vecto to the point whee the foce of kinetic fiction acts and the diection of F k, which is 90. (c) We can find how fa the ball slides by fist finding v using a and finding ω using α, then finding the time the ball slides by setting v Rω : v v i + at v i µ k gt slide ω ω i + αt 5µ kg 2R t slide (ω i 0) v i µ k gt slide R( 5µ kg 2R t slide) (µ k g µ kg)t slide 7 2 µ kgt slide t slide 2 v i 7 µ k g m/s 7 (0.2)(9.8 m/s 2 ).8 s The distance the ball slides is then given by d v i t + 2 at2 v i t µ kg 2 t2 slide (8.5 m/s)(.8 s) (0.2)(9.8 m/s2 ) (.8 s) m 2 Putty Rotation axis (2b) Two 2.00 kg balls ae attached to the ends of a thin od of negligible mass that is 50.0 cm long. The od is fee to otate in a vetical plane without fiction about a hoizontal axis though its cente. With the od initially hoizontal, a 50.0 g wad of putty dops onto one of the balls, hitting it with a speed of 3.00 m/s and sticking to it. (a) What is the angula speed of the system just afte the putty hits? (b) What is the atio of the kinetic enegy of the system afte the collision to that of the putty just befoe the collision? (c) Though what angle will the system otate until it momentaily stops?
8 (a) The angula momentum of the od+balls+putty will be conseved in this inelastic collision. Just befoe the collision, all the angula momentum belongs to the putty. To find this initial angula momentum, we need to choose a point about which we will calculate the angula momentum I chose the point about which the od can otate, i.e., the cente of the od. The initial angula momentum of the falling putty is then L i putty p putty sin φ putty m putty v putty sin φ (0.250 m)( kg)(3.00 m/s) sin(90) kg m 2 /s Hee, φ is the angle between and p: putty p Rotation axis The angula momentum afte the collision will be L f Iω, whee I is the total moment of inetia fo the balls and putty. Teating the two balls like point masses: I f 2[M ball R 2 ] + m putty R 2 [4M ball + m putty ]R 2 [4(2.0 kg) kg](0.250 m) 2 ] kg m 2 L i L f putty m putty v putty I f ω f kg m 2 /s (0.503 kg m 2 )ω f ω f ad/s (b) The kinetic enegy befoe the collision is puely due to the putty: KE i 2 m puttyv 2 putty 2 (0.050 kg)(3.00 m/s) J The kinetic enegy just afte the collision is KE f 2 Iω2 2 (0.503 kg m2 )( ad/s) J Thus, the atio of the kinetic enegies just afte and just befoe the collision is KE f /KE i J/0.225 J In othe wods, about 99.4% of the initial kinetic enegy is lost. (c) The od will momentaily stop when it gets to a point whee all its initial kinetic enegy has been tansfomed into gavitational potential enegy. We can choose any height fo h 0, so let s choose the initial hoizontal height of the two balls. When the balls ae not hoizontal, one will have a gavitational potential enegy of +mgh, while the othe will have a gavitational PE of mg( h) mgh. The sum of these two is obviously zeo, just as it was befoe the putty hit. This will always be tue  the gavitational potential enegies of the two balls will always add to zeo. Theefoe, it is the mass and height of the putty that will detemine whee the od stops. The system will acceleate as it swings downwads, attaining its maximum speed when the putty is at the bottom. The maximum height attained by the putty will be given by m putty gh max 2 Iω2 h max ( m putty g )( 2 Iω2 ) ( (0.050 kg)(9.8 m/s 2 ( J) ) h max m 2.86 mm
9 h max Theefoe, the angle though which the system swings befoe it momentaily stops is 80 plus acsin( /0.25) L a 3 a 4 a 2 a (3b) Fou identical bicks of length L ae stacked on top of one anothe as shown, such that pat of each extends beyond the one beneath. Find in tems of L the maximum values of the ovehangs a, a 2, a 3, a 4, and h such that the stack is in equilibium. It may seem odd, but the best way to go about this is to stat at the top and wok ou way down. Looking at the top bick, in ode fo it to be in equilibium, thee must be no net foce and no net toque acting on it. If we imagine stating with the top bick so that its ight side is flush with the ight side of the bick below, then slowly moving it ove to the ight, it can emain in equilibium as long as its cente of mass is located ove the bick below  as soon as its cente of mass moves out beyond the edge of the bick below, the top bick will feel a net toque and fall ove. Theefoe, the maximum distance fo a is a L/2. Now look at the top two bicks. They will fall if thei cente of mass lies beyond the edge of the next bick down. So let s find the cente of mass of the two top bicks. It is convenient to choose the ight edge of the second bick down as x 0. In this case, the cente of mass of the two bicks will be x 2,com m(0)+m( L/2) L/4 2m whee m is the mass of each of the bicks. The cente of mass of the top two bicks is a distance L/4 fom the ight edge of the second bick down so, and this is the futhest ove the bicks can be and be in equilibium: a 2 L/4. Now continue with the top thee bicks, using the ight edge of the thid bick down as x 0. The cente of mass of the thid bick is located at x L/2 elative to its ighthand edge. The second bick s cente of mass is located a distance L/4 to the left of the ighthand edge of the thid bick, o at x L/4. The cente of mass of the top bick is located a distance a 2 L/4 to the ight of the ighthand edge of the thid bick, o at x L/4. Theefoe: x 3,com m( L/4)+m(L/4)+m( L/2) L/6 3m The cente of mass of the top thee bicks is a distance L/6 fom the ight edge of the thid bick down so, and this is the futhest ove the bicks can be and still be in equilibium: a 3 L/6. Caefully caying out the same pocedue fo all fou bicks yields a 4 /8. Thus, the maximum value fo h is h a + a 2 + a 3 + a 4 25L/24! Note that it is not necessay to calculate the COM positions fo each bick sepaately as I did above. Once you have the COM of the top two bicks, you can then subsequently teat them as a single object with mass 2m sitting at the location of the COM, and so foth. h
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