Solutions for Physics 1201 Course Review (Problems 10 through 17)

Transcription

1 Solutions fo Physics 101 Couse Review (Poblems 10 though 17) 10) a) Hee, thee is no fiction between any of the sufaces in contact, so we do not need to be concened with the nomal foces on the blocks. Since the only hoizontal foce on m 1 would be due to fiction, in this case the uppe block is not acceleated; thus, a 1 = 0. In the absence of fiction, the only hoizontal foce acting on m is the applied foce F = 6 N. ; hence, the acceleation of the lowe block is a = m F = 6 N. 4.6 m. 1.3 kg.. The lowe block will ultimately be pulled out fom unde the uppe one. b) Since the contact sufaces between the blocks and between the lowe block and the tabletop ae hoizontal, the nomal foces on the blocks ae simply N 1 = m 1 g and N = ( m 1 + m ) g. The static fictional foce acting between the blocks is then f s µ s N 1 = µ s m 1 g. So the two blocks will move togethe as a unit povided that this limit applies. (Since the tabletop is fictionless, N is not impotant to know and contact with the uppe block is the only souce of fiction.) The hoizontal foce equation fo the lowe block is F f s = m a, while that fo the uppe block is f s = m 1 a 1. If the two blocks ae to stay togethe, thei acceleations must match, implying that a 1 = f s m 1 = a = F f s possible value fo the static fictional foce f smax applied foce is given by m. When we apply the maximum = µ s N 1, we find that the limit fo the µ s m 1 g m 1 = F max µ s m 1 g m F max = µ s (m 1 + m ) g 0.45 ( kg.) (9.81 m. ) 8.83 N.

2 Since F max > 6 N., a hoizontal applied foce at this level still pemits the blocks to stay togethe. Because they move as a unit, thei mutual acceleation is F 6 N. a = m 1 + m = 3.00 m. ( kg.). Note that this means that the static fictional foce hee is f s = m 1 a = ( 0.7 kg. ) ( 3.00 m./ ) =.10 N., which is cetainly less than f smax = µ s m 1 g = 0.45 (0.7 kg.) (9.81 m. ) 3.09 N. As a check, we find that a = F f s N m. m 1.3 kg. = a. Fo an applied hoizontal foce F = 10 N. > F max, the blocks will not move togethe and the kinetic fictional foce f k = µ k m 1 g must be used in ou analysis. Upon eplacing f s with f k in the hoizontal foce equations fo each block, we now have a 1 ' = f k = µ k m 1 g = µ k g 0.3 (9.81 m. m. ) 3.14 m 1 m 1 and a ' = F f k m 10.0 N. 0.3 (0.7 kg.) (9.81 m. ) 1.3 kg N. 1.3 kg m.. The lowe block will be pulled out fom unde the uppe one, but hee the uppe block will move fowad a bit fist, which was not the case in the situation we teated in pat (a) above. c) Nothing else about this system is changed if we apply the foce to the lowe block with a sping, instead of a cod. In ode fo the two blocks to continue moving togethe, the foce applied must not exceed F max 8.83 N. Hooke s Law tells us that the estoing foce of a sping is given by F sp = k Δx, whee Δx is the displacement of the sping fom equilibium. A maximum allowed foce of 8.83 N. equies that the maximum pemitted displacement fo this sping be (Δ x ) max = F max 8.83 N m. This will set the limit on the k 7 N./m. amplitude fo the pai of blocks oscillating at the end of this sping. 11) a) The thee hoizontal foces acting on the ca ae the applied foce fom the engine, F A, the kinetic fictional foce between the ties and the oad, f k, and the dag foce fom the ai, F D. Since the vehicle is on a level oad hee, the nomal foce on it fom the oad suface is N = Mg, so the kinetic fictional foce is f k = µ k Mg.

3 With the ca taveling at constant speed, the net hoizontal foce on the ca is F netx = F A + f k + F D = 0 F netx = F A f k F D = 0. The net powe applied to the ca will then also bezeo: P net = F netx v = F A v + f k v + = F A v f k v F D v = 0. F D v = F A v cos 0 o + f k v cos 180 o + F D v cos 180 o We ae told that the powe being povided by the engine is P A = F A v = 95,000 W. At a speed of v = 65 mi m. 1 h m. h. mi. 3600, the potion of this powe that is being used to ovecome ai dag is P D = F D v = F A v f k v = P A µ k Mgv = 95,000 W 0.05 (900 kg.) (9.81 m. m. ) (9.05 ) 95, W 88,600 W. To maintain the ca at this speed, to ovecome ai dag. 88,600 W 95,000 W of the engine s powe is being used b) If the dag foce is popotional to the squae of the vehicle s speed, then at 80 F D ' mph, the dag is = v' 80 mph = F D v times stonge than it is at 65 mph 65 mph. The powe necessay to ovecome ai dag at this highe speed, howeve, is P D ' = F D 'v' P D F D v = v' v v' 3 v = 80 mph times geate than at 65 mph. This gives 65 mph us P D P D ,600 W 165,00 W. The total powe the engine must now supply is P A = P D + P f 165, W 171,600 W ( 30 hp ). c) With the ca now moving up an incline, the nomal foce fom the oad suface becomes N = Mg cos θ, making the kinetic fictional foce f k = µ k Mg cos θ. In addition to the foces aleady descibed, the weight foce on the vehicle now also has a component paallel to the incline, W = Mg sin θ.

4 The net foce on the ca is now F net = F A + F D + f k ' + W = 0, so the net powe becomes P net = F A v F D v ( µ k Mg cos θ ) v ( Mg sin θ ) v = 0. Since the ca is on a 1% gade, tan θ = 0.1 sin θ , cos θ We have seen that the powe exeted by ai dag is P D 88,600 W at 9.05 m./ (65 mph) and that it is popotional to v 3, so we can wite P D = 88,600 v 3 W. The powe which must be 9.05 deliveed by the engine in ode fo the vehicle to maintain a speed v on the 1% upwad incline is then P A = F A v = P D + (µ k Mg cos θ ) v + (Mg sin θ ) v 88, 600 v 3 + (0.05) (900 kg.) (9.81 m. m. ) (0.999) v + (900 kg.) ( ) (0.1191) v v v v v v. We ae going to keep the powe fom the engine at the same level it had in pat (b) above, so we need to solve fo v the equation v v = 171,600. It is not vey convenient to solve this last equation diectly (we can, of couse, use gaphing softwae, as will be discussed below), but we don t need to esot to tial-and-eo completely eithe. We know that this amount of powe on level gound allows the ca to tavel at 80 mph ( 35.8 m./ ), so we might expect that on a 1% climb, the ca might move, say, 10% o so slowe. Let s make a fist guess that the solution is v = 3 m./ Since the cubic tem in the equation, v 3, changes much moe apidly than the linea tem, 171v, we ll simply set this tem to fo the pesent and solve the following fo v : v = v = 171, v 3 = 130,900 v 3 36,00 v 33.1 m., which is petty close to ou initial guess. If we adjust the linea tem to accommodate this new value, we find v = v = 171,600 v 33.0 m.. Ou solution has stabilized to one decimal place, so we may stop hee. A moe pecise solution, using gaphing softwae to find the x-intecepts fo the function v v 171,600, gives us v 3.98 m./, so ou esult is acceptable.

5 As a check, we can calculate the individual powe tems: powe exeted against dag: P A ,900 W powe exeted against fiction: P f W powe exeted against gavity: P W ,700 W total powe to be deliveed by engine: 171,800 W, ageeing with the intended total value to about 0.1%. d) Without the applied foce fom the engine, and assuming that thee is no intenal fiction in the dive tain of the ca (a bit unealistic), the net foce on the vehicle acting paallel to the downwad incline is F net ' = W + F D + f k ' = 0 F net ' = W F D f k ' = 0, making the net powe P net = ( Mg sin θ ) v F D v ( µ k Mg cos θ ) v = 0. Each of these tems has the same value as it did in pat (c) above, giving us 105 v v v 83.3v v 3 = 0, which is much easie to solve fo v than the powe equation we found in the pevious pat. We can facto this as v ( v ) = 0, so eithe v = 0 (the uninteesting solution, since the powe tems ae of couse all zeo when the ca is paked) o v = 0 v 30.3 v 15. m./ (34 mph). In a eal vehicle, intenal fiction would make the downhill coasting speed somewhat lowe than this. 1) a) The cental event in this physical aangement is the elastic collision between the two blocks. The combination of linea momentum consevation with consevation of kinetic enegy leads to the esult that the elative velocity of the masses befoe the collision is the opposite of the elative velocity aftewads, that is, v 1 v = ( v 1 v ), which may be eaanged moe conveniently as v 1 + v 1 = v + v. Fo ou collision, this gives us v + v = 0 + V. This povides a elationship among the velocities, but tells us nothing else. We will need to uncove the est by examining the enegies of the blocks.

6 One way to analyze this would be to pefom a detailed study of all the enegy changes fo the blocks at each significant event. Howeve, we can save ouselves some touble by noticing a featue of the motion of m 1. Upon leaving the sping launche at the top of its amp, m 1 has a kinetic enegy equal to the potential enegy U sp that had been stoed in the sping. When it eaches the bottom of the amp to collide with m, m 1 has a kinetic enegy U sp plus the gavitational potential enegy it acquied by descending 35 cm., since this amp is fictionless (so the angle of this incline o any cuvatue that it has become ielevant). Afte the collision, m 1 has just enough kinetic enegy to come to est as it eaches its launche; hence, it has lost an amount of enegy U sp. But because the collision with m is elastic, this must be the amount of enegy tansfeed to m. We can thus conclude that afte the collision, has a kinetic enegy of m ½ m V = U sp = ½ k 1 ( Δx ) 1 ½ ( 100 N./m. ) ( 0.11 m. ) J. Retuning to m1, we have said that its kinetic enegy befoe the collision was ½ m 1 v = U sp + m 1 gh = ( 0.1 kg. ) ( 9.81 m./ ) ( 0.35 m. ) J J. and afte the collision, the kinetic enegy became ½ m 1 v = m 1 gh J. So we can calculate that " v = (U sp + m 1 gh ) % $ # m ' 1 & 1/ v' = " # m 1 gh 1/ $ ' + # J.. m ( m 1 has evesed diection, so v % & 1 ( ) * " - 0 * ".6, 0.1 kg. / m. becomes negative). The elative-velocity equation then tells us that 1/ ( *, + ) J. 0.1 kg. - /. 1/ ( 4.36 m. and v + v = 0 + V V (.6 ) 1.73 m./* * ounding fom a lage numbe of significant figues Since we know the kinetic enegy of m afte the collision, we can now solve fo the value

7 m = K ' V = U sp V J. (1.74 m. ) 0.40 kg. (This can also be detemined using linea momentum consevation.) While much of the infomation in this pat of the Poblem may seem to have been extaneous, it will be quite useful fo the next potion. b) In this new situation, m comes to est just as it makes contact with the sping bumpe on its amp. So afte the collision, m has only enough kinetic enegy to climp the amp and ovecome its fiction. The wok m must pefom against gavity is m g ( h 1 + h ) = ( 0.40 kg. ) ( 9.81 m./ ) ( m. ) 0.96 J. The length of incline which exets fictional foce on m is given by sin 9 o = h L L = h sin 9 o 0.05 m m. The wok which m must pefom against kinetic fiction is then µ k N L = µ k ( m g cos 9º ) L ( 0.17 ) ( 0.40 kg. ) ( 9.81 m./ ) ( ) ( 0.30 m. ) 0.1 J. Thus, m must have depated fom the collision with m 1 with a kinetic enegy of J., which means that its velocity afte the collision in this situation J. 1/ was V 1.59 m kg.. We now need to find the speed of m 1 befoe this collision, in ode to detemine how it was launched. The elative-velocity equation tells us v + v = 0 + V 1.59 m./ As we ae eally only inteested in v, we will solve this equation fo v and eliminate it fom the linea momentum consevation equation: m 1 v + m 0 = m 1 v + m V ( kg. ) v ( kg. ) ( 1.59 v m./ ) + ( 0.40 kg. ) ( 1.59 m./ ) v kg. m./ 0.00 v v 4.00 m./ So the kinetic enegy of m 1 befoe the collision was ½ m 1 v ½ ( kg. ) ( 4.00 m./ ) J.

8 As in pat (a), this enegy is the sum of the sping launche s stoed potential enegy and the gavitational potential enegy eleased as m 1 dops by 35 cm. Since this latte amount has aleady been found to be J., the launche must have held a potential enegy of J J. Using the fomula fo the (ideal) sping potential enegy in the launche, the initial compession of the sping in this situation would have been 1/ U sp ' = 1 k J. 1 (Δx') J. Δx' 100 N m. o 9.6 cm. m. 13) a) In a simple pendulum, it is idealized that all of the pendulum s mass is concentated in a mass-point at one end. Fo a ealistic o physical pendulum, howeve, we must take into account that the mass is distibuted ove the body of the pendulum. If we deflect the pendulum fom its vetical equilibium by an angle θ, the gavitational toque acting to estoe it to equilibium will be in the opposite diection with a magnitude of CM Mg sin θ, whee CM is the distance fom the pivot point of the pendulum to its cente-of-mass. Since the pendulum tuns igidly, the toque is τ = I tot α = I tot d θ = Mg dt CM sin θ. If the angula displacement fom equilibium is small ( θ < 0.3 adians ), then sin θ θ (in adians) and we can wite I tot d θ dt Mg CM θ d θ dt + Mg CM I tot θ 0, which is a diffeential equation fo simple hamonic motion. Fo this pendulum, ω = Mg CM I tot, so it will oscillate with a peiod of T = π ω = π I tot Mg CM. If we apply this fomula to the simple pendulum, I tot = M l and CM = l, since all the mass is at one end; we then obtain the familia esult T = π M l Mgl = π l g.

9 Fo this longcase clock (sometimes called a gandfathe s clock ) pendulum, the od has mass m and length L and is pivoted fom one end, with the bob being a disk of mass M and adius R centeed at a distance d below the pivot. The distance of the cente-ofmass fom the pivot is then CM = m ( 1 L ) + M d m + M (.000 kg.) ( m.) kg. d kg d m. The total moment of inetia of the pendulum is that of a od pivoting about one end plus a disk evolving about a point outside its cente-of-mass. The moment of inetia fo the unifom od is then 1 3 ml. By the paallel-axis theoem, the moment inetia of the heavy pendulum bob is ½ MR + Md. Thus, the total moment of inetia of the pendulum is given by I tot = 1 3 ml + 1 MR + M d 1 3 (.000 kg.) (1.100 m.) + 1 (5.000 kg.) (0.080 m.) + (5.000 kg.) d d kg. m. We want to adjust this pendulum so that the peiod is vey pecisely equal to.000 seconds at sea level on the Eath s equato. So we will need to move the bob to the position fo which T = π I tot M tot g CM M tot is the total mass of the pendulum d π kg. m. ( kg.) ( m. ) ( d m.) 1/ d d.000 π d d 5d d d 1.04 m. only positive solution b) Themal contaction has caused the linea dimensions of this pendulum to be educed by 1/1000 ( 0.1% ). The effect on the bob itself is insignificant, but the effect on the length of the od and the distance of the bob fom the pivot both must be consideed. The length of the od is now L = L m. This bings the bob close to the pivot by a popotional amount, so d = d m m. The expession fo the distance fom the pivot to the cente-of-mass becomes CM ' = m ( 1 L') + M d' (.000 kg.) ( m.) kg m m. m + M kg.

10 and that fo the total moment of inteia is I tot ' = 1 3 ml' + 1 MR + M d' a 0.1% eduction of R has negligible effect on the middle tem 1 3 (.000 kg.) ( m.) + 1 (5.000 kg.) (0.080 m.) + (5.000 kg.) (1.041 m.) kg. m. This gives the peiod of the pendulum in this situation as kg. m. T ' π (7.000 kg.) ( m. ) ( m.) 1/ In a week of 7 days /day = 604,800 seconds, the pendulum will make 604,800 T ' 604, ,553 oscillations, coesponding to 30, ,106 seconds egisteed by the clock. The clock would then be 605, , seconds ( 5.09 minutes ) fast. To estoe the pendulum to keeping accuate time, the bob would need to be moved to a distance D fom the pivot given by D π kg. m. (7.000 kg.) ( m. ) ( D m.) 1/.000 5D D D m. only positive solution The bob should theefoe be moved downwad away fom the pivot by D d m m. ( 1. mm.). To make such fine adjustments to the position of the bob, it was often fitted with a calibated venie scew, indicated the numbe of minutes pe day of coection. It is this sensitivity to envionmental changes that led people to see non-mechanical methods of timekeeping.

11 14) We know that this sping-mass system (ideally) obeys the simple hamonic motion diffeential equation, m d (Δ x ) dt + k(δ x ) = 0, fo which the solution fo the displacement fom equilibium is Δx = A sin( ωt + φ ), with ω = k. At a moment we m will call t = 0, the displacement is Δx = 0.37 m. (the block stats, say, to the left of the oigin), the velocity is v =.64 m./ (the block is moving to the ight), and the acceleation is a = 1.90 m./ (the foce on the block is to the ight). The sping constant is given as k = 64 N/m. If we diffeentiate the displacement equation with espect to time, we obtain v = dx dt = A ω cos (ωt + φ ), a = dv dt = A ω ω [ sin (ωt + φ ) ] = A ω sin (ωt + φ ). At time t = 0, this gives us ( Δx ) 0 = A sin ( ω 0 + φ ) = A sin φ = 0.37 m., v 0 = Aω cos ( ω 0 + φ ) = Aω cos φ =.64 m./, and a 0 = Aω sin ( ω 0 + φ ) = Aω sin φ = 1.90 m./. We can now calculate a 0 = A ω sin φ (Δ x ) 0 A sin φ = ω 1.90 m m. ω ad. ω.66 ad. ; peiod of oscillato: T = π ω π ad..66 ad..77 ; the mass of the block is given by ω = k m m = k ω ω (Δ x ) 0 v 0 = ω A sin φ Aω cos φ ad. (.66 = tan φ ) ( 0.37 m.).64 m. 64 N. m ad. 1.5 kg. ; Since ( Δx ) 0 < 0, so sin φ < 0 also, and v 0 > 0, so cos φ > 0 as well. This tells us that φ is in quadant IV, and thus that the phase angle is given simply by the calculato value to φ tan 1 ( ) ad. (o ad.). Finally, we can detemine the amplitude of oscillation fom A = (Δ x ) 0 sin φ = A sin φ sin φ 0.37 m. sin ( 0.308) 0.37 m m. The equation fo the displacement fom equilibium as a function of time is theefoe Δx = 1.18 sin (.66 t ) m. The total mechanical enegy of this spingmass oscillato is given by E = ½ k A ½ ( 64 N./m. ) ( 1.18 m. ) J.

12 15) Sphee A can be teated as a sphee of adius R and density ρ, with an added, embedded sphee with adius ½ R and a density of ρ (giving that egion a total density of 3ρ ). Sphee B has a cavity in it of the same size, so we will wok with it as a sphee of adius R and density ρ, with the added sphee having a density of ρ (poducing a egion with density zeo). We will then wok out the popeties of these two sphee in paallel calculations. a) total mass This fist calculation will make it clea why we choose to beak up these asymmetic sphees in the peculia way descibed above. We have the lage sphee with adius R and density ρ, which has a mass of M = ρ 4π 3 R 3 ; a small excess-density sphee with adius ½ R, density ρ, and a mass of m A = ρ 4π 3 1 R 3 = ρ 4π 3 8 R 3 = ρ π 3 R 3 1 o 4 M ; and a small negative- 3 density sphee with adius ½ R, density ρ, and a mass of m B = ρ 4π 3 1 R = ρ 4π 3 8 R 3 = ρ π 6 R 3 o 1 8 M. We can now put these esults togethe to find 4π the total mass of sphee A as M A = M + m A = ρ 3 + π R 3 = ρ 5π 3 3 R 3 4π (so M = 4m A and M A = 5m A ) and M B = M + m B = ρ 3 π R 3 = ρ 7π 6 6 R 3 (so M = 8m B and M B = 7m B ). These additional elationships will be helpful in the pats of the poblem to follow. b) moment of inetia about symmety axis The axis which passes though both the cente of the lage sphee and the cente of the small sphee may be egaded as the symmety axis fo both sphee A and sphee B.

13 Since the centes-of-mass of the lage and small sphee lie on this axis, the moment of inetia of the assemblage about this axis will simply be the sum of the cente-of-mass moments of inetia fo the unifom-density sphees. Thus, fo sphee A, we have I A1 = 5 M R + 5 m A ( 1 R ) = 5 (4m A ) R + 5 m A ( 1 R ) while fo sphee B, we find = ( ) m A R = m A R = M A R = 0.34 M A R, I B1 = 5 M R + 5 m B ( 1 R ) = 5 ( 8m B ) R + 5 m B ( 1 R ) = ( ) m B R = m B R = M B R 0.443M B R. We see that sphee A, having a geate density of mass concentated nea this otation axis, has a lowe coefficient of otational inetia ( 0.34 ) than does sphee B ( ), whee the cavity causes moe of its mass to be distibuted fathe away fom this otation axis. c) moment of inetia about pependicula axis though geometical cente Hee, we will use a otational axis though the cente of the lage sphee and pependicula to the line we ve called the symmety axis. We can thus still use the cente-of-mass moment of inetia fo the lage sphee, but we need to invoke the paallel-axis theoem to wok out the alteed moment fo the small sphee. This equies the addition of a tem which is the mass of the object times the squae of the distance fom its cente-of-mass to the otation axis. The new esult fo sphee A is

14 I A = 5 M R + 5 m A ( 1 R ) + m A ( R 3 ) "paallel axis"tem = 5 (4m A ) R m A ( 1 R ) = ( ) m AR = 39 0 m AR = M AR = 0.39 M A R, while fo sphee B, we find I B1 = 5 M R + 5 m B ( 1 R ) + m B ( R 3 ) "paallel axis"tem = 5 ( 8m B ) R m B ( 1 R ) = ( ) m B R = 57 0 m B R = M B R M B R. Relative to this otation axis, thee is fa less diffeence between the coefficients of otational inetia of sphees A and B. d) location of cente of mass When an object has a unifom density, its cente- ofmass will be in the position of its geometical cente. Since we ae teating sphees A and B as composite objects, made up of two individual sphees of constant density, the centes fo A and B will lie along the symmety axis connecting the centes of the lage sphee and the small sphee. So we can identify the symmety axis as the x-axis and make the cente of the lage sphee x = 0 ; this places the cente of the small sphee at x = ½ R. This pemits us to calculate the position of the cente-of-mass of sphee A as x A = and that of sphee B as 0 M + ( 1 R ) m A M + m A = 1 m A M A R = m A 5m A R = 1 10 R, x B = 0 M + ( 1 R ) m B M + m B = 1 m B M B R = m B ( 7m B ) R = 1 14 R. e) moment of inetia about pependicula axis though cente-of-mass We again choose otation axes pependicula to the symmety axes of ou sphees A and B, as we did in pat (c), but now these axes pass though the centes-of-mass we have just detemined.

15 As befoe, we apply the paallel axis theoem to find this moment of inetia fo sphee A to be I A3 = [ 5 M R + M x 13 A "paallel axis"tem ] + [ 5 m A ( 1 R ) + m A ( 1 R x A ) ] "paallel axis"tem = 5 (4m A ) R + (4m A ) ( 1 10 R ) + 5 m A 1 4 R + m A ( 1 R 1 10 R ) ] = ( ) m A R = m A R = M A R = 0.38 M A R, and fo sphee B, I B3 = [ 5 M R + M x 13 B "paallel axis"tem ] + [ 5 m B ( 1 R ) + m B ( 1 R x B ) ] "paallel axis"tem = 5 ( 8m B ) R + ( 8m B ) ( 1 14 R ) + 5 m B 1 4 R + m B ( 1 R [ 1 14 R]) ] = ( ) m BR = m BR = M BR 0.40M B R. 16) a) Since the only hoizontal foce acting on the block is static fiction with the tuntable suface, this must be the souce of the centipetal foce that keeps the block tuning about the otation axis of the tuntable. So we have F c = mv R = mω R = f s µ s N. Because the suface is hoizontal, N = mg ; hence, ω R µ s g.

16 The tuntable is otating at a fequency of f = 45 ev./min., so its angula speed is ω = π f = π ad. ev. 45 ev. 1 min ad. min. 60. With the block placed at R = 0.06 m. fom the otation axis, the coefficient of static fiction necessay to hold the block in place is given by µ s ω R (4.71 ad. g ) 0.06 m m b) In ode fo static fiction to hold the block in place at the edge of the tuntable (4.71 ad. platte, whee R = 0.15 m., we now equie µ s ) 0.15 m m c) Fo a fixed coefficient of static fiction, we can eaange the foce inequality to find the maximum distance at which an object can be situated fom the otation axis and still be held in place by static fiction, R µ s g ω. With µ s maximum distance fom the otation axis would be given by = 0.3 fo this tuntable, the R m. (4.71 ad. ) 0.10 m. (10. cm.). d) While the tuntable s angula speed is inceasing, thee will be both a centipetal acceleation a C = v T R = ω R and a tangential acceleation a T = αr, whee v T is the instantaneous tangential velocity of the tuntable at adius R and α is the angula acceleation; both of the these acceleations ae in the plane of the tuntable suface. Since centipetal acceleation points along a adius of the cicle of motion and the tangential acceleation points along a tangent to that cicle (hence the name), a C and a T ae pependicula. So the total acceleation of a point on the tuntable suface is the magnitude of the esultant acceleation vecto, a tot = a C + a T. The static fiction must be able to povide this total acceleation to keep the block moving along with the tuntable, so we equie a tot µ s g. The tuntable acceleates fom est ( ω 0 = 0 ) to a final angula speed ω f = 4.71 ad./ (45 pm) at a unifom ate α in a time T, o α = ω f ω 0 T = ω f T.

17 The tangential acceleation is then a T = αr = ω f R. The geatest total acceleation T will theefoe occu as the tuntable eaches its final angula speed, at which time a tot = (ω f R ) + ( ω 1/ f T R ). So the equiement fo static fiction to hold the block in place on the acceleating tuntable gives us (ω f R ) + ( ω f T R ) 1/ µ s g ( ω f T R ) (µ s g ) (ω f R ) ω f R T (µ s g ) (ω f R ) ω f R (µ s g ) (ω f R ) T T ( m. ) (4.71 ad. ) (0.06 m.) ([ 4.71 ad. ] 0.06 m. ) T Thus, the minimum time in which the tuntable may acceleate fom est to its opeating angula ate of 45 pm without causing the block to slip is seconds fo unifom angula acceleation. (Note that the mass of the block does not ente into any of the esults in this Poblem.) 17)

18 a) The hanging sping scale eads weight by egisteing the tension T in its sping. With the scale connected to the block by a cod, this tension can be detemined fom the fee-body diagam fo the block fom T + B Mg = 0 T = Mg B, whee B is the buoyancy foce, equal to the weight of a volume of wate which is the same as the volume of the block, so B = ρ w V g = ρ w M ρ g = ρ w b ρ Mg, with ρ b w being the density of wate and ρ b being the density of the block. So the eading on the sping scale is T = Mg ρ w ρ Mg = 1 ρ w b ρ Mg. b The platfom scale eads weight by egisteing the nomal foce N, which is povided by an intenal mechanism of the scale (fequently, a sping system). This foce can be detemined fom the fee-body diagam fo the scale as N M w g M B g B = 0 N = ( M w + M B )g + B, whee M w is the mass of wate and M B is the mass of the beake. The buoyancy foce entes into this foce equation because the fluid pessue which pushes on the block also pushes on the walls of the beake. Fo ou poblem, the block has mass M = kg., the beake, M B = kg., and the wate, M w = 1.36 kg. The eadings fo aluminum and magnesium blocks ae density ( gm./cm. 3 ) buoyancy foce, B aluminum gm. cm gm. cm kg m N magnesium gm. cm gm. cm kg m N sping scale eading, T = Mg B aluminum (0.84 kg m. ) N magnesium (0.84 kg m. ) N platfom scale eading, N = ( M w + M B )g + B aluminum ([ kg.] 9.81 m. ) N magnesium ([ kg.] 9.81 m. ) N

19 b) As this is a physical aangement in static equilibium, the net foce on the od is zeo; thus, T L + T R + B Mg = 0. We ae told that the od has a length of 75 cm. and a coss-sectional aea of 8.0 cm., so its volume is 600 cm. 3 The buoyancy foce is equal to the weight of an equal volume of wate, hence B = ρ w V g = (1 gm. cm. 3 ) 1 kg. (600 cm. 3 ) (9.81 m gm. ) 5.89 N. The foce equation then gives us T L + T R = Mg B = (.8 kg. ) ( 9.81 m./ ) N. We can also analyze the toques acting on the od. If we place the pivot point at the left end of the od, then we can neglect the tension T L, since it will have a momentam of zeo length. As fo the othe toques, toque due to weight of od: τ W = ( 5 cm. ) Mg sin 90º toque due to buoyancy foce: τ B = + ( ½ 75 cm. ) B sin 90º toque due to tension in ight-hand cod: τ TR = + ( 75 cm. ) T sin 90º

20 Fo this static equilibium, τ W + τ B + τ TR = 5 Mg B + 75 T = 0 T R = (5 cm.) (.8 kg.) (9.81 m. ) (37.5 cm.) (5.89 N ) 75 cm. 6.1 N. Fom the foce equation above, we find that T L = 1.58 T R N. Wee the ight-hand cod disconnected o emoved, its tension contibution would become zeo and ou foce equation would now be T L = Mg B 1.58 N. The cente-of-mass of the od would then need to be in a diffeent location in ode fo the od to emain static and level. Let us call x this new distance of the cente of mass fom the left end of the od. The toque equation is then τ W + τ B = x Mg B = 0 x = (37.5 cm.) (5.89 N ) (.8 kg.) (9.81 m. ) 8.04 cm. -- G.Ruffa oiginal notes developed duing evised: Septembe 010