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

Solutions for Physics 1201 Course Review (Problems 10 through 17)10)a) Here, there is no friction between any of the surfaces in contact, so we do notneed to be concerned with the normal forces on the blocks. Since the only horizontal forceon m 1would be due to friction, in this case the upper block is not accelerated; thus, a 1=0 . In the absence of friction, the only horizontal force acting on m 2is the applied force F= 6 N. ; hence, the acceleration of the lower block is a 2 =m F = 6 N. ≈ 4.62 m.2 1.3 kg. sec. 2 .The lower block will ultimately be pulled out from under the upper one.b) Since the contact surfaces between the blocks and between the lower block andthe tabletop are horizontal, the normal forces € on the blocks are simply N 1= m 1g and N 2= ( m 1+ m 2) g . The static frictional force acting between the blocks is thenf s≤ µ sN 1= µ sm 1g . So the two blocks will move together as a unit provided that this limitapplies. (Since the tabletop is frictionless, N 2is not important to know and contact withthe upper block is the only source of friction.)The horizontal force equation for the lower block is F – f s= m 2a 2, while that forthe upper block is f s= m 1a 1. If the two blocks are to stay together, their accelerationsmust match, implying thata 1 = f sm 1= a 2 = F − f spossible value for the static frictional forceapplied force is given byµ s m 1 gm 1f smaxm 2€= F max − µ s m 1 g⇒ Fm € max = µ s ⋅ (m 1 + m 2 )⋅ g2. When we apply the maximum= µ s N 1 , we find that the limit for the≈ 0.45 ⋅ (0.7 + 1.3 kg.)⋅ (9.81 m.sec. 2 ) ≈ 8.83 N. €€

€€Since F max> 6 N. , a horizontal applied force at this level still permits the blocks tostay together. Because they move as a unit, their mutual acceleration isF6 N.a =m 1+ m=≈ 3.00 m.2 (0.7 + 1.3 kg.) sec.2 . Note that this means that the staticfrictional force here is f s= m 1a = ( 0.7 kg. ) ( 3.00 m./sec. 2 ) = 2.10 N. , which is certainlyless than f smax= µ s m 1 g = 0.45 ⋅ (0.7 kg.)⋅ (9.81 m.sec. ) ≈ 3.09 N. 2 As a check, we findthat a 2 = F − f s 6.00 − 2.10 N.≈ ≈ 3.00 m.m 2 1.3 kg.sec. = a 2 .€For an applied horizontal force F = 10 N. > F max, the blocks will not move togetherand the kinetic frictional force f k= µ km 1g must be used in our analysis. Upon replacingf swith f kin the horizontal force equations for each block, we now havea 1 ' =f k= µ k m 1 g = µ k g ≈ 0.32 ⋅ (9.81 m.m.) ≈ 3.14m 1 m 1sec.2sec. 2 and€a 2 ' = F − f km 2≈10.0 N. − 0.32 ⋅ (0.7 kg.)⋅ (9.81 m.sec. 2 )1.3 kg.€≈10.0 − 2.20 N.1.3 kg.≈ 6.00 m.sec. 2 .The lower block will be pulled out from under the upper one, but here the upper block willmove forward a bit first, which was not the case in the situation we treated in part (a) above.€c) Nothing else about this system is changed if we apply the force to the lower blockwith a spring, instead of a cord. In order for the two blocks to continue moving together, theforce applied must not exceed F max≈ 8.83 N. Hooke’s Law tells us that the restoring force of aspring is given by F sp= k · Δx , where Δx is the displacement of the spring from equilibrium.A maximum allowed force of 8.83 N. requires that the maximum permitted displacement forthis spring be (Δ x ) max = F max 8.83 N.≈ ≈ 0.123 m. This will set the limit on thek 72 N./m.amplitude for the pair of blocks oscillating at the end of this spring.11)€€a)rThe three horizontal forces acting on the car are the applied forcerfrom theengine, F A , the kinetic frictional force between the tires and the road, f k , and the dragforce from the air,€rF D . Since the vehicle is on a level road here, the normal force on itfrom the road surface is N = Mg , so the kinetic frictional force is fk= µ kMg .€

€€With the car traveling at constant speed, the net horizontal force on the car isrF netx = F rA + f rk + F rD = 0 ⇒ F netx = F A − f k − F D = 0 . The net power applied tothe car will then also bezero:P net = F rnetx ⋅ v r =rF A ⋅ r v + rf k ⋅ r v += F A v − f k v − F D v = 0 .rF D ⋅ r v =rF A v cos 0 o+ rf k v cos 180 o + rF D v cos 180 oWe are told that the power being provided by the engine is P A= F Av = 95,000 W . At€ a speed of v = 65 mi. ⋅ 1609 m. 1 hr.⋅ ≈ 29.05 m.hr. mi. 3600 sec. sec. , the portion of this power that isbeing used to overcome air drag is€€P D = F D v = F A v − f k v = P A − µ k Mgv= 95,000 W − 0.025 ⋅ (900 kg.) (9.81 m. m.) (29.05sec.2 sec. )≈ 95,000 − 6410 W ≈ 88,600 W .€To maintain the car at this speed,to overcome € air drag.88,600 W95,000 W ≈ 0.933 of the engine’s power is being used€b) If the drag force is proportional to the square of the vehicle’s speed, then at 80F € D'mph, the drag is = v'22⎛⎛ 80 mph⎞⎞=F D v 2 ⎜⎜ ⎟⎟ ≈ 1.515 times stronger than it is at⎝⎝ 65 mph⎠⎠65 mph. The power necessary to overcome air drag at this higher speed, however, isP D '= F D 'v'P DF€ Dv = ⎛⎛ v'2 ⎞⎞⎜⎜⎝⎝ v 2 ⎟⎟ ⋅ v'3⎠⎠ v = ⎛⎛ 80 mph⎞⎞⎜⎜ ⎟⎟ ≈ 1.864 times greater than at 65 mph. This gives⎝⎝ 65 mph⎠⎠us P D’ ≈ 1.864 P D≈ 1.864 · 88,600 W ≈ 165,200 W . The total power the engine mustnow supply is P A’ = P D’ + P f≈ 165,200 + 6410 W ≈ 171,600 W ( 230 hp ) .c) With the car now moving up an incline, the normal force from the road surfacebecomes N’ = Mg cos θ , making the kinetic frictional force f k’ = µ kMg cos θ .In addition to the forces already described, the weight force on the vehicle now also has acomponent parallel to the incline, W ||= Mg sin θ .

The net force on the car is nowrF net|| = F rA +F rD + f rk ' +rW ||= 0 , so the netpower becomes P net= F Av − F Dv − ( µ kMg cos θ ) v − ( Mg sin θ ) v = 0 . Since the car ison a 12% grade, tan θ = 0.12 ⇒ sin θ ≈ 0.1191 , cos θ ≈ 0.9929 . We have seen thatthe power exerted by air drag is P€ D≈ 88,600 W at 29.05 m./sec. (65 mph) and that it is⎛⎛proportional to v 3 , so we can write P D = 88,600 ⎜⎜ v ⎞⎞⎟⎟ 3 W . The power which must be⎝⎝ 29.05⎠⎠delivered by the engine in order for the vehicle to maintain a speed v on the 12% upwardincline is thenP A= F A v = P D + (µ k Mg€cos θ ) v + (Mg sin θ ) v€⎛⎛≈ 88,600 ⎜⎜ v ⎞⎞⎟⎟ 3 + (0.025) (900 kg.) (9.81 m.m.) (0.9929) v + (900 kg.) (9.81⎝⎝ 29.05⎠⎠) (0.1191) vsec. 2 sec. 2≈ 3.614 v 3 + 219.2 v + 1052 v ≈ 3.614 v 3 + 1271v .€We are going to keep the power from the engine at the same level it had inpart (b) above, so we need to solve for v the equation 3.614 v 3 + 1271 v = 171,600 .€ It is not very convenient to solve this last equation directly (we can, of course, use graphingsoftware, as will be discussed below), but we don’t need to resort to trial-and-errorcompletely either. We know that this amount of power on level ground allows the car totravel at 80 mph ( ≈ 35.8 m./sec. ), so we might expect that on a 12% climb, the car mightmove, say, 10% or so slower. Let’s make a first guess that the solution isv = 32 m./sec. Since the cubic term in the equation, 3.614 v 3 , changes much more rapidlythan the linear term, 1271v , we’ll simply set this term to 1271 · 32 for the present andsolve the following for v :3.614 v 3 + 1271⋅ 32 = 3.614 v 3 + 40670 = 171,600 ⇒ 3.614 v 3 = 130,900 ⇒ v 3 ≈ 36,200 ⇒ v ≈ 33.1sec.m. ,€which is pretty close to our initial guess. If we adjust the linear term to accommodate thisnew value, we find€3.614 v 3 + 1271⋅ 33.1 = 3.614 v 3 + 42070 = 171,600 ⇒ v ≈ 33.0sec.m. . €Our solution has stabilized to one decimal place, so we may stop here. A more precisesolution, using graphing software to find the x-intercepts for the function3.614 v 3 + 1271 v − 171,600 , gives us v ≈ 32.98 m./sec. , so our result is acceptable.

As a check, we can calculate the individual power terms:power exerted against drag: P A≈ 3.614 · 33.0 3 ≈ 129,900 Wpower exerted against friction: P f≈ 219.2 · 33.0 ≈ 7230 Wpower exerted against gravity: P W≈ 1052 · 33.0 ≈ 34,700 Wtotal power to be delivered by engine: 171,800 W ,agreeing with the intended total value to about 0.12% .d) Without the applied force from the engine, and assuming that there is no internalfriction in the drive train of the car (a bit unrealistic), the net force on the vehicle actingparallel to the downward incline isrF net|| ' =rW || +F rD + f rk ' = 0 ⇒ F net|| ' = W || − F D − f k ' = 0 ,making the net power P net= ( Mg sin θ ) v − F Dv − ( µ kMg cos θ ) v = 0 .€€Each of these terms has the same value as it did in part (c) above, giving us1052 v − 3.614 v 3 − 219.2 v ≈ 832.3v − 3.614 v 3 = 0 ,which is much easier to solve for v than the power equation we found in the previous part.We can factor this as v ⋅ (832.3 − 3.614 v 2 ) = 0 , so either v = 0 (the uninterestingsolution, since the power terms are of course all zero when the car is parked) or832.3 − 3.614 v 2 = 0 ⇒ v 2 ≈ 230.3 ⇒ v ≈ 15.2 m./sec. (34 mph) .In a real vehicle, internal friction would make the downhill coasting speed somewhat lowerthan this. €€12)a) The central event in this physical arrangement is the elastic collision between thetwo blocks. The combination of linear momentum conservation with conservation ofkinetic energy leads to the result that the relative velocity of the masses before the collisionis the opposite of the relative velocity afterwards, that is, v 1− v 2= −( v 1’ – v 2’ ) , which maybe rearranged more conveniently as v 1+ v 1’ = v 2+ v 2’ . For our collision, this gives us v+ v’ = 0 + V . This provides a relationship among the velocities, but tells us nothing else.We will need to uncover the rest by examining the energies of the blocks.

One way to analyze this would be to perform a detailed study of all the energychanges for the blocks at each significant event. However, we can save ourselves sometrouble by noticing a feature of the motion of m 1. Upon leaving the spring launcher at thetop of its ramp, m 1has a kinetic energy equal to the potential energy U spthat had beenstored in the spring. When it reaches the bottom of the ramp to collide with m 2, m 1has akinetic energy U spplus the gravitational potential energy it acquired by descending 35 cm.,since this ramp is frictionless (so the angle of this incline or any curvature that it hasbecome irrelevant). After the collision, m 1has just enough kinetic energy to come to restas it reaches its launcher; hence, it has lost an amount of energy U sp. But because thecollision with m 2is elastic, this must be the amount of energy transferred to m 2. We canthus conclude that after the collision,has a kinetic energy ofm 2½ m 2V 2 = U sp= ½ k 1( Δx ) 1 2 ≈ ½ ( 100 N./m. ) ( 0.11 m. ) 2 ≈ 0.605 J.Returning to m1 , we have said that its kinetic energy before the collision was½ m 1v 2 = Usp+ m 1gH = 0.605 + ( 0.1 kg. ) ( 9.81 m./sec. 2 ) ( 0.35 m. )≈ 0.605 + 0.343 J. ≈ 0.948 J.and after the collision, the kinetic energy became ½ m 1v’ 2 = m 1gH ≈ 0.343 J.So we can calculate that"v = 2 (U sp + m 1 gH ) %$# m '1 &1/ 2v' = " 2 # m 1 gH1/ 2$ ' + 2 # 0.343 J. .m( m 1has reversed direction, so v’%&1 () * " - 0 * " 2.62, 0.1 kg. /sec.m.becomes negative). The ! relative-velocity equation then tells us that1/ 2(*,+2 ) 0.948 J.0.1 kg.-/.1/ 2( 4.36 m.sec.and!v + v’ = 0 + V ⇒ V ≈ 4.36 + ( −2.62 ) ≈ 1.73 m./sec.** rounding from a larger number of significant figuresSince we know the kinetic energy of m 2after the collision, we can now solve for the value

m 2 = 2K 2'V 2= 2U spV 2 ≈2 ⋅ 0.605 J.(1.74 m.sec. )2≈ 0.402 kg.(This can also be determined using linear momentum conservation.) While much of theinformation in this part of the Problem may seem to have been extraneous, it will be quiteuseful € for the next portion.b) In this new situation, m 2comes to rest just as it makes contact with the springbumper on its ramp. So after the collision, m 2has only enough kinetic energy to climp theramp and overcome its friction. The work m 2must perform against gravity is m 2g · ( h 1+h 2) = ( 0.402 kg. ) ( 9.81 m./sec. 2 ) ( 0.025 + 0.05 m. ) ≈ 0.296 J. The length of inclinewhich exerts frictional force on m 2is given bysin 9 o = h 2L⇒ L =h 2sin 9 o ≈0.05 m.0.156≈ 0.320 m.€The work which m 2must perform against kinetic friction is thenµ kN 2L = µ k · ( m 2g cos 9º ) · L≈ ( 0.17 ) ( 0.402 kg. ) ( 9.81 m./sec. 2 ) ( 0.988 ) ( 0.320 m. ) ≈ 0.212 J.Thus, m 2must have departed from the collision with m 1with a kinetic energy of0.296 + 0.212 ≈ 0.508 J. , which means that its velocity after the collision in this situation⎛⎛ 2 ⋅ 0.508 J. ⎞⎞1/ 2was V ≈ ⎜⎜ ⎟⎟ ≈ 1.59 m.⎝⎝ 0.402 kg. ⎠⎠ sec..€We now need to find the speed of m 1before this collision, in order to determinehow it was launched. The relative-velocity equation tells us v + v’ = 0 + V≈ 1.59 m./sec. As we are really only interested in v , we will solve this equation for v’and eliminate it from the linear momentum conservation equation:m 1v + m 2 · 0 = m 1v’ + m 2V⇒ ( 0.100 kg. ) · v ≈ ( 0.100 kg. ) ( 1.59 – v m./sec. ) + ( 0.402 kg. ) ( 1.59 m./sec. )≈ 0.159 − 0.100 v + 0.640 kg.−m./sec.⇒ 0.200 v ≈ 0.799 ⇒ v ≈ 4.00 m./sec.So the kinetic energy of m 1before the collision was½ m 1v 2 ≈ ½ ( 0.100 kg. ) ( 4.00 m./sec. ) 2 ≈ 0.799 J.

As in part (a), this energy is the sum of the spring launcher’s stored potential energy and thegravitational potential energy released as m 1drops by 35 cm. Since this latter amount hasalready been found to be 0.343 J. , the launcher must have held a potential energy of 0.799 –0.343 J. ≈ 0.456 J. Using the formula for the (ideal) spring potential energy in the launcher,the initial compression of the spring in this situation would have been1/ 2⎛⎛ ⎞⎞U sp ' = 1 2 k 2 ⋅ 0.456 J.1 (Δx')2 ≈ 0.456 J. ⇒ Δx' ≈ ⎜⎜ ⎟⎟⎜⎜⎝⎝100 N. ≈ 0.096 m. or 9.6 cm.⎟⎟m. ⎠⎠13)€€a) In a “simple” pendulum, it is idealized that all of the pendulum’s mass isconcentrated in a mass-point at one end. For a realistic or “physical” pendulum, however,we must take into account that the mass is distributed over the body of the pendulum. Ifwe deflect the pendulum from its vertical equilibrium by an angle θ , the gravitationaltorque acting to restore it to equilibrium will be in the opposite direction with a magnitudeof r CM · Mg · sin θ , where r CMis the distance from the pivot point of the pendulum to itscenter-of-mass. Since the pendulum turns rigidly, the torque isτ = I tot α = I tot ⋅ d 2 θ= − Mgrdt 2CM sin θ . If the angular displacement from equilibrium issmall ( θ < 0.3 radians ), then sin θ ≈ θ (in radians) and we can writeI tot ⋅ d 2 θdt 2 ≈ − Mgr CM θ ⇒ d 2 θdt 2+ Mgr CMI totθ ≈ 0 ,which is a differential equation for simple harmonic motion. For this pendulum,ω 2= €Mgr CMI tot, so it will oscillate with a period ofT = 2π ω = 2π ⋅ I totMgr CM.€If we apply this formula to the simple pendulum, I tot= M l 2 and r CM= l , since all themass is at one end; we then obtain the familiar result€T = 2π ⋅M l 2Mgl= 2π ⋅lg .€

For this longcase clock (sometimes called a “grandfather’s clock”) pendulum, the rodhas mass m and length L and is pivoted from one end, with the bob being a disk of massM and radius R centered at a distance d below the pivot. The distance of the center-ofmassfrom the pivot is thenr CM = m ⋅ ( 1 2 L ) + M ⋅ dm + M≈ (2.000 kg.) ⋅ ( 1 ⋅1.100 m.) + 5.000 kg. ⋅ d22.000 + 5.000 kg.≈ 0.1571 + 0.7143 d m.€The total moment of inertia of the pendulum is that of a rod pivoting about one end plus adisk revolving about a point outside its center-of-mass. The moment of inertia for the1uniform rod is then3 mL2 . By the parallel-axis theorem, the moment inertia of the heavypendulum bob is ½ MR 2 + Md 2 . Thus, the total moment of inertia of the pendulum isgiven by€I tot = 1 3 mL2 + 1 MR 2 + M d 22€€≈ 1 3 (2.000 kg.) (1.100 m.)2 + 1 2 (5.000 kg.) (0.080 m.)2 + (5.000 kg.) d 2 ≈ 0.82267 + 5d 2 kg.⋅ m. 2We want to adjust this pendulum so that the period is very precisely equal to 2.000seconds at sea level on the Earth’s equator. So we will need to move the bob to the positionfor whichT = 2π ⋅I totM totgr CMMtotis the total mass of the pendulum€⎡⎡⎤⎤⎢⎢0.82267 + 5d≈ 2π ⋅2 kg.⋅m. 2⎥⎥⎢⎢(2.000 + 5.000 kg.) (9.78033 m. ) (0.1571 + 0.7143 d m.)⎥⎥⎣⎣ ⎢⎢sec.2 ⎦⎦ ⎥⎥1/ 2≈ 2.000 sec.⇒€0.82267 + 5d 210.7554 + 48.9026 d ≈ ⎛⎛ 2.000 ⎞⎞⎜⎜ ⎟⎟⎝⎝ 2π ⎠⎠2⇒ 0.82267 + 5d 2 ≈ 1.08975 + 4.95487 d€€⇒ 5d 2 − 4.95487 d − 0.26708 ≈ 0 ⇒ d ≈ 1.0422 m.only positive solutionb) Thermal contraction has caused the linear dimensions of this pendulum to bereduced by 1/1000 ( 0.1% ) . The effect on the bob itself is insignificant, but the effect onthe length of the rod and the distance of the bob from the pivot both must be considered.The length of the rod is now L’ = 0.999 · L ≈ 1.0989 m. This brings the bob closerto the pivot by a proportional amount, so d’ = 0.999 · d ≈ 0.999 · 1.0422 m. ≈ 1.0412 m.The expression for the distance from the pivot to the center-of-mass becomesr CM ' = m ⋅ ( 1 L') + M ⋅ d'2≈ (2.000 kg.) ⋅ ( 1 ⋅1.0989 m.) + 5.000 kg. ⋅1.0412 m.2≈ 0.90070 m.m + M7.000 kg.€

and that for the total moment of interia isI tot ' = 1 3 mL'2 + 1 2 MR2 + M d' 2a 0.1% reduction of R hasnegligible effect on the middle term€≈ 1 3 (2.000 kg.) (1.0989 m.)2 + 1 2 (5.000 kg.) (0.080 m.)2 + (5.000 kg.) (1.0412 m.) 2 ≈ 6.24154 kg.⋅ m. 2This gives the period of the pendulum in this situation as€⎡⎡⎤⎤⎢⎢6.24154 kg.⋅m.T ' ≈ 2π ⋅2⎥⎥⎢⎢(7.000 kg.) (9.78033 m. ) (0.90070 m.)⎥⎥⎣⎣ ⎢⎢sec.2 ⎦⎦ ⎥⎥1/ 2≈ 1.99899 sec.In a week of 7 days · 86400 sec./day = 604,800 seconds, the pendulum will make€604,800T '≈ 604,8001.99899≈ 302,553 oscillations,corresponding to 2 · 302,553 ≈ 605,106 seconds registered by the clock. The clockwould then be 605,106 − 604,800 ≈ 306 seconds ( 5.09 minutes ) fast.€To restore the pendulum to keeping accurate time, the bob would need to be movedto a distance D from the pivot given by⎡⎡⎤⎤⎢⎢0.82105 + 5D2π ⋅2 kg.⋅m. 2⎥⎥⎢⎢(7.000 kg.) (9.78033 m. ) (0.15699 + 0.7143 D m.)⎥⎥⎣⎣ ⎢⎢sec.2 ⎦⎦ ⎥⎥1/ 2≈ 2.000 sec.€⇒ 5D 2 − 4.95477 D − 0.26794 ≈ 0 ⇒ D ≈ 1.0424 m. only positive solution€The bob should therefore be moved downward away from the pivot byD – d’ ≈ 1.0424 − 1.0412 m. ≈ 0.0012 m. ( ≈ 1.2 mm.) .To make such fine adjustments to the position of the bob, it was often fitted with acalibrated vernier screw, indicated the number of minutes per day of correction. It is thissensitivity to environmental changes that led people to see non-mechanical methods oftimekeeping.

14) We know that this spring-mass system (ideally) obeys the simple harmonic motiondifferential equation, m d 2 (Δ x )dt 2 + k(Δ x ) = 0 , for which the solution for thedisplacement from equilibrium is Δx = A sin( ωt + φ ) , with ω =k. At a moment wemwill call t = 0 , the displacement is Δx = −0.37 m. (the block starts, say, to the left of theorigin), the velocity € is v = 2.64 m./sec. (the block is moving to the right), and theacceleration is a = 1.90 m./sec. 2 (the force on the block is to the right). The spring€constant is given as k = 64 N/m.If we differentiate the displacement equation with respect to time, we obtainv = dxdt= A ⋅ ω ⋅ cos (ωt + φ ) ,a = dvdt= A ⋅ ω ⋅ ω ⋅[ − sin (ωt + φ ) ] = − A ω 2 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 = 2.64 m./sec. , anda 0= −Aω 2 sin ( ω · 0 + φ ) = −Aω 2 sin φ = 1.90 m./sec. 2 .We can now calculatea 0= −A ω 2 sin φ(Δ x ) 0A sin φ= − ω 2 ≈1.90 m.sec. 2−0.37 m.⇒ω 2 ≈ 5.135 rad.sec. 2⇒ω ≈ 2.266 rad.sec.;€period of oscillator:€T = 2π ωthe mass of the block is given byω ⋅ (Δ x ) 0v 0= ω ⋅ A sin φAω cos φ€≈2π rad.2.266 rad.sec.≈ 2.77 sec. ;€ω 2 = k m ⇒ m = kω 2rad.(2.266= tan φ ≈ sec. ) (−0.37 m.)2.64 m.sec.≈64 N.m.5.135 rad.2sec. 2 ≈ 12.5 kg. ;≈ − 0.318 .Since ( Δx ) 0< 0 , so sin φ < 0 also, and v 0> 0 , so cos φ > 0 as well. This tells usthat φ is in quadrant IV , and thus that the phase angle is given simply by the calculator€value to φ ≈ tan −1 ( −0.318 ) ≈ −0.308 rad. (or 5.976 rad.). Finally, we can determine theamplitude of oscillation fromA = (Δ x ) 0sin φ= A sin φsin φ≈−0.37 m.sin (−0.308)≈−0.37 m.−0.313≈ 1.18 m.The equation for the displacement from equilibrium as a function of time istherefore Δx = 1.18 sin ( 2.266 t − 0.308 ) m. The total mechanical energy of this springmassoscillator is given by E = ½ k A 2 ≈ ½ ( 64 N./m. ) ( 1.18 m. ) 2 ≈ 44.56 J €.

15) Sphere A can be treated as a sphere of radius R and density ρ , with an added,embedded sphere with radius ½ R and a density of 2ρ (giving that region a totaldensity of 3ρ ). Sphere B has a cavity in it of the same size, so we will work with it as asphere of radius R and density ρ , with the added sphere having a “density” of −ρ(producing a region with density zero). We will then work out the properties of thesetwo sphere in parallel calculations.€a) total mass – This first calculation will make it clear why we choose to “breakup” these asymmetric spheres in the peculiar way described above. We have the “largesphere” with radius R and density ρ , which has a mass of M =ρ ⋅ 4π 3 R 3 ;a “small excess-density sphere” with radius ½ R , density 2ρ , and a mass ofm A= 2 ρ ⋅ 4π 3 ⋅ ⎛⎛ 12 R ⎞⎞⎜⎜ ⎟⎟ 3 = ρ ⋅ 2 ⋅ 4π⎝⎝ ⎠⎠ 3 ⋅ 8 ⋅ R 3 = ρ ⋅ π 3 ⋅ R 3 1or4M ; and a “small negativedensitysphere” with radius ½ R , density 2ρ , and a mass of m B€= − ρ ⋅ 4π 3 ⋅ ⎛⎛ 12 R ⎞⎞⎜⎜ ⎟⎟ 3⎝⎝ ⎠⎠€ = − ρ ⋅ 4π3 ⋅ 8 ⋅ R 3 = − ρ ⋅ π 6 ⋅ R 3 or − 1 8M . We can now put these results together to find⎛⎛ 4πthe total mass of sphere A as M A= M + m A= ρ ⋅3 + π ⎞⎞⎜⎜⎝⎝ 3€⎟⎟ ⋅ R 3 = ρ ⋅ 5π ⎠⎠ 3 R 3⎛⎛ 4π(so M = 4m Aand M A= 5m A) and M B= M + m B= ρ ⋅3 − π ⎞⎞⎜⎜ ⎟⎟ ⋅ R 3 = ρ ⋅ 7π ⎝⎝ 6 ⎠⎠ 6 R 3(so M = −8m Band M B= −7m B). These € additional relationships will be helpful in theparts of the problem to follow.b) moment of inertia about symmetry axis – The axis which passes through boththe center of the “large sphere” and the center of the “small sphere” may be regarded as thesymmetry axis for both sphere A and sphere B .€

Since the centers-of-mass of the large and small sphere lie on this axis, the moment ofinertia of the assemblage about this axis will simply be the sum of the center-of-massmoments of inertia for the uniform-density spheres. Thus, for sphere A, we haveI A1= 2 5 M R 2 + 2 5 m A ⋅ ( 1 2 R )2 = 2 5 ⋅ (4m A ) R 2 + 2 5 m A ⋅ ( 1 2 R )2= ( 8 5 + 2 20 ) ⋅ m A R 2 = 1710 m A R 2 = 1750 M A R 2 = 0.34 M A R 2 ,€while for sphere B, we find€ I B1= 2 5 M R 2 + 2 5 m B ⋅ ( 1 2 R )2 = 2 5 ⋅ (− 8m B ) R 2 + 2 5 m B ⋅ ( 1 2 R )2= (− 16 5 + 2 20 ) ⋅ m B R 2 = − 3110 m B R 2 = 3170 M B R 2 ≈ 0.443M B R 2 .€We see that sphere A, having a greater density of mass concentrated near this rotation axis,has a lower coefficient of rotational inertia ( 0.34 ) than does sphere B ( 0.443 ), where thecavity € causes more of its mass to be distributed farther away from this rotation axis.c) moment of inertia about perpendicular axis through geometrical center –Here, we will use a rotational axis through the center of the large sphere andperpendicular to the line we’ve called the symmetry axis.We can thus still use the center-of-mass moment of inertia for the large sphere, but weneed to invoke the “parallel-axis theorem” to work out the altered moment for the smallsphere. This requires the addition of a term which is the mass of the object times thesquare of the distance from its center-of-mass to the rotation axis. The new result forsphere A is

I A2= 2 5 M R 2 + 2 5 m A ⋅ ( 1 2 R )2 + m A ⋅ ( 11 424 R 3)2"parallel−axis"term= 2 5 ⋅ (4m A ) R 2 + 7 5 m A ⋅ ( 1 2 R )2€= ( 8 5 + 720 ) ⋅ m AR 2 = 3920 m AR 2 = 39100 M AR 2 = 0.39 M A R 2 ,while for sphere B, we findI€ B1= 2 5 M R 2 + 2 5 m B ⋅ ( 1 2 R )2 + m B ⋅ ( 11 424 R 3)2"parallel−axis"term= 2 5 ⋅ (− 8m B ) R 2 + 7 5 m B ⋅ ( 1 2 R )2€= (− 16 5 + 7 20 ) ⋅ m B R 2 = − 5720 m B R 2 = 57140 M B R 2 ≈ 0.407 M B R 2 .Relative to this rotation axis, there is far less difference between the coefficients ofrotational inertia of spheres A and B .€d) location of center of mass – When an object has a uniform density, its center- ofmasswill be in the position of its geometrical center. Since we are treating spheres A and Bas “composite objects”, made up of two individual spheres of constant density, the centersfor A and B will lie along the symmetry axis connecting the centers of the large sphere andthe small sphere. So we can identify the symmetry axis as the “x-axis” and make the centerof the large sphere x = 0 ; this places the center of the small sphere at x = ½ R . Thispermits us to calculate the position of the center-of-mass of sphere A as0 ⋅ M + ( 1x A = 2 R ) ⋅ m 1A= 2 ⋅ m AR =M + m AM Am A2 ⋅ 5m AR = 110 R ,and that of sphere B as€x B =0 ⋅ M + ( 1 2 R ) ⋅ m BM + m B=12 ⋅ m BM BR =m B2 ⋅ (−7m B) R = − 114 R .€e) moment of inertia about perpendicular axis through center-of-mass –We again choose rotation axes perpendicular to the symmetry axes of our spheres A and B,as we did in part (c), but now these axes pass through the centers-of-mass we have justdetermined.

As before, we apply the “parallel axis theorem” to find this moment of inertia forsphere A to beI A3= [ 2 5 M R 2 2+ M ⋅ x123A"parallel−axis"term] + [ 2 5 m A ⋅ ( 1 2 R )2 + m A ⋅ ( 1 R − x A ) 214 42444 3]"parallel−axis"term€= 2 5 ⋅ (4m A ) R 2 + (4m A )⋅ ( 110 R )2 + 2 5 m A ⋅ 1 4 R 2 + m A ⋅ ( 1 2 R − 110 R )2 ]= ( 8 5 + 4100 + 2 20 + 16100 ) ⋅ m A R 2 = 1910 m A R 2 = 1950 M A R 2 = 0.38 M A R 2 ,€and for sphere B,€ I B3= [ 2 5 M R 2 2+ M ⋅ x123B"parallel−axis"term] + [ 2 5 m B ⋅ ( 1 2 R )2 + m B ⋅ ( 1 R − x B ) 214 42444 3]"parallel−axis"term€€€16)= 2 5 ⋅ (− 8m B ) R 2 + (− 8m B )⋅ (− 114 R )2 + 2 5 m B ⋅ 1 4 R 2 + m B ⋅ ( 1 2 R − [− 114 R])2 ]= (− 16 5 − 8196 + 2 20 + 64196 ) ⋅ m BR 2 = − 19770 m BR 2 = 197490 M BR 2 ≈ 0.402M B R 2 .a) Since the only horizontal force acting on the block is static friction with theturntable surface, this must be the source of the centripetal force that keeps the blockturning about the rotation axis of the turntable. So we haveF c = mv 2R = mω 2 R = f s ≤ µ s N .Because the surface is horizontal, N = mg ; hence,ω 2 R ≤ µ s g .€€

€The turntable is rotating at a frequency of f = 45 rev./min. , so its angular speed isω = 2π f = 2π rad.rev.⋅ 45 rev. 1 min.⋅ ≈ 4.71 rad.min. 60 sec. sec. . With the block placed at R = 0.06 m.from the rotation axis, the coefficient of static friction necessary to hold the block in placeis given by µ s ≥ ω 2 R (4.71rad.g≈ sec. )2 ⋅ 0.06 m.9.81 m. ≈ 0.136 .sec. 2b) In order for static friction to hold the block in place at the edge of the turntable(4.71rad.platter,€where R = 0.15 m. , we now require µ s ≥ sec. )2 ⋅ 0.15 m.9.81 m. ≈ 0.339 .sec. 2c) For a fixed coefficient of static friction, we can rearrange the force inequality tofind the maximum distance at which an object can be situated from the rotation axis and€still be held in place by static friction, R ≤ µ s gω 2 . With µ = 0.23 for this turntable, thesmaximum distance from the rotation axis would be given byR ≈0.23 ⋅ 9.81 m.€sec. 2(4.71sec. rad. ≈ 0.102 m. (10.2 cm.) .)2€acceleration€d) While the turntable’s angular speed is increasing, there will be both a centripetal€ a C = v T 2R = ω 2 R and a tangential acceleration a T= αR , where v Tis theinstantaneous tangential velocity of the turntable at radius R and α is the angularacceleration; both of the these accelerations are in the plane of the turntable surface. Sincecentripetal acceleration points along a radius of the circle of motion and the tangentialacceleration points along a tangent to that circle (hence the name), a Cand a Tareperpendicular. So the total acceleration of a point on the turntable surface is the magnitudeof the resultant acceleration vector,€a tot = a C 2 + a T2 . The static friction must be able toprovide this total acceleration to keep the block moving along with the turntable, so werequirea tot ≤ µ s g .The turntable accelerates from rest ( ω 0= 0 ) to a final angular speedω f= 4.71 rad./sec. (45 rpm) at a uniform rate α in a time T , orα = ω f− ω 0T= ω fT.€

The tangential acceleration is then⎡⎡a tot = (ω 2 f R ) 2 + ( ω fT R ⎤⎤⎢⎢⎣⎣)2 ⎥⎥€ ⎦⎦⎛⎛a T = αR = ω f ⎞⎞⎜⎜ ⎟⎟ R . The greatest total acceleration⎝⎝ T ⎠⎠will therefore occur as the turntable reaches its final angular speed, at which time1/ 2place on the accelerating turntable gives us. So the requirement for static friction to hold the block in€⎡⎡(ω 2 f R ) 2 + ( ω fT R ⎤⎤⎢⎢⎣⎣)2 ⎥⎥⎦⎦1/ 2≤ µ s g ⇒ ( ω fT R )2 ≤ (µ s g ) 2 − (ω f 2 R ) 2€⇒ ω f 2 R 2T 2 ≤ (µ s g ) 2 − (ω f 2 R ) 2 ⇒ω f 2 R 2(µ s g ) 2 − (ω f 2 R ) 2 ≤ T 2€⇒ T 2 ≥(0.23 ⋅ 9.81 m.sec. 2 )2(4.71 rad.sec. )2 ⋅ (0.06 m.) 2− ([ 4.71 rad.sec. ]2 ⋅ 0.06 m. ) 2 ≈ 0.0241 sec. 2⇒T ≥ 0.155 sec.€Thus, the minimum time in which the turntable may accelerate from rest to its operatingangular rate of 45 rpm without causing the block to slip is 0.155 seconds for uniform€ angular acceleration. (Note that the mass of the block does not enter into any of the resultsin this Problem.)17)

€a) The hanging spring scale “reads weight” by registering the tension T in itsspring. With the scale connected to the block by a cord, this tension can be determinedfrom the free-body diagram for the block from T + B – Mg = 0 ⇒ T = Mg – B , where Bis the buoyancy force, equal to the weight of a volume of water 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 ρb ⎠⎠wbeing thedensity of water and ρ bbeing the density of the block. So the reading on the spring scaleρ ⎞⎞ ⎡⎡ ⎛⎛wisρ⎟⎟ ⋅ Mg = 1 − ρ ⎞⎞ ⎤⎤w⎢⎢ ⎜⎜b ⎠⎠⎝⎝ ρ⎟⎟ ⎥⎥ ⋅ Mg .⎣⎣ b ⎠⎠ ⎦⎦⎛⎛T = Mg − ⎜⎜⎝⎝€The platform scale “reads weight” by registering the normal force N , which isprovided by an internal mechanism of the scale (frequently, a spring system). This forcecan be determined from the free-body diagram for the scale asN − M wg − M Bg − B = 0 ⇒ N = ( M w+ M B)g + B ,where M wis the mass of water and M Bis the mass of the beaker. The buoyancy forceenters into this force equation because the fluid pressure which pushes on the block alsopushes on the walls of the beaker.For our problem, the block has mass M = 0.840 kg. , the beaker, M B= 0.095 kg. ,and the water, M w= 1.36 kg. The readings for aluminum and magnesium blocks aredensity ( gm./cm. 3 )buoyancy force, Baluminum 2.65⎛⎛⎜⎜⎜⎜⎝⎝1.00 gm.cm. 32.65 gm.cm. 3⎞⎞⎟⎟⎟⎟⎠⎠⋅ 0.84 kg. ⋅ 9.81m.sec. 2≈ 3.11 Nmagnesium 1.74€⎛⎛1.00 gm. ⎞⎞⎜⎜ cm. 3 ⎟⎟⎜⎜ 1.74 gm. ⎟⎟⎝⎝ ⎠⎠cm. 3⋅ 0.84 kg. ⋅ 9.81m.sec. 2≈ 4.74 Nspring scale reading, T = Mg − Baluminummagnesiumaluminummagnesium€€€€(0.84 kg. ⋅ 9.81 m. ) − 3.11 ≈ 5.13 Nsec.2(0.84 kg. ⋅ 9.81 m. ) − 4.74 ≈ 3.50 Nsec.2platform scale reading, N = ( M w+ M B)g + B([1.36 + 0.095 kg.] ⋅ 9.81 m. ) + 3.11 ≈ 17.38 Nsec.2([1.36 + 0.095 kg.] ⋅ 9.81 m. ) + 4.74 ≈ 19.01 Nsec.2€

) As this is a physical arrangement in static equilibrium, the net force on therod is zero; thus, T L+ T R+ B − Mg = 0 .We are told that the rod has a length of 75 cm. and a cross-sectional area of 8.0 cm. 2 , soits volume is 600 cm. 3 The buoyancy force is equal to the weight of an equal volume ofwater, hence B = ρ w V g = (1 gm.cm. 3 ) ⎛⎛ 1 kg. ⎞⎞⎜⎜ ⎟⎟ (600 cm. 3 ) (9.81 m.⎝⎝ 1000 gm. ⎠⎠sec. 2 ) ≈ 5.89 N .The force equation then gives usT L+ T R= Mg − B = ( 2.8 kg. ) ( 9.81 m./sec. 2 ) − 5.89 ≈ 21.58 N .€We can also analyze the torques acting on the rod. If we place the pivot point atthe left end of the rod, then we can neglect the tension T L, since it will have a momentarmof zero length. As for the other torques,torque due to weight of rod: τ W= − ( 25 cm. ) · Mg · sin 90ºtorque due to buoyancy force: τ B= + ( ½ · 75 cm. ) · B · sin 90ºtorque due to tension in right-hand cord: τTR= + ( 75 cm. ) · T r · sin 90º

For this static equilibrium, τ W+ τ B+ τTR= −25 Mg + 37.5 B + 75 T r= 0⇒ T R =(25 cm.) ⋅ (2.8 kg.) ⋅ (9.81 m. ) − (37.5 cm.) ⋅ (5.89 N )2sec.75 cm.≈ 6.21 N .From the force equation above, we find that T L= 21.58 − T R≈ 15.37 N .€ Were the right-hand cord disconnected or removed, its tension contributionwould become zero and our force equation would now be T L= Mg − B ≈ 21.58 N .The center-of-mass of the rod would then need to be in a different location in order forthe rod to remain static and level. Let us call x this new distance of the center of massfrom the left end of the rod. The torque equation is thenτ W + τ B = − x ⋅ Mg + 37.5 ⋅ B = 0 ⇒ x =(37.5 cm.) ⋅ (5.89 N )(2.8 kg.) ⋅ (9.81 m.sec. 2 ) ≈ 8.04 cm.€-- G.Ruffaoriginal notes developed during 2007-09revised: September 2010