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Motion of Masses and Moments of Inertia, Exams of Applied Mathematics

Problems related to the motion of masses and the calculation of moments of inertia. The problems involve finding the speed and acceleration of masses, the moments of inertia of circular discs, and the forces acting on masses. The document also includes calculations using integrals.

Typology: Exams

2012/2013

Uploaded on 02/20/2013

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Download Motion of Masses and Moments of Inertia and more Exams Applied Mathematics in PDF only on Docsity! Coimisiún na Scrúduithe Stáit State Examinations Commission LEAVING CERTIFICATE 2008 MARKING SCHEME APPLIED MATHEMATICS HIGHER LEVEL 1. (a) A ball is thrown vertically upwards with an initial velocity of 39.2 m/s. Find (i) the time taken to reach the maximum height (ii) the distance travelled in 5 seconds. 5 5 5 5 20 m 3.83 9.44.78 distance total m 9.4 )1(9.40 :secondfifth m 4.78 )16(9.4)4(2.39 (ii) s4 )(8.92.390 (i) 2 2 1 2 2 1 = += = += += = −= += = −= += ftuts ftuts t t ft uv Page 2 1. (b) Two particles P and Q, each having constant acceleration, are moving in the same direction along parallel lines. When P passes Q the speeds are 23 m/s and 5.5 m/s, respectively. Two minutes later Q passes P, and Q is then moving at 65.5 m/s. Find (i) the acceleration of P and the acceleration of Q (ii) the speed of P when Q overtakes it (iii) the distance P is ahead of Q when they are moving with equal speeds. ( ) ( ) m.52512301755 distance 123060 2 1 2 1)60(5.5 175560 24 5 2 1)60(23 s 60 2 155 24 523 V V m/s 48 )120( 24 523 )( m/s 24 5 )120( 2 1)120(234260 P m 4260 )120( 2 1 2 1)120(5.5 m/s 2 1 )120(5.55.65 Q )( 2 2 QP 2 2 2 2 1 2 2 2 1 2 =−= =⎟ ⎠ ⎞ ⎜ ⎝ ⎛+= =⎟ ⎠ ⎞ ⎜ ⎝ ⎛+= =⇒+=+ = = += += = += += = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛+= += = += += Q P S S tt .t (iii) at uvii a a ftuts ftuts f f ft uvi 5 5 5 5 5 5 30 Page 3 2. (a) Two straight roads cross at right angles. A woman C, is walking towards the intersection with a uniform speed of 1.5 m/s. Another woman D is moving towards the intersection with a uniform speed of 2 m/s. C is 100 m away from the intersection as D passes the intersection. Find (i) the velocity of C relative D D C (ii) the distance of C from the intersection when they are nearest together. South 53.13East :direction m/s .52 :magnitude 2 5.1 2 0 0 1.5 )( ° −= −= += += ji VVV jiV jiVi DCCD D C rr rrr rrr rrr 5 5 5 25 5 5 C D 2.5 100 53.13 X m 64 36100 on intersecti thefrom C of distance m 36241.5 travelsC timeIn this s 24 5.2 1353cos100 e tim )( = −= =× = ° = = . V CX ii CD r Page 4 3 (b) A particle is projected down an inclined plane with initial velocity u m/s. The line of projection makes an angle of °θ2 with the inclined plane and the plane is inclined at °θ to the horizontal. The plane of projection is vertical and contains the line of greatest slope. The range of the particle on the inclined plane is g ku 2 sin θ . Find the value of k. { } ( ){ } { } ( ){ } 4 sin4 sincossin4 sinsincos4 sin2sinsincos4 sinsin2sin2cos4 sin4sin 2 1sin42cos Range sin4or cos 2sin2 cos2sin 0 0 2 22 2 32 2 322 2 2 2 2 2 2 1 =⇒ = += += +−= += ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ = =⇒ −= = k g u g u g u g u g u g u.g g uu g u g u t .tg.t u rj θ θθθ θθθ θθθθ θθθθ θθθθ θ θ θ θθ 5 5 5 5 5 25 Page 7 4. (a) The diagram shows a light inextensible string having one end fixed, passing under a smooth movable pulley A of mass m kg and then over a fixed smooth light pulley B. The other end of the string is attached to a particle of mass kg. 1m The system is released from rest. Show that the upward acceleration of A is ( ) mm gmm + − 1 1 4 2 . ( ) ( ) ( ) ( ) mm gmmf fmm mg gm mfmg T fm T gm f mmg T f mT gm f m f + − = +=− =− =− =− =− = = 1 1 11 11 11 1 4 2 42 2 422 2 2 2 ofon accelerati Downward A ofon accelerati Upward 5 5 5 5 5 5 30 1m kg B A Page 8 2 m 4m 4 (b) Particles of mass 2m and m are connected by a light inextensible m string which passes over a smooth pulley at the vertex of a wedge-shaped block, one particle resting on each of 30° 30° the smooth faces. The mass of the wedge is 4m and the inclination of each face to the horizontal is 30°. The wedge rests on a smooth horizontal surface and the system is released from rest. (i) Show, on separate diagrams, the forces acting on the wedge and on the particles. (ii) Find the acceleration of the wedge. (i) ( ) ( ) { } { } 19 3 2 38 2 3 30sin3830cos 830sin30osmg30sin230osmg2 8 430sin30sin30cos30cos 30sin30osmg 30sin230osmg2 2 )( 21 21 2 1 gq qqg mqmqmg mqmqcmqc mq RR mq RRTT q m cR m qm Rcm ii =⇒ += += =+−− =− =−+− =− =− 5,5 5 5 20 mg 2mg S 4mg T 1R 1R 2R 2R T T T Page 9 6. (a) A particle of mass 5 kg is suspended from a fixed point by a light elastic string which hangs vertically. The elastic constant of the string is 500 N/m. The mass is pulled down a vertical distance of 20 cm from the equilibrium position and is then released from rest. (i) Show that the particle moves with simple harmonic motion. (ii) Find the speed and acceleration of the mass 0.1 seconds after it is released from rest. ( ) 2 2 22 22 00 m/s 8.10 10806.0100 on accelerati m/s 68.1 10806.02.010 10806.01cos2.0 cos 2.0 amplitude )( 10 with 0about S.H.M. 100 5 500onAccelerati 500 50055 )(5005inc. of dirn.in Force :position Displaced 100 or 500 55 5 and :position mEquilibriu )( = = = = −= −= == = = ==⇒ −=−= −= −−= +−= ==⇒ == x xav tax ii ωx xx x xgg xdg x gg k gd g T kd T i ω ω ω 5 5 5 5 25 5 Page 12 6 (b) A and B are two fixed pegs, A is 4 m vertically A above B. A mass m kg, connected to A and B by two light inextensible strings of equal length, is describing a horizontal circle with uniform 4 m angular velocity ω . For what value of ω will the tension in the upper string be double the tension in the lower string? B 2 3 2 3 3 sinsin3 sinsin2 2 2 cos coscos2 2 2 2 2 g mmg mT mT mrTT mgT mgT mgT mgTT =⇒ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = =+ = =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = =− ω ω ω θωθ ωθθ θ θθ l l l l l l 25 A θ 2T 5 5 5 5 5 4 m T mg θ B Page 13 7. (a) One end of a uniform ladder, of weight W, rests against a rough vertical wall, and the other end rests on rough horizontal ground. The coefficient of friction at the ground is 4 1 and at the wall is 2 1 . The ladder makes an angle α with the horizontal and is in a vertical plane which is perpendicular to the wall. The ladder is on the point of slipping. Find αtan . ( ) ( ) ( ) ( ) 4 7tan 1tan tan tan cos cossin : systemfor about moments Take 16 9 4 1 1118 1 2 1 14 1 12 1 1 2 1 12 1 14 1 =⇒ =+ =++ =+ = =+ =+ = α α α α α αα RRRR RWR R WR o WRR RR l ll 25 5 5 5 5 5 ∟ α ∟ α R21 R 1R 1 1 R W o 4 Page 14 8 (b) Masses of 4 kg and 6 kg are suspended from the ends of a light inextensible string which passes over a pulley. The axis of rotation of the pulley is horizontal, perpendicular to the pulley, and passes through the centre of the pulley. The moment of inertia of the pulley is 0.08 kg m2 and its radius is 20 cm. The particles are released from rest and move 4 kg 6 kg vertically. When each mass has acquired a speed of 1 m/s, find (i) the common acceleration of the masses (ii) the tensions in the vertical portions of the string. ( )( ) ( )( ) ( )( ) N 7.45or 3 14 6 44 N 49or 5 6 66 (ii) 6 601 2 3 26 PEin Loss KEin Gain 2 46 PEin Loss 6 1416508.0 KEin Gain (i) 11 22 22 2 2 12 2 12 2 1 2 22 12 12 12 2 1 g T ggT g T gTg ga g a asuv g h gh gh ghgh vmvmI =⇒⎟ ⎠ ⎞ ⎜ ⎝ ⎛=− =⇒⎟ ⎠ ⎞ ⎜ ⎝ ⎛=− =⇒+= += = = = = −= = ++= ++= ω 5 5 5 5 5 5 30 Page 17 9. (a) A uniform rod, of length 2 m and relative p density 9 7 , is pivoted at one end p and is free to move about a horizontal axis through p. The other end of the rod is immersed in water. The rod is in equilibrium and is inclined to the vertical as shown in the diagram. Find the length of the immersed part of the rod. ( ) m. 06.1 028369 2 2 14 W9 sin1sin 2 2 14 W9or 9 7 2 part immersed oflength Let the 2 =⇒ =+− =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = = l ll ll l l l l W WB W B θθ 5 5 5 5 20 R p B θ W Page 18 9 (b) A cylinder contains water to a height of 20 cm. A solid body of mass 0.06 kg is placed in the cylinder. It floats and the water level rises to 24 cm. 24 cm 25 cm The body is then completely submerged and tied by a string to the bottom of the cylinder. The water level rises to 25 cm. Find (i) the relative density of the body (ii) the tension in the string (iii) the radius of the cylinder. ( ) ( ) ( ) ( ) ( ) cm. 18.2or m 0218.0 0.0015 0015.0 06.005.0800 06.0 06.0 )( N 147.0or 015.0 06.0 8.0 06.0 )( 8.0 100005.0100004.0 A cylinder of area :sectional-crossLet )( 2 = = = = = = = += += = = = = r r A gAg gVg gWiii gT gTg WTBii s gsAgA WB i π ρ 5 5 5 5 5 5 30 Page 19 Some Alternative Solutions 2 (a) ( ) ( )( ) m 64 36100 on intersecti thefrom C of distance m 36241.5 travelsC timeIn this 24 if 0 85.15.1100 2 2 45.1100 )( 222 = −= =× = = +−−= +−= t tt dt dxx ttxii x 2t 100-1.5t 5 5 2 (b) ( ) 3 92 and 2 133 3 13 2 13 3 13 2 13 3 6 and 53 3 5 5 2 5 5 2 − = + =⇒ −= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −+⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = += −= = −==⇒ −= −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ += += −= = xaaw jix jwaiw VVV jwiwV jaV xav jix jviav VVV jvivV iaV MWMW WM M MWMW WM M rr rr rrr rrr rr rr rr rrr rrr rr 5 5 5 5 Page 22 5 (b) ( ) ( ) ( ) ( ) ( ) ( ) 3 1tan 1 21 1 1 2 tan tan1 1tan 45an tan 1 2 1 22 2 u tan 1 22 045cos NEL 045cos PCM 1 21 21 e e e e t e eu euv u e v v mv mv m um − + = − + − −= + − = −= − = − = −=⇒ −−=− +=+ α α θ θ θα θ 5 5 5 5 5 25 Page 23 7 (b) ( ) ( ) 3 2 2 3 2 32 2 2Y :ABfor Babout Moments 322222 22 2 2 22Y 2 2 :system for Cabout Moments =⇒ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= = =⇒=+ =⇒ = = +=++ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++⎟ ⎠ ⎞ ⎜ ⎝ ⎛ μ μ μ WX RX WRWRY WX WY W WWYYX WW X ll ll l l l l 25 5 5 5 5 5 45° A B C Y X W W R R μ Page 24
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