Mechanics 3. Revision Notes

Transcription

1 Mechanics 3 Revision Notes June 6

2 M3 JUNE 6 SDB

3 Mechanics 3 Futhe kinematics 3 Velocity, v, and displacement, x Foces which vay with speed... 4 Reminde a = v dv dx... 4 Elastic stings and spings 5 Hooke s Law... 5 Elastic stings... 5 Elastic spings... 5 Enegy stoed in an elastic sting o sping Impulse and wok done by vaiable foces 9 Impulse of a vaiable foce... 9 Wok done by a vaiable foce Newton s Law of Gavitation Newton s law of gavitation... Finding k in F = kx Simple hamonic motion, S.H.M. 3 The basic S.H.M. equation x = ω x... 3 x = a sin ωt and x = a cos ωt... 3 Peiod and amplitude... 3 v = ω (a x )... 3 Hoizontal spings o stings... 5 Vetical stings o spings Motion in a cicle 7 Angula velocity... 7 Acceleation... 7 Altenative poof... 8 Motion in a hoizontal cicle... 8 Conical pendulum... 9 Banking... 9 Inside an inveted vetical cone...

4 7 Motion in a cicle Motion in a vetical cicle... Poof that a = v fo vaiable speed... Fou types of poblem... i Vetical motion of a paticle attached to a sting... ii Vetical motion of a paticle inside a smooth sphee... 3 iii Vetical motion of a paticle attached to a igid od... 4 iv Vetical motion of a paticle on the outside of a smooth sphee Centes of mass 6 Cente of mass of a lamina... 6 Cente of mass of a secto... 9 Cente of mass of a cicula ac... 3 Standad esults fo cente of mass of unifom laminas and acs... 3 Centes of mass of compound laminas... 3 Cente of mass of a solid of evolution... 3 Cente of mass of a hemispheical shell method a Cente of mass of a hemispheical shell method b Cente of mass of a conical shell Cente of mass of a squae based pyamid Standad esults fo cente of mass of unifom bodies Centes of mass of compound bodies Cente of mass of a hemispheical shell method Tilting and hanging feely Tilting Hanging feely unde gavity... 4 Body with point mass attached hanging feely... 4 Hemisphee in equilibium on a slope... 4 M3 JUNE 6 SDB

5 Futhe kinematics Velocity, v, and displacement, x. We know that v = dx dt dv = x, and a = dt v = a dt and x = v dt Note: dx dt = d x dt = x = x is the ate of incease of x, theefoe it must always be measued in the diection of x inceasing. Fo the same eason d x = x must also be measued in the dt diection of x inceasing. x is the displacement fom in the positive x-axis diection, P x You must mak x and x in the diections shown x x Example: A paticle moves in a staight line and passes a point,, with speed 5 m s - at time t =. The acceleation of the paticle is given by a = t 6 m s. Find the distance moved in the fist 6 seconds afte passing. Solution: x P 5 x x x = v = x dt = t 6 dt = t 6t + c ; v = 5 when t = c = 5 v = x = t 6t + 5 x = x dt = t 6t + 5 dt = 3 t3 3t + 5t + c x = when t = c = x = 3 t3 3t + 5t. Fist find when v =, t = o 5. The paticle will change diection at each of these times. t = x = t = x = 3 t = 5 x = 8 3 paticle moves fowads 3 paticle moves backwads 3 paticle moves fowads 3 fom t = to fom t = to 5 fom t = 5 to 6 t = 6 x = 6 total distance moved is 5 3 m. M3 JUNE 6 SDB 3

6 Foces which vay with speed Reminde a = v dv dx a = dv = dx dv dv = v dt dt dx dx Example: n joining a motoway a ca of mass 8 kg acceleates fom ms to 3 ms. The engine poduces a constant diving foce of 4 newtons, and the esistance to motion at a speed of v ms is 9v newtons. Find how fa the ca tavels while acceleating. Solution: In this case the ca is always tavelling in the same diection. Res F = ma 4.9v = 8 v dv.9v X dx 3 = 8 v 4 9v dv dx X = (8 8) [ln (4 9v ] 3 X = ln 39 = the ca tavels a distance of 4 m, to 3 S.F. R 8g 4 4 M3 JUNE 6 SDB

7 Elastic stings and spings Hooke s Law Elastic stings The tension T in an elastic sting is T = λx, whee l is the natual (unstetched) length of the l sting, x is the extension and λ is the modulus of elasticity. l x T When the sting is slack thee is no tension. Elastic spings The tension, o thust, T in an elastic sping is T = λx, whee l is the natual length of the l sping, x is the extension, o compession, and λ is the modulus of elasticity. In a sping thee is tension when stetched, and thust when compessed. l Tension (stetched) T x T x l Thust (compessed) Example: An elastic sting of length.6 metes and modulus of elasticity 3 N is stetched between two hoizontal points, P and Q, which ae a distance.4 metes apat. A paticle of mass m kg is then attached to the midpoint of the sting, and ests in equilibium,.5 metes below the line PQ. Find the value of m. Solution: By symmety, the tensions in each half of the sting will be equal. Each half has natual length l = 8 m, and modulus of elasticity λ = 3 N. P θ T θ 5 L θ T θ Q Pythagoas PL = 3 extension in each half, x, = 5 m mg T = λx l = = 8 75 Res T sin θ = mg = mg m = g = = 5 to S.F. M3 JUNE 6 SDB 5

8 Example: Two light stings, S and S, ae joined togethe at one end only. ne end of the combined sting is attached to the ceiling at, and a mass of 3 kg is attached to the othe, and allowed to hang feely in equilibium. The moduli of S and S ae 75 N and N, and thei natual lengths ae 5 cm and 4 cm. Find the distance of the 3 kg mass below. Solution: 5 S x As the stings ae light, we can ignoe thei masses and assume that the tensions in the two stings ae equal. (The tension is assumed to be constant thoughout the length of the combined sting.) Res Fo S, T = 3g T = 3g = λx l = 75x 5 x = g 5 Fo S, S 4 x T T = 3g = λx l = x 4 x + x = x = 3g = 94 g 3g Distance of 3 kg mass below, is =.94 = m to S.F. Example: A box of weight 49 N is placed on a hoizontal table. It is to be pulled along by a light elastic sting with natual length 5 cm and modulus of elasticity 5 N. The coefficient of fiction between the box and the table is 4. If the acceleation of the box is cm s and the sting is pulled hoizontally, what is the length of the sting? Solution: F R T l = 5 λ = 5 µ = x Res R = 49 Box moving F = F max = µ R = 4 49 = 9 6 Res NL, T F = 5 T = 6 m = = 5 Hooke s Law T = 5 x 5 = 6 x = 68 the length of the sting is = 8 = m to 3 S.F. 6 M3 JUNE 6 SDB

9 Example: Two elastic spings, S and S, ae joined at each end, so that they ae side by side. The bottom end of the combined sping is placed on a table, and a weight of 6 N is placed on the top. The moduli of S and S ae 8 N and N, and thei natual lengths ae 5 cm and 6 cm. Find the distance of the 6 N weight above the table. Solution: λ = 8, l = 5, and λ =, l = 6. The spings will have the same compessed length, but thei compessions, x and x, will diffe. Res T + T = 6 I 5 x T T x 6 Hooke s Law T = 8x, T = x 5 6 I and II 6 x + 5 x = 6 3 II 6 and compessed lengths equal 5 x = 6 x x = x 6(x ) x = 6 98 x = 76 x = weight is.6 x = 3673 m = 37 cm above the table, to S.F. Enegy stoed in an elastic sting o sping Fo an elastic sting the tension is given by T = λx, when the extension is x. If the sting is l extended by a futhe small amount, δx, then the wok done δ W T δx Total wok done in extending fom x = to x = X is appoximately T δx X and, as δx, the total wok done, W = T W = λx l extension of X. X dx = λx dx l is the wok done in stetching an elastic sting fom its natual length to an X Similaly W = λx l is the wok done in stetching (o compessing) an elastic sping fom its natual length to an extension (o compession) of x. This expession, λx, is also called the Elastic Potential Enegy, o E.P.E., of an elastic sping o sting. l M3 JUNE 6 SDB 7

10 Example: An elastic sping, with natual length 3 cm and modulus of elasticity 4 N, is lying on a ough hoizontal table, with one end fixed to the table at A. The sping is held compessed so that the length of the sping is 4 cm. A teddy bea of mass kg is placed on the table at the othe end of the sping, and the sping is eleased. If the fiction foce is 5 N, find the speed of the teddy bea when the length of the sping is 9 cm. Solution: At a length of 4 m the compession x = 3 4 = 6 and the enegy stoed, E.P.E., is 4 6 = 5 J. 3 At a length of 8 m the compession x = 3 9 = and the enegy stoed, E.P.E., is 4 = 7 J, 3 enegy eleased by the sping is 5 7 = 45 J. The initial speed of the teddy bea is, and let its final speed be v m s. Wok done by the sping is 45 J, which inceases the K.E. Wok done by fiction is 5 4 = J, which deceases the K.E. Final K.E. = Initial K.E. + enegy eleased by sping wok done by fiction v = + 45 = 45 v = 45 = 3 speed of the teddy bea is cm s, to S.F. Example: A climbe is attached to a ope of length 5 m, which is fixed to a cliff face at a point A, 4 metes below him. The modulus of elasticity of the ope is 98 N, and the mass of the climbe is 8 kg. The gound is 8 m below the point, A, to which the ope is fixed. The climbe falls (oh dea!). Will he hit the gound? Solution: nly an idiot would conside what happens at the moment the ope becomes tight! Assume the gound is not thee how fa would he fall befoe being stopped by the ope. In this case both his initial and final velocities would be, and let the final extension of the ope be x m. Loss in P.E. = mgh = 8 g ( x) = 8g (9 + x), which inceases K.E. and so is positive. Wok done in stetching ope, E.P.E., = 98x = 98 x 5 Final K.E. = Initial K.E. + Loss in P.E. E.P.E. = + 8g (9 + x) 98 x x 8x 7 x = (o negative) The climbe would fall. m if thee was no gound, so he would hit the gound m below, but not going vey fast M3 JUNE 6 SDB

11 3 Impulse and wok done by vaiable foces Impulse of a vaiable foce A paticle of mass m moves in a staight line unde the influence of a foce F(t), which vaies with time. In a small time δt the impulse of the foce δi F(t) δt and the total impulse fom time t to t is t I F(t) δt and as δt, the total impulse is I = t t F(t) dt Also, F(t) = ma = m dv t F(t) t dt dt = V m dv U t I = F(t) dt = mv mu t which is the familia impulse = change in momentum equation. t Example: When a golf ball is hit, the ball is in contact with the club fo 8 seconds, and ove that time the foce is modelled by the equation F = kt( 8 t) newtons, whee k = 4 3. Taking the mass of the golf ball to be 45 gams, and modelling the ball as a paticle, find the speed with which the ball leaves the club. Soution: F(t) = kt( 8 t), U =, V =?, m = 45 I = 8 45V = F(t) dt = mv mu 8 kt( 8 t) = k 4t 8 3 t3 = V = 75 9 m s (o 73 km h ) dt F(t) M3 JUNE 6 SDB 9

12 Wok done by a vaiable foce. A paticle of mass m moves in a staight line unde the influence of a foce G(x), which vaies with time. ve a small distance δx the wok done by the foce δw G(x)δx and the total wok done in moving fom a displacement x to x is W Gx) δx x and as δx, the total wok done is W = x x G(x) dx Also, G(x) = ma = m dv x x G(x) W = x x dt V dx = mv U dx = m dv dv = mv dt dx dx dv G(x) dx = mv mu which is the familia wok - enegy equation. x Example: A paticle of mass 5 kg moves on the positive x-axis unde the action of a vaiable foce 4 newtons, diected away fom. The paticle passes though a point x metes fom, with velocity 8 m s in the positive x-diection. It expeiences a constant esistance foce of 6 newtons. Find the speed of the paticle when it is 5 metes fom. Solution: 6 4 x 5 x x 8 m s v m s The wok done by the esistance is 6 3 = 8 J 5 x The wok done by the foce is 4 Deceases K.E. so negative dx = 4 5 = J. Inceases K.E. so positive x Final K.E. = Initial K.E. wok done by esistance + wok done by foce 5V = = J V = 4 m s. M3 JUNE 6 SDB

13 4 Newton s Law of Gavitation Tycho Bahe made many, many obsevations on the motion of planets. Then Johannes Keple, using Bahe s esults, fomulated Keple s laws of planetay motion. Finally Si Isaac Newton poduced his Univesal Law of Gavitation, fom which Keple s laws could be deived. Newton s law of gavitation The foce of attaction between two bodies of masses M and M is diectly popotional to the poduct of thei masses and invesely popotional to the squae of the distance, d, between them:- F = GM M d whee G is a constant known as the constant of gavitation. Howeve the Edexcel A-level couse does not use the full vesion of this law, but states that the foce on a body at a distance x m fom the cente of the eath is invesely popotional to the distance of the body fom the cente of the eath, F = k x. Note that the body must lie on the suface of the eath o above. Finding k in F = k x. Model the eath as a sphee, adius R metes. The foce on a body x metes fom the cente of the eath is F = k x The foce on a paticle of mass m at the suface of the eath is m R F = k R But we know that the foce on m is mg, towads the cente of the eath, k R = mg k = mgr This is so easy that you should wok it out evey time It can be shown that the foce of gavitation of a sphee acting on a paticle lying outside the sphee, acts as if the whole mass of the sphee was concentated at its cente. M3 JUNE 6 SDB

14 Example: A space ocket is launched with speed U fom the suface of the eath whose adius is R. Find, in tems of U, g and R, the speed of the ocket when it has eached a height of R. The foce on a body x m fom the cente of the eath is F = k x Solution: Fistly, when the ocket is at a height of R, it is 3R fom the cente of the eath. At the suface of the eath, taking the mass of the ocket as m, k R = mg k = mgr Gavitational foce at a distance of x fom the cente of the k mg R eath is x = x Wok done by gavity = = 3R mg R x R 3R R mg R x dx = mgr Deceases K.E. so negative 3 Final K.E. = Initial K.E. wok done against gavity mv = mu 3 mgr R 3R V = U 4 3 gr M3 JUNE 6 SDB

15 5 Simple hamonic motion, S.H.M. The basic S.H.M. equation x = ω x If a paticle, P, moves in a staight line so that its acceleation is popotional to its distance fom a fixed point, and diected towads, then x = ω x and the paticle will oscillate between two points, A and B, with simple hamonic motion. The amplitude of the oscillation is A = B = a. Notice that x is maked in the diection of x inceasing n the diagam, and, since ω positive, x is negative, so the acceleation acts towads. A a x a x P is B x = a sin ωt and x = a cos ωt Solving x = ω x, A.E. is m = ω m = ±iω G.S. is x = λ sin ω t + µ cos ω t If x stats fom, x = when t =, then x = a sin ω t and if x stats fom B, x = a when t =, then x = a cos ω t Peiod and amplitude Fom the equations x = a sin ωt and x = a cos ω t we can see that the peiod, the time fo one complete oscillation, is T = π ω. The peiod is the time taken to go fom B A, o fom B A B and that the amplitude, maximum distance fom the cental point, is a. v = ω (a x ) x = ω x, and emembe that x = v dv v dv = dx ω x v dv = ω x dx v = ω x + c But v = when x is at its maximum, x = ±a, c = a ω v = ω x + a ω v = ω (a x ) dx M3 JUNE 6 SDB 3

16 Example: A paticle is in simple hamonic motion about. When it is 6 metes fom its speed is 4 m s, and its deceleation is.5 m s. Find the amplitude of the oscillation, and the geatest speed as it oscillates. Find also the time taken to move a total distance of 6 m stating fom the futhest point fom. Solution: We ae told that v = 4 and x = 5 when x = 6 x = ω x 5 = 6ω ω = 5 = 5 taking positive value v = ω (a x ) 6 = 5 (a 6 ) a = taking positive value Stating fom the futhest point fom, we use x = a cos ω t = cos 5t The paticle stats at x = + so when the paticle has moved 6 metes, x = 6 6 = cos 5t t = accos( 6) = 4 43 seconds to 3 S.F. 4 M3 JUNE 6 SDB

17 Hoizontal spings o stings Example: Two identical spings, of natual length l and modulus λ, ae joined at one end, and placed on a smooth, hoizontal table. The two ends of the combined sping ae fixed to two points, A and B, a distance l apat. A paticle of mass m is attached to the spings at the midpoint of AB; the paticle is then displaced a distance a towads B and eleased. (a) (b) (c) Show that the paticle moves unde S.H.M. Find the peiod of the motion. Find the speed of the paticle when it has moved though a distance of.5a. Solution: A good diagam is essential. x A T T x B l l (a) Conside the mass at a displacement of x fom. Note that you cannot wok fom x = a. T = λx l and is a tension: T = λx l Res, F = ma λx = mx l and is a thust x = λ λ x, which is the equation of S.H.M., with = ml ml ω as we ae dealing with spings λ, m and l ae all positive (Note that the diagam still woks when the paticle is on the left of. x will be negative, and so both T and T will be negative, and will have become thust and tension espectively.) a (b) The peiod is T = π ω = π ml λ (c) When the paticle has moved.5a, it is on the left of and x =.5a v = ω (a x ) v = λ ml a (.5a) = 3λ ml a v = 3λ ml a M3 JUNE 6 SDB 5

18 Vetical stings o spings In these poblems you diagam should show clealy the natual length, l the extension, e, to the equilibium position, E the extension fom the equilibium position to the point P, x. Example: A mass of m hangs in equilibium at the end of a vetical sting, with natual length l and modulus λ. The mass is pulled down a futhe distance a and eleased. Show that, with cetain estictions on the value of a which you should state, the mass executes S.H.M. Solution: In the equilibium position, E, l Res T e = λe = mg l Afte a futhe extension of x, the paticle is at P, e x E mg T e P mg T x Res NL, mg T = mx mg λ(e+x) l = mx mg λe λx l l = mx x = λ lm x λe since = mg l which is S.H.M., with ω = λ. lm The amplitude will be a, and, since this is a sting, the mass will pefom full S.H.M. only if a e. Note If a > e the mass will pefom S.H.M. as long as the sting emains taut; when the sting is not taut, the mass will move feely unde gavity. If a sping is used then the mass will pefom S.H.M. fo any a (as long as the mass does not ty to go above the top of the sping). 6 M3 JUNE 6 SDB

19 6 Motion in a cicle Angula velocity A paticle moves in a cicle of adius with constant speed, v. Suppose that in a small time δ t the paticle moves though a small angle δθ, then the distance moved will be δ s = δθ and its speed v = δs δt = δθ δt and, as δ t, v = dθ dt dθ dt = θ = θ is the angula velocity, usually witten as the Geek lette omega, ω and so, fo a paticle moving in a cicle with adius, its speed is v = ω δθ δ s ω Example: Find the angula velocity of the eath, and the speed of a man standing at the equato. The equatoial adius of the eath is 6378 km. Solution: The eath otates though an angle of π adians in 4 hous π ω = = = ad s - to 3 S.F. A man standing at the equato will be moving in a geat cicle speed v = ω = = 464 m s to 3 S.F. Acceleation δθ v v δv v v δθ A paticle moves in a cicle of adius with constant speed, v. Suppose that in a small time δ t the paticle moves though a small angle δθ, and that its velocity changes fom v to v, then its change in velocity is δ v = v v, which is shown in the second diagam. The lengths of both v and v ae v, and the angle between v and v is δθ. isosceles tiangle δ v = v sin δθ v δθ = v δθ, since sin h h fo h small δv δt v δθ δt as δ t, acceleation a = dv dt = v dθ dt = v θ M3 JUNE 6 SDB 7

20 But θ = ω = v a = v = ω Notice that as δθ, the diection of δ v becomes pependicula to both v and v, and so is diected towads the cente of the cicle. The acceleation of a paticle moving in a cicle with speed v is a = ω = v, and is diected towads the cente of the cicle. Altenative poof If a paticle moves, with constant speed, in a cicle of adius and cente, then its position vecto can be witten cos θ = = sin θ θ sin θ cos θ θ Paticle moves with constant speed θ = ω is constant since is constant sin θ = ω speed is v = ω, and is along the tangent since. = cos θ cos θ θ = ω = ω sin θ θ cos θ sin θ = ω acceleation is ω (o v ) diected towads. in opposite diection to Motion in a hoizontal cicle Example: A blob of mass of 3 kg is descibing hoizontal cicles on a smooth, hoizontal table. The blob does evolutions each minute. An elastic sting of natual length 6 metes and modulus of elasticity 7 newtons is attached at one end to a fixed point on the table. The othe end is attached to the blob. Find the full length of the sting. Solution: Let the extension of the sting be x. λ = 7, l = 6, m = 3 ω = π 6 = π 3 ad s- Res NL, T = mω = 3( 6 + x) π 3 = ( 6 + x) π Hooke s Law ( 6 + x) π 3 T = 7 x 6 = x = x 6π + xπ = 36x x = 6π 36 π = full length of sting is 6 + x = 87 to 3 S.F. 3.6 T ω x 8 M3 JUNE 6 SDB

21 Conical pendulum Example: An inextensible light sting is attached at one end to a fixed point A, and at the othe end to a bob of mass 3kg. The bob is descibing hoizontal cicles of adius 5 metes, with a speed of 4 m s. Find the angle made by the sting with the downwad vetical. Solution: Acceleation = v = 4 Res NL, T sin θ = 3 3 Res T cos θ = 3g Dividing tan θ = 3 θ = 47 4 o to D.P. = 3, = 3 3g = 8843 A θ T g Banking Example: A ca is tavelling ound a banked cuve; the adius of the cuve is m and the angle of banking with the hoizontal is o. If the coefficient of fiction between the tyes and the oad is 6, find the maximum speed of the ca in km h -. Solution: v R F mg Fo maximum speed (i) the fiction must be acting down the slope and (ii) the fiction must be at its maximum, µr. F = 6R I Res (pependicula to the acceleation) R cos = F sin + mg II Res, NL, I and III m v F cos + R sin = m v = R ( 6 cos + sin ) IV I and II mg = R (cos 6 sin ) V IV V v g = ( 6 cos + sin ) (cos 6 sin ) v = m s - = km h = 8 km h to S.F. III M3 JUNE 6 SDB 9

22 Inside an inveted vetical cone Example: A paticle is descibing hoizontal cicles on the inside of an upside down smooth cone (dunce s cap), at a height h above the vetex. Find the speed of the paticle in tems of g and h. Solution: At fist, it seems as if thee is not enough infomation. Put in lettes and hope fo the best! Let the angle of the cone be θ, the adius of the cicle in which the paticle is moving, the nomal eaction R and the mass of the paticle be m. Res NL, R cos θ = m v Res R sin θ = mg Dividing cot θ = But cot θ = h h = v g v = gh v = gh v g h θ R v mg M3 JUNE 6 SDB

23 7 Motion in a cicle Motion in a vetical cicle When a paticle is moving unde gavity in a vetical cicle, the speed is no longe constant. The altenative poof, given a few pages ealie, can easily be modified to show that the acceleation towads the cente is still v, although thee will be a component of the acceleation along the tangent (pependicula to the adius) see below. Poof that a = v fo vaiable speed If a paticle moves in a cicle of adius and cente, then its position vecto can be witten cos θ = sin θ = sin θ θ cos θ θ = cos θ θ sin θ θ sin θ θ + cos θ θ sin θ = θ cos θ = θ cos θ sin θ + θ sin θ cos θ since is constant Fom this we can see that the speed is v = θ = ω, and is pependicula to the adius since. = We can also see that the acceleation has two components θ = ω = v towads the cente and θ pependicula to the adius opposite diection to which is what we should expect since v = θ,and is constant. In pactice we shall only use a = ω = v, diected towads the cente of the cicle. Fou types of poblem i) A paticle attached to an inextensible sting. ii) A paticle moving on the inside of a smooth, hollow sphee. iii) A paticle attached to a od. iv) A paticle moving on the outside of a smooth sphee. Types i) and ii) ae essentially the same: the paticle will make complete cicles as long as it is moving fast enough to keep T o R, whee T is the tension in the sting, o R is the nomal eaction fom the sphee. Types iii) and iv) ae simila when the paticle is moving in the uppe semi-cicle, the thust fom a od coesponds to the eaction fom a sphee. Howeve the paticle will at some stage leave the suface of a sphee, but will always emain attached to a od. Fo a od the paticle will make complete cicles as long as it is still moving at the top the thust fom the od will hold it up if it is moving slowly. Don t foget the wok-enegy equation it could save you some wok. M3 JUNE 6 SDB

24 i Vetical motion of a paticle attached to a sting Example: A small ball, B, of mass 5 gams hangs fom a fixed point,, by an inextensible sting of length 5 metes. While the ball is in equilibium it is given a hoizontal impulse of magnitude 5 N s. (a) (b) (c) (d) Find the initial speed of the ball. Find the tension in the sting when the sting makes an angle θ with the downwads vetical. Find the value of θ when the sting becomes slack. Find the geatest height eached by the ball above the lowest point. Solution: (a) I = mv mu 5 = v v = m s-. (b) Suppose that the paticle is moving with speed v at P. Res NL, T g cosθ = Gain in P.E. = g ( 5 5cosθ) v 5 a = v.5 Fom the wok-enegy equation v v = g 5( cosθ) v = 5g + 5g cosθ I θ.5 T P ( 5g + 5g cosθ ) T = g cosθ + 5 g cosθ + g + g cosθ = T = 4 7 cosθ + g Notice that this still descibes the situation when θ > 9 o, since cosθ will be negative. (c) The sting will become slack when thee is no tension T = 4.7 cosθ +. = cosθ = 4 7 θ = = 33 9 o to the neaest tenth of a degee. M3 JUNE 6 SDB

25 (d) path of ball P v 43.9 P 33.9 A A At the geatest height, the speed will not be zeo, so we cannot use enegy to get staight to the final answe. Theefoe we need to stop and stat again. We know that v = 5g + 5g cosθ, fom I, and that cosθ = at P, v = 7 initial vetical component of velocity is u = 7 cos θ final vetical component of velocity =, and g = 9 8 Using v = u + as we get s = The height of P above A is 5 5 cosθ = the geatest height of the ball above A is 4 7 m to S.F. 4 7 ii Vetical motion of a paticle inside a smooth sphee Example: A paticle is moving in a vetical cicle inside a smooth sphee of adius a. At the lowest point of the sphee, the speed of the paticle is U. What is the smallest value of U which will allow the paticle to move in complete cicles. Solution: Suppose the paticle is moving with speed v when it eaches the top of the sphee, and that the nomal eaction of the sphee on the paticle is R. Res NL, R + mg = m v a Fo the paticle to emain in contact with the sphee (i.e. to make complete cicles), R v ag Fom the lowest point, A, to the top, the gain in P.E. is m g a = mga The wok-enegy equation gives v v R+mg a mv = mu mga U = v + 4ga 5ag since v ag Note that if U = 5ag the paticle will still be moving at the top (v = ag), and so will make complete cicles Fo complete cicles, U 5ag. Note that the method is exactly the same fo a paticle attached to a sting, eplacing the eaction, R, by the tension, T. a A M3 JUNE 6 SDB 3

26 iii Vetical motion of a paticle attached to a igid od Example: A paticle is attached to a igid od and is moving in a vetical cicle of adius a. At the lowest point of the cicle, the speed of the paticle is U. What is the smallest value of U which will allow the paticle to move in complete cicles. Solution: As long as the paticle is still moving at the top of the cicle, it will make complete cicles. Let v be the speed of the paticle at the top of the cicle. If the paticle is moving slowly (v < ag see pevious example), the foce in the od will be a thust, T, and will pevent it fom falling into the cicle. If v =, it will stop at the top, fo complete cicles v > Fom the lowest point, A, to the top the gain in P.E. is m g a = mga The wok-enegy equation gives T v v a A mg a mv = mu mga U = v + 4ga > 4ag since v > Fo complete cicles, U > ag. 4 M3 JUNE 6 SDB

27 iv Vetical motion of a paticle on the outside of a smooth sphee Example: A smooth hemisphee of adius a is placed on hoizontal gound. A small bead of mass m is placed at the highest point and then dislodged. θ is the angle made between the line joining the cente of the hemisphee to the bead with the upwad vetical. (a) Find the foce of eaction between the bead and the hemisphee, in tems of m, g, a and θ. (b) Find the value of θ when the bead leaves the suface of the hemisphee. (c) Find the speed with which the bead stikes the gound. Solution: Path of bead v a R a θ mg v (a) When the angle is θ, assuming the bead is still in contact with the sphee, P.E. lost = mg(a a cosθ) Wok-enegy equation mv = + mga( cosθ) v = ga( cosθ)..i Res NL, mg cosθ R = m v R = mg cosθ m v a a II I and II R = mg(3cosθ ) R = mg cosθ mg( cosθ) (b) R can neve be negative, and so the bead will leave the hemisphee when R = cosθ = 3 (c) θ = 48. o to the neaest tenth of a degee. The only foce doing wok as the paticle falls fom the top of the hemisphee to the gound is gavity. Note that R is always pependicula to the path and so does no wok. P.E. lost = mga, w is speed with which the paticle hits the gound Wok-enegy equation gives mw = + mga w = ag M3 JUNE 6 SDB 5

28 8 Centes of mass When finding a cente of mass Centes of mass depend on the fomula Mx = m i x i, o simila. Remembe that lim δx f(x i )δx = f(x) dx. Cente of mass of a lamina Example: A unifom lamina is bounded by the paabola y = x and the line x = 4, and has suface density ρ. By symmety y =. y x i y i y = x ) To find the mass of the lamina, M x M = Aea density 4 = ρ x dx x = 4 = 4 3 ρx3 4 3 = ρ 3 δx ) To find x, fist choose an element with constant x co-odinate thoughout. Take a stip paallel to the y-axis, a distance of x i fom the x-axis and width δx. This stip is appoximately a ectangle of length y i and width δx Aea of stip y i δx mass of stip = m i y i ρ δx 4 m i x i 4 y i ρx i δx We know that y = x and we let δx 4 m i x i 4 = y i ρx i δx 4 ρx 3 x = m ix i M = 8 5 ρ 3 3 ρ = 5 = 4 cente of mass of the lamina is at ( 4, ). dx = 4 5 ρx5 4 = 8 5 ρ 6 M3 JUNE 6 SDB

29 Example: A unifom lamina is bounded by the x- and y-axes and the pat of the cuve y = cos x fo which x π. Find the coodinates of its cente of mass. y Solution: The figue shows the lamina y = cos x and a typical stip of width δx and height cos x, with suface density ρ. ) To find the mass. M = ρ π cos x dx π = ρ[sin x] = ρ ) To find x, fist choose an element with constant x co-odinate thoughout. Take a stip paallel to the y-axis, a distance of x i fom the x-axis and width δx. This stip is appoximately a ectangle of length y i and width δx mass of typical stip = m i y i ρ δx π m i x i π y i ρx i δx We know that y = cos x and we let δx π m i x i π x = m ix i M π ρ x cos x dx = ρ π = ρ ( π ) ρ = π integating by pats 3) To find y we can use the same stips, because the cente of mass of each stip is appoximately y i fom the x-axis; we can now conside each stip as a point mass, m i y i ρ δx, at a distance y i fom the x-axis. x i y i δx π x π m i y i π y i ρ y i δx We know that y = cos x and we let δx π m i y i π y = π m iy i 8 = ρ π M ρ ρ cos x dx = ρ π any fool can do this integal 8 = π 8 cente of mass is at π, π 8 M3 JUNE 6 SDB 7

30 Example: A unifom lamina occupies the closed egion bounded by the cuve y = x, the line y = x and the x-axis. Find the coodinates of its cente of mass. Solution: y (, ) y = x y = x (x, y) (x, y) y ) To find the mass, M. The aea = aea of tiangle + aea unde cuve M = ρ ( + x dx) = 7 ρ 6 which I am too lazy to do! ) To find y. The typical stip is appoximately a ectangle of length x x and height δy, with a constant y-coodinate. The mass of the stip is m i = ρ (x x )δy. But x = y (lies on the cuve y = x), and x = y (lies on y = x) m i = ρ ( y i y i )δy m i y i lim m iy i δy y = ρ ( y i y i )y i δy = ρ( y y)y dy = M m iy i = 5ρ 7ρ 6 = ρ you ought to do this one! 8 M3 JUNE 6 SDB

31 3) To find x. The cente of mass of the typical stip is (x + x ) fom the y-axis (mid-point of stip) and m i = ρ (x x )δy as befoe. we can now conside each stip as a point mass, m i ρ (x x )δy, at a distance (x + x ) fom the y-axis m i x i = ρ (x x )δy (x + x ) But (x x ) (x + x ) = x x = ( y ) y = 4 5y + y 4 and the limits go fom to because the δy means we ae summing in the y diection. m i x i = ρ (4 5y + y 4 ) δy lim m ix i = δy ρ (4 5y + y 4 ) x = M m ix i = 9ρ 5 7ρ 6 = the cente of mass is at 38 35, 5 4. Cente of mass of a secto dy = 9 5 ρ In this case we can find a nice method, using the esult fo the cente of mass of a tiangle. We take a secto of angle α and divide it into many smalle sectos. Mass of whole secto = M = α ρ = αρ Conside each small secto as appoximately a tiangle, with cente of mass, G, along the median fom. 3 Woking in pola coodinates fo one small secto, m i = ρ δθ P = G x i cos θ 3 3 α lim m i x i δθ θ= α = 3 3 ρ sin α = ρ cos θ dθ 3 α x = m ix i M = 3 3 ρ sin αρ By symmety, y = = sin 3α α x i θ δθ G P (, θ ) sin cente of mass is at 3α, M3 JUNE 6 SDB 9

32 Cente of mass of a cicula ac Fo a cicula ac of adius which subtends an angle of α at the cente. The length of the ac is α mass of the ac is M = α ρ Fist divide the ac into seveal small pieces, each subtending an angle of δθ at the cente. The length of each piece is δθ m i = ρ δθ We now think of each small ac as a point mass at the cente of the ac, with x-coodinate x i = cosθ α lim m i x i δθ θ= α = ρ sinα α = ρ cos θ dθ α x = m ix i M By symmety, y = = ρ sin αρ sin cente of mass is at α, = sin α α θ x i δθ δθ Standad esults fo cente of mass of unifom laminas and acs Tiangle Semi-cicle, adius Secto of cicle, adius, angle α Cicula ac, adius, angle α 3 4 3π of the way along the median, fom the vetex. fom cente, along axis of symmety sin α 3α sin α α fom cente, along axis of symmety fom cente, along axis of symmety 3 M3 JUNE 6 SDB

33 Centes of mass of compound laminas The secet is to fom a table showing the mass, o mass atio, and position of the cente of mass fo each component. Then use x = m ix i M to find the cente of mass of the compound body., y = m iy i M Example: A semi-cicle of adius is cut out fom a unifom semi-cicula lamina of adius. Find the position of the cente of mass of the esulting shape. Solution: By symmety the cente of mass will lie on the axis of symmety, A. The mass of the compound shape is M = (4π π )ρ = 3 π ρ A and the cente of mass of a semi-cicle 4 is fom the cente. 3π compound shape + small semi-cicle = lage semi-cicle Mass Distance above 3 π ρ 3 y π ρ y + π ρ π ρ 4 3π π ρ 4 3π 8 3π = π ρ 8 3π y = 8 9 The cente of mass lies on the axis of symmety, at a distance of 8 fom the cente. 9 M3 JUNE 6 SDB 3

34 Cente of mass of a solid of evolution Example: A machine component has the shape of a unifom solid of evolution fomed by otating the egion unde the cuve y = 9 x, x, about the x-axis. Find the position of the cente of mass. Solution: y 9 y = 9 x x 9 Mass, M, of the solid = ρ π y 9 dx = ρ π (9 x) M = 8 ρπ. The diagam shows a typical thin disc of thickness δx and adius y = 9 x. Mass of disc ρπ y δx = ρπ (9 x) δx Note that the x coodinate is the same (nealy) fo all points in the disc 9 m i x i ρπ (9 x i )x i δx 9 lim m ix i = ρπ (9 x)x δx x = m 43 ix i M = ρπ 8 ρπ = 3 By symmety, y = x δx dx = 43 ρπ dx the cente of mass is on the x-axis, at a distance of 3 fom the oigin. 3 M3 JUNE 6 SDB

35 Cente of mass of a hemispheical shell method a This method needs techniques fo finding the suface aea of a solid of evolution fom FP3. Peliminay esult Take a small section of a cuve of length δs and the coesponding lengths δx and δy, as shown in the diagam. A vey small section of cuve will be nealy staight, δs and we can fom a tiangle. δy δs δx + δy δx δs δx + δy δx and as δx, ds dx = + dy dx Mass of shell Let the density of the shell be ρ, adius In the xy-plane, the cuve has equation x + y = x + y dy = dy dx dx = x y y i y x i δs (x i, y i ) x ds dx = + dy dx = y + x y Take a slice pependicula to the x-axis though the point (x i, y i ) to fom a ing with ac length δs. Aea of the ing π y i δs mass of ing m i π y i ρ δs Total mass πy i ρ δs Total mass M = lim δs πy i ρ δs = πyρ ds M = πyρ ds dx M = πρ dx = πyρ y + x y dx dx = πρ x = πρ M3 JUNE 6 SDB 33

36 To ind m i x i = π y i ρ δs x i lim π y i ρ δs x i = πρ yx ds δs dx dx = πρ yx y + x x = m ix i M = πρ 3 πρ = y dx = πρ x = πρ 3 the cente of mass is on the line of symmety at a distance of fom the cente. Cente of mass of a hemispheical shell method b This method is simila to method a but does not need FP3 techniques, so is suitable fo people who have not done FP3 (I think it is pefeable to method see late). Mass of shell Let the density of the shell be ρ, adius Take a slice pependicula to the x-axis though the point (x i, y i ) to fom a ing with ac length δθ, and cicumfeence πy. This can be flattened out to fom a ectangle of length πy and height δθ y i y δθ (x i, y i ) δθ θ x Aea of the ing mass of ing m i π ρ y i δθ x i Total mass πyρ δθ Total mass M = lim δθ πyρ δθ = πyρ dθ But y = sin θ π M = π sin θ ρ dθ = πρ cos θ To ind m i x i = π y i ρ δθ x i π = πρ lim π y i ρ δs x i = πρ yx δθ But x = cos θ and y = sin θ π dθ 34 M3 JUNE 6 SDB

37 π m i x i = πρ 3 sin θ cos θ π dθ = πρ 3 cos θ = πρ 3 x = m ix i M = πρ 3 πρ = the cente of mass is on the line of symmety at a distance of fom the cente. Cente of mass of a conical shell To find the cente of mass of a conical shell, o the suface of a cone, we divide the suface into small sectos, one of which is shown in the diagam. x G P We can think the small secto as a tiangle with cente of mass at G, whee G = 3 P. This will be tue fo all the small sectos, and the x-coodinate, x, of each secto will be the same G A the x-coodinate of the shell will also be x As the numbe of sectos incease, the appoximation gets bette, until it is exact, and as G = P then G = A (simila tiangles) 3 3 the cente of mass of a conical shell is on the line of symmety, at a distance of of the 3 height fom the vetex. M3 JUNE 6 SDB 35

38 Cente of mass of a squae based pyamid A squae based pyamid has base aea A and height h The cente of mass is on the line of symmety volume = Ah 3 x i mass M = 3 Ahρ Take a slice of thickness δ x at a distance x i fom The base of the slice is an enlagement of the base of the pyamid with scale facto x i h atio of aeas is x i h aea of base of slice is x i A h mass of slice m i = δx h lim m i x i δx x= h = x3 h Aρ dx = 4 h Aρ x = m ix i M = 4 h Aρ 3 Ahρ = 3 h 4 The cente of mass lies on the line of symmety at a distance 3 h fom the vetex. 4 δ x h The above technique will wok fo a pyamid with any shape of base. The cente of mass of a pyamid with any base has cente of mass 3 of the way along the line 4 fom the vetex to the cente of mass of the base (consideed as a lamina). Thee ae moe examples in the book, but the basic pinciple emains the same: find the mass of the shape, M choose, caefully, a typical element, and find its mass (involving δx o δy) fo solids of evolution about the x-axis (o y-axis), choose a disc of adius y and thickness δ x, (o adius x and thickness δ y). find m i x i o m i y i let δx o δy, and find the value of the esulting integal x = m M ix i, y = m M iy i 36 M3 JUNE 6 SDB

39 Standad esults fo cente of mass of unifom bodies Solid hemisphee, adius Hemispheical shell, adius Solid ight cicula cone, height h Conical shell, height h 3 8 3h h 4 3 fom cente, along axis of symmety fom cente, along axis of symmety fom vetex, along axis of symmety fom vetex, along axis of symmety Centes of mass of compound bodies This is vey simila to the technique fo compound laminas. Example: A solid hemisphee of adius a is placed on a solid cylinde of height a. Both objects ae made fom the same unifom mateial. Find the position of the cente of mass of the compound body. Solution: By symmety the cente of mass of the compound body, G, will lie on the axis of symmety. The mass of the hemisphee is π 3 a3 ρ at G, and the mass of the cylinde is π a aρ = π a 3 ρ at G mass of the compound shape is M = 8 3 π a3 ρ, G = 3a 8, and G = a Now daw up a table G G G a a Body hemisphee + cylinde = compound body Mass Distance above 3 π a3 ρ π a 3 ρ 3a 8 a 8 3 π a3 ρ y π 3 a3 ρ 3a 8 y = a 3 + π a 3 ρ ( a) = 8 3 π a3 ρ y cente of mass is at G, below, whee G = 3 a, on the axis of symmety. M3 JUNE 6 SDB 37

40 Cente of mass of a hemispheical shell method Note: if you use this method in an exam question which asks fo a calculus technique, you would have to use calculus to pove the esults fo a solid hemisphee fist. The best technique fo those who have not done FP3 is method b. We can use the theoy fo compound bodies to find the cente of mass of a hemispheical shell. Fom a hemisphee with adius + δ we emove a hemisphee with adius, to fom a hemispheical shell of thickness δ and inside adius. minus equals adius + δ Mass cente of mass above base π ( + δ ) 3 ρ 3 π 3 3 ρ π ( + δ ) 3 ρ π ρ 3 ( + δ) 3 y 8 8 π ( + δ ) 3 ρ 3 ( + δ) π ρ 3 = { π ( + δ ) 3 ρ π ρ}y πρ 4 ( δ 4 ) = 3 δ δ y 3 πρ (3 + 3 δ 3 ) y ignoing (δ ) and highe and as δ, y = The cente of mass of a hemispheical is on the line of symmety, fom the cente. 38 M3 JUNE 6 SDB

41 Tilting and hanging feely Tilting Example: The compound body of the pevious example is placed on a slope which makes an angle θ with the hoizontal. The slope is sufficiently ough to pevent sliding. Fo what ange of values of θ will the body emain in equilibium. Solution: The body will be on the point of tipping when the cente of mass, G, lies vetically above the lowest cone, A. Cente of mass is a = a a fom the base G θ 43 3 a At this point tan θ = a 43a 3 = 3 43 A a θ = θ The body will emain in equilibium fo θ 36 7 o to the neaest o. M3 JUNE 6 SDB 39

42 Hanging feely unde gavity This was coveed in M. Fo a body hanging feely fom a point A, you should always state, o show clealy in a diagam, that AG is vetical this is the only piece of mechanics in the question! Body with point mass attached hanging feely The best technique will pobably be to take moments about the point of suspension. Example: A solid hemisphee has cente, adius a and mass M. A paticle of mass M is attached to the im of the hemisphee at P. The compound body is feely suspended unde gavity fom. Find the angle made by P with the hoizontal. Solution: As usual a good, lage diagam is essential. Let the angle made by P with the hoizontal be θ, then GL = θ. We can think of the hemisphee as a point mass of M at G, whee G = 3a 8. The pependicula distance fom to the line of action of Mg is L = 3a sin θ, 8 and the pependicula distance fom to the line of action of Mg is K = a cosθ L θ θ G Mg 3a 8 θ a K Mg P Taking moments about Mg 3a 8 sin θ = Mg a cosθ tanθ = 4 3 θ = 53 o. 4 M3 JUNE 6 SDB

43 Hemisphee in equilibium on a slope Example: A unifom hemisphee ests in equilibium on a slope which makes an angle of o with the hoizontal. The slope is sufficiently ough to pevent the hemisphee fom sliding. Find the angle made by the flat suface of the hemisphee with the hoizontal. Solution: Don t foget the basics. The cente of mass, G, must be vetically above the point of contact, A. If it was not, thee would be a non-zeo moment about A and the hemisphee would not be in equilibium. B θ G BGA is a vetical line, so we want the angle θ. A must be pependicula to the slope (adius tangent), and with all the 9 o angles aound A, AG = o. Let a be the adius of the hemisphee o a A 7 o o then G = 3a 8 and, using the sine ule sin GA a = sin 3a 8 Clealy GA is obtuse GA = o 4 9 BG = = 4 9 GA = 4 9 θ = = 65 8 o to the neaest o. M3 JUNE 6 SDB 4

44 Index Acceleation v dv/dx, 3 x dv/dt, 3 Angula velocity, 7 Centes of mass bodies hanging feely, 4 bodies with mass attached, 4 cicula ac, 3 compound bodies, 37 compound laminas, 3 conical shell, 35 hemisphee on slope, 4 hemispheical shell method a, 33 hemispheical shell method b, 34 hemispheical shell method, 38 laminas, 6 pyamids, 36 secto of cicle, 9 solids of evolution, 3 standad esults fo laminas and acs, 3 standad esults fo unifom bodies, 37 tilting bodies, 39 Foce impulse of vaiable foce, 9 vaying with speed, 4 Gavitation Newton's law, Hooke s Law elastic spings, 5 elastic stings, 5 enegy stoed in a sting o sping, 7 Impulse vaiable foce, 9 Motion in a cicle acceleation towads cente, 7 angula velocity, 7 banking, 9 conical pendulum, 9 hoizontal cicles, 8 inveted hollow cone, vetical cicles, vetical cicles at end of a od, 4 vetical cicles at end of a sting, vetical cicles inside a sphee, 3 vetical cicles on outside of a sphee, 5 Simple hamonic motion a sin ω t and a cos ω t, 3 amplitude, 3 basic equation, 3 hoizontal stings o spings, 5 peiod, 3 v = ω (a - x ), 3 vetical stings o spings, 6 Wok vaiable foce, 4 M3 JUNE 6 SDB