Thomas Whitham Sixth Form Mechanics in Mathematics

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1 Thomas Whitham Sixth Form Mechanics in Mathematics 6/0/00 Unit M Rectilinear motion with constant acceleration Vertical motion under gravity Particle Dynamics Statics

2 . Rectilinear motion with constant acceleration t = 0 t a u v s Remember that s is a displacement, is directed, and should be shown with one arrow head. v u at s ut at v u as s u vt Example A cyclist moves along a straight line passing through points O, A and B with constant acceleration. seconds after passing O he is at A where OA = 9m and after a further 4 seconds he is at B where AB = 6m. Find his constant acceleration, his speed at O and his speed at B. a t t = t = 6 u 9m A v B s ut at O A O B 9 u a 4 6u 8a 8 a 4m a.m / s, u m / s

3 v u at O B v. 6 m / s Properties of the velocity/time graph Gradient = acceleration Area under graph = distance travelled Example A train starting from rest is uniformly accelerated during the first ½ km of its run, maintains its acquired speed for the next ½ km, and is then brought to rest with uniform retardation in the last ¼ km. The time for the whole journey is minutes. Find the acceleration in the first part of the run and the retardation in the final stage. velocity v t t t time Area v 00 t t v 000 t v 00 t v 0 t v 00 t v tv tv 000 v t t t 000

4 v v v 0m / s t 00 and t 0 Gradient st stage acceleration = m / s rd stage deceleration =. Vertical motion under gravity 0 0 m / s The assumption here will be that the motion is unrestricted, so that the constant acceleration formulae can be used with g usually given as having magnitude 9.8 m/s Example A ball is thrown vertically upwards with speed 4.7 m/s from a platform 9.6m above level ground. Find the time for the ball to reach the ground (g = 9.8 m/s ) State any assumption that you make. 4.7m/s Notice here how the displacement is shown 9.8m/s from the platform downwards and NOT 9.6m from ground upwards. This is very important. t

5 s at ut t 9.8t 4 t t 4 0 t t 0 Assume no air resistance during motion. t s. Particle Dynamics Applying Newton You need never write F ma, but if you insist on doing so, understand that the F is the resultant force which produces the acceleration. Example A man of mass 7kg carrying, in his right hand, a suitcase of mass 0kg steps into a lift. Calculate (i) the reaction on his feet as the lift accelerates upwards at.m/s (ii) the force on his right hand due to the suitcase as the lift decelerates at.m/s whilst moving upwards. (g = 9.8 m/s ) (i) Model the man plus suitcase as a particle of mass 8kg R 8.m/s N R 8 8. R 960.N

6 (ii) F.m/s This diagram shows the forces acting on the suitcase. N 98 F 0. F 8N 98 Example A particle of mass kg is placed on a rough plane ( 0. ) (inclination tan 4 with the horizontal) and on release from rest accelerates down the plane at.m/s. Find the air resistance acting on the particle assuming it to be constant. Take g = 9.8 m/s.m/s N F+R Perpendicular to plane N 9.6cos. 68 Sliding N F N down plane 9.6sin.6 R. R.984N

7 Example 6 A particle of mass m is held on a smooth plane which is inclined at 0 to the horizontal. It is connected by a light inelastic string passing over a smooth pulley at the top of the plane to a particle of mass M which hangs freely. The two parts of the string lie in one vertical plane. On release, the particle on the plane accelerates up the plane at g m/s. Find the ratio of M to m, and the force on the pulley in terms of M and g. g m/s N T T g m/s N mass M Mg T M. g Mass m 0 T mgsin 0 m. g 0 Mg mg mg 9 M 0 Mg mg m M : m : Mg

8 The resultant force on the pulley is given by R T cos 0 Mg Mg A further assumption used here is that there was no air resistance. Hookes Law For a string or spring of natural length l and modulus of elasticity the tension T in the string or spring when extended by e is given by e T l so long as e and l have the same units, the units of will be Newtons. Play safe and always express e and l in metres. Be aware that extended length means For a spring, the corresponding law when it is compressed by an amount c is T c l l e where T is the thrust in the spring. This law will apply to light elastic strings or springs which will return to their natural length when not under tension or compression. Example A particle of mass 00g rests on a rough horizontal plane with coefficient of friction equal to 7 4. One end of an elastic string, modulus of elasticity equal to 7N and of natural length 0cm, is fastened to the particle. The other end of the string is pulled horizontally until the particle is just on the point of sliding. Find the extension of the string in this position. (g = 9.8 m/s )

9 8 T N T Vertical N 4. 9 Limiting Equilibrium F N 4 7 F F. 8 Horizontal T F T. 8 HL T e l 7e.8 0. e 0.m 0cm Impulse and Momentum If a constant force P N acts on a particle for a time t s then the Impulse I Ns of the force is given by I Pt The Impulse of the force is also given by the change in momentum which it produces. Example A particle of mass 0.kg is moving along a straight line with speed m/s when it is struck a blow which reverses its direction and propels it with speed m/s. Find the Impulse of the blow, and the magnitude of the force given that contact is for 0.04 seconds. Before After I m/s m/s

10 Impulse/Mom I Ns 9 The force P is given by P P 40N Pt I Direct collision of two particles In general, two equations to write down (i) (ii) Linear momentum conserved Newton s experimental law which states that Speed of separation = e x speed of approach Where e is the coefficient of restitution and 0 e Inelastic collision Perfectly elastic collision Example Before Particles A and B of masses m and km respectively are moving towards each other along the same straight line, and each with speed u. The coefficient of restitution at the collision is equal to ½. Find the range of values of k for which the particles move in opposite directions after the collision. u u After x m km y Mom kmy mx mu kmu

11 ky x u ku NEL y x u u Adding equations y x u 0 ky y u ku y k u k k y u k k x u k follows Requiring x 0 and y 0 k 0 and k 0 k and k i.e. k Direct collision of a particle with a plane Newton s experimental law states that, speed of separation is equal to e x speed of approach. Example A ball is dropped from a height of. m onto a horizontal floor and rebounds to a height of 0.4 m. Find the coefficient of restitution of the ball, and find the impulse of the force on the ball at the bounce given that the mass of the ball is 0grammes (g=9.8m/s )

12 v u as falling v g..m v g v I 0.4m ev rising 0 e v g 0. 4 e v 0. 8g e g 0. 8g Impulse/mom I mev mv mev mv mv e Ns e 0.6 e Statics Particle Statics Example ABCD is a rectangle with AB = 8m, AD = 6m. The point P lies on DC where DP =.m Forces acting at A are 7N along AB, 0N in the direction CA, N along AP and 6N along AD. Find the magnitude of the resultant force and the angle it makes with AB.. D P C Sorting out the geometry first! 6 A 8 6 B tan

13 sin tan sin and. 6 and cos 4 cos Suitable directions for resolution will be along AB and along AD D P C 6 0 A 7 B ResAB ResAD X 7 0cos sin Y 6 cos 0sin 6 0

14 R R 4 R 4 0N 60 4 tan 4 tan Summary of technique Sort out geometry of figure. Find the sum of resolved parts in two suitable directions at right angles; Combine using the rectangle law. Example A particle P of weight W is in free suspension at the end of a light inextensible string fastened to a fixed point O. A force is now applied to the particle, always in the direction perpendicular to OP, until the string makes 0 with the downward vertical through O. In this position, find in terms of W, the magnitude of the force and the tension in the string. O 0 T F Resolve Perpendicular string F W sin 0 0 F W Resolve Parallel to string T W cos 0 0 W 0 T W

15 4 In general Choose suitable directions at 90 and equate to zero the sum of the resolved parts of forces in these directions. Example The diagram shows a particle of mass m on a rough inclined plane, coefficient of friction. It is at the end of a light inextensible string which is being pulled at an angle to the line of greatest slope of the plane which is at to the horizontal. Find the least force which will just drag the particle up the plane. First recall that the contact force between particle and plane is resolved into two components, for sake of brevity called frictional force [F] and normal reaction [N] Also recall that at all times F N When sliding, or on the point of sliding, F N F N. Otherwise Finally just move means on the point of moving i.e. limiting friction has been T reached. N Perpendicular to plane F N T sin mg cos Parallel to plane mg

16 T cos F mgsin Lim Friction F N T cos mgsin mg cos T sin T cos sin mg cos sin cos sin T mg cos sin Rigid Body Statics Examples involving bodies in equilibrium under parallel forces are usually in the form of horizontally supported beams. Any variations are straightforward. A standard technique involves taking moments about a suitable point (once) and resolving vertically. Example A uniform beam of mass kg is supported in a horizontal position by means of two equal vertical wires. One wire is attached to an end of the beam and the other to a point ¼ the length of the beam from the other end. Find the tensions in the two wires. T a a a g T A Mom A g a T a Vertical T 0g T T g T g

17 6 Centre of gravity / centre of mass The centre of gravity of a rigid body is the point through which the weight acts. Many are determined by symmetry E.g. Uniform rods, rectangular laminae, circular laminae, cuboids. For a uniform triangular lamina the centre of gravity is the way down any one of the medians. a a G For point masses placed in the x/y plane, the formula for the centre of mass is mx x, m y This formula is used in finding the centre of mass of composites and remainders. Recall here that, for a rigid body in free suspension, the centre of gravity will lie vertically below the point of suspension. Example This uniform lamina is to be freely suspended from A. Find the position of the centre of gravity and the angle which AB will make with the downward vertical. A 8cm B my m 6cm D cm C

18 7 O G 6 x y Area Mass Centre of Gravity Triangle 9 cm m (, ) Rectangle 0 cm 0m (., ) Composite - m x, y x y mx 6m m m m my 6m 0m m m 6 6 OG will be vertical and y 6 6 tan tan x 6 6 Example The following 6cm x 4cm rectangular lamina has a circular hole punched out, of diameter cm. The centre of the hole is cm from one of the shorter sides and cm from both long sides of the rectangle. To find the coordinates of the centre of gravity.

19 y 6 8 y by symmetry 4 O x Area Mass Centre of Gravity Rectangle (composite) 4 4m x Circle m 4 Remainder - 4m-m d x mx m 4m 7 4 d 7 4 d 4 4m m d 4m 4 NOTE THE TECHNIQUE HERE put the circle back, and it forms a composite with the remainder. The composite is the original rectangle.

20 9 Notes

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