Semester 1 Revision Last modified: 05/06/2018
Contents Links Motion with Uniform Acceleration Equations Method Example Forces Equations Method Example Static Equilibrium Equations Method Example Energy Equations Method Example Momentum Equations Method Example
Motion with Uniform Acceleration For motion with uniform (i.e. constant) acceleration in one dimension, the motion quantities are connected by the equations: v = u + at s = ut + 1 2 at2 v 2 = u 2 + 2as Contents where u = initial velocity (i.e. at t = 0), v = velocity at time t, a = (constant) acceleration and s = displacement at time t. In two dimensions, the above equations apply in EACH dimension: v x = u x + a x t s x = u x t + 1 2 a xt 2 v 2 x = u 2 x + 2a x s x v y = u y x + a y t s y = u y t + 1 2 a yt 2 v 2 y = u 2 y + 2a y s y
Uniform Acceleration: Method Contents Draw a clear, labelled diagram. Think carefully about where your initial and final points are. List what you know. What are you trying to find? Which of the above equations connects this with what you know? Apply this equation. Exam Tip: Be careful with plus and minus signs. Remember that u, v, s and a are vectors, so direction is important. A lot of students forget and put a positve value instead of a negative (most often we see this with s).
Uniform Motion: Example Contents Bob the bird is standing on the lawn, happily eating worms. 3 m away from Bob, is the 2 m high garden wall. Fluffy the cat is on top of this wall, and decides that Bob would make a tasty snack. If Fluffy jumps horizontally, what speed does she need to jump at to catch Bob? First draw a diagram (sadly, there are no extra marks for artistic excellence!): V =? g = 10 m/s 2 2 m 3 m In this problem: t = 0 is when the cat jumps, and the final position is t = T when the cat arrives at the bird s position.
Contents In this problem and any other where the motion is in TWO dimensions, the best approach is to think about the x and y motions separately. List the known and unknown quantities: (all in SI units, note the MINUS signs on a y and s y ) x y a x = 0 a y = 10 u x = V u y = 0 v x =? v y =? s x = 3 s y = 2 t = T t = T We are aiming to calculate V which only appears in the x column. Using s x = u x t + 1 2 a xt 2 gives 3 = VT V = 3/T, so we need to calculate T in order to determine T. To do this we use the info about the y motion and the equation s y = u y t + 1 2 a yt 2 which gives 2 = 0 + 1 2 ( 10)T 2 2 T = 5 = 0.63 s. Therefore V = 3/T = 3/0.63 = 4.74 m/s. This problem is typical - we used the information about the y motion to find the time, which we then used in the x motion. In most problems we have to either do this, or the reverse - using x information to find the time, then using this in the y equations.
Forces: Equations Contents Newton s Second Law: The resultant force (i.e. sum of all forces) acting on a mass m is proportional to the acceleration a of the mass. F = ma Newton s Third Law: If there is a force F AB of object A acting on another object B, then there will be an equal and opposite force F BA of B acting on A: F AB = F BA Kinetic friction = µ K R Max static friction = µ S R
Forces: Method Contents Draw a clear, labelled diagram. Think carefully about what the motion, and thus the acceleration a, will be. Mark this on your diagram. If there is more than one object, then draw in a for each one. For EACH object, draw in ALL of the forces acting on it. If there is more than one object, it is a good idea to redraw a separate diagram for each object. Be careful not to leave any forces out: Weight (always DOWN) Tension (always AWAY from the object, always ALONG the string) Reaction (if the object is in contact with another object/surface) Friction (ONLY if there is a Reaction and µ 0 ) For EACH object, choose x and y directions to make calculations as simple as possible. This means minimizing the number of vectors needing to be resolved into components.
Contents For EACH object, apply Newton s Second Law (in both x and y directions) to obtain a set of equations involving several unknowns (usually, one of the unknowns will be a). Solve these equations. Exam Tips: Draw LARGE diagrams, to make everything clear, and to leave space for force components if required. Remember, actually solving the equations is generally only worth a small number of marks, so don t waste too much time with algebra.
Forces: Example Contents Two blocks of masses M and m, where M > m are connected by a massless string over a massless, frictionless pulley as shown below. Use Newton s laws to determine expressions for the acceleration a of the blocks, and T, the tension in the string. M µ m θ
Contents Clearly, in this problem, block M will fall DOWN, pulling block m UP the plane. The magnitudes of the accelerations will be equal. Draw a diagram, adding the accelerations of the blocks. a a M µ m There are TWO blocks involved here so we must think about the forces on the blocks separately. θ
Contents First draw a new diagram for block M and carefully add forces, one at a time. T Tension Weight There are only two forces, weight and tension, and so we know that F = Ma = Mg T down We ve chosen the down direction to be positive to make the equation simpler. Nearly always, we should choose the x-axis to be in the direction of a. Mg Mg
Contents We repeat this procedure for the other mass, again carefully drawing in the forces one by one. Tension Normal reaction T T R Weight mg mg Friction mg T a R x fr = µr mg y Again we choose the axes so that a is in the x direction. Then, using Newton s Second Law: F = ma and x F = 0 y
Contents With this choice of axes, only one force - weight - needs to be resolved into components. a Use the diagram to complete these equations: T R x F = ma = T mg sin θ µr mg cos θ y x θ F = 0 = R mg cos θ mg sin θ fr = µr y mg We now have THREE equations in THREE unknowns: a, T and R. Ma = Mg T (1) ma = T mg sin θ µr (2) 0 = R mg cos θ (3)
Contents The next steps are straightforward algebra. From equation (3), R = mg cos θ which we substitute into (2): ma = T mg sin θ µmg cos θ (4) Adding equations (1) and (4) now gives: a(m + m) = Mg mg sin θ µmg cos θ a = g(m m(sin θ + µ cos θ)) (M + m) (5) To get T, rearrange equation (1) and substitute a from equation (5): T = Mg Ma =... = mmg(1 + sin θ + µ cos θ) (M + m)
Static Equilibrium: Equations Contents Forces acting at different points on an object will tend to cause a rotation of that object. A force F acting at a displacement r away from a pivot point X creates a torque τ around X: τ = r F X r θ F so the magnitude of the torque is τ = rf sin θ The same force will produce different torques about different pivot points. In this course, we mostly encounter torque in Static Equilibrium problems. Conditions for Static Equilibrium: F = 0 F = 0 and F = 0 x y τ = 0 for any pivot point X X
Static Equilibrium: Method Contents Decide which object we are considering - usually a ladder/beam etc. Draw a diagram of this object, showing all forces acting on it. Be careful not to leave any out or add extra ones. If two objects are attached (by a hinge for example), then the direction of the reaction force is unknown - break it into x and y components. If an object is resting or leaning on a surface, then the reaction will be NORMAL, and there may also be friction. Use the conditions for static equilibrium to obtain THREE equations involving these forces. As in the forces problems, choose x and y directions to make calculations as simple as possible. Usually, the traditional directions are best. Any choice of pivot will work, but try to make a smart choice to simplify equations. Usually, this is a point with an unknown reaction acting. As you have a choice, it is important to clearly indicate what your choice is. Solve these equations
Static Equilibrium: Example Contents Bob the bird narrowly escapes from Fluffy, and flies to a seemingly safe position on a uniform beam, hinged at one end to a wall, and supported at the other by a rope as shown below. Still hungry, Fluffy climbs up after Bob. Bob is not worried however, as he topped his Physics class at Bird College, and has calculated that the rope will break before Fluffy can reach him. Is Bob correct? Calculate the distance x that Fluffy reaches at the moment the rope breaks. (Tmax = 50 N). 3 m M Fluffy = 1.5 kg M Bob = 250 g M beam = 4 kg x 1 m 4 m
Contents The main object we are looking at is the beam. We will only care about forces on the beam, not on the wall or the cat, or the bird. Draw a diagram, showing all forces: First the weights, not just of the beam, but any objects supported by the beam. Remember for the uniform beam, weight acts at the centre. 2 m 3 m x W Fluffy = 1.5g W beam = 4g W Bob = 0.25g
Contents Next the tension in the rope. 4 m T 1.5g 4g 0.25g Finally, we know there must be a reaction at the wall, but don t know exactly what direction it is in. It will make our calculations easier if we immediately break it up into x and y components. R y θ T R x 1.5g 4g 0.25g
Contents So our completed diagram looks like (after resolving T into x and y components): R y T T sin θ = 3 5 T H R x T cos θ = 4 5 T 1.5g 4g 0.25g Add forces in x direction: x Add forces in y direction: y F = 0 = R x 4 5 T F = 0 = R y + 3 5T 1.5g 4g 0.25g
Contents The best choice of pivot for calculating torques is at the hinge H. Make sure you specify this in your working. Take anticlockwise torque to be positive (if you prefer, clockwise will also work): H τ = 0 = +4( 3 T ) x(1.5g) 2(4g) 3(0.25g) 5 Rearranging this, and putting g = 10 gives: 15x = 12 5 T 87.5 x = 4 25 T 35 6 x max = 4 25 T max 35 6 = 4 35 50 25 6 = 2.17 m Which is less than Bob s position at 3 m along the beam. So Bob did calculate correctly and will be able to laugh at Fluffy plummeting to the ground after the rope breaks. Note that in this problem, the two force equations aren t actually needed because of our clever choice of pivot point, but in general all three equations will be needed.
Energy: Equations Contents Work W done by a force F acting through a displacement s is: W = F s The energy of a system is conserved: Estart + E input = E end + E loss Where the total energy can consist of: Kinetic energy = 1 2 mv 2 (for EVERY mass in the system) Potential energy (gravitational) = mgh (for EVERY mass) Potential energy (spring) = 1 2 kx 2 E input and E loss are the work done by forces external to the system, ading or removing energy. The commonest external force we see is friction, which always removes energy from the system.
Energy: Method Contents Choose a suitable system - usually all the moving objects. Think carefully about what you will use as your start and end positions. Draw TWO diagrams showing these positions. Calculate the total (kinetic + potential) energy at each of these positions. If there is friction acting, calculate the work done by the friction in between the start and end positions. Put these results together into the energy conservation formula. Solve this equation for the required quantity. Exam Tip: The words conservation of energy should appear in your answer
Energy: Example Contents Two masses are connected by a massless string over a massless. frictionless pulley as shown. Initially the blocks are at rest, and the spring is unstretched. If the block 2M is allowed to fall, it will drop a distance d before momentarily coming to rest. Determine an expression for d in terms of the other parameters given. µ M k 2M
Contents The correct choices of start and end positions should be clear. The start position is when both blocks are at rest with the spring unstretched, before they are released. The end position occurs after 2M has dropped distance d and both blocks are again momentarily at rest. v = 0 M M µ µ d 2M d 2M v = 0 start end
Contents Calculate the energy at the start position: Nothing is moving kinetic energy = 0 The spring is at its natural length, so spring potential energy = 0 If we take this position as our reference point for gravitational potential energy, then this also = 0 Estart = 0 Calculate the energy at the end position: Nothing is moving (again!) kinetic energy = 0 The spring is stretched by a distance d, so spring potential energy = 1 kd 2 2 Block M has moved sideways only, so has not changed its gravitational potential energy. Block 2M has moved down a distance d from the reference point, so must have negative potential energy = 2Mgd. E end = 1 2 kd 2 2Mgd
Contents Calculate the work done by friction: First we need to work out what the friction force acting on block M is, which requires us to first find the reaction R. Drawing in the forces on the block and using Newton s second law gives us R = Mg and thus that the friction fr = µr = µmg. T R = Mg Mg fr = µr = µmg The distance this force acts through is d, so the work done by the frictional force (i.e. E loss ) is (µmg)d Finally, put the pieces together: Estart = E end + E loss 0 = 1 2 kd 2 2Mgd + µmgd 1 kd = (2 µ)mg 2 d = 2(2 µ)mg/k
Momentum: Equations Contents The total momentum p = mv of a system is conserved during a collision or other interaction: p before = p after If we know how much kinetic energy is lost in a collision, then we can form an extra equation: KE before = KE after + KE loss In an elastic collision, the loss of kinetic energy is zero: KE before = KE after
Momentum: Method Contents The logic is very similar to energy problems: Choose a suitable system - usually all the moving objects. Decide on your before and after positions Calculate the total momentum of the system before and after. Remember, momentum is a VECTOR. If told the collision is elastic, calculate KE before and after also. Use momentum conservation, and if elastic, KE conservation to obtain equations. Solve equations. Exam Tips: The words conservation of momentum should appear in your answer Don t forget momentum is a vector!
Momentum: Example Contents Two masses, m and 2m, collide as shown below. After the collision they form two new objects, each with a mass 3 2 m. Calculate the speed V B and direction θ B of the object B. Also, determine whether or not the collision is elastic. u = 5 m/s m rest 2m y x V A = 1 m/s A M A = M B = 3 2 m α = tan 1 ( 4 3 ) θ B V B =? B (a) before collision (b) after collision
Contents This collision occurs in two dimensions, so we need to consider both x and y components of momentum. Before Collision: p x = 5m p y = 0 After Collision: ( ) ( ) ( ) ( 3 3 3 3 p x = 2 m + 5 2 m (V B ) x p y = 2 m ) ( 4 5 ) ( ) 3 + 2 m (V B ) y Conservation of Momentum: x direction: 5 m = 9 10 m + 3 2 m(v B ) x (V B ) x = 2.73 m/s y direction: 0 = 12 10 m + 3 2 m(v B ) y (V B ) y = 0.8 m/s
Contents Thus we have: V B = 2.73i 0.8j m/s The question asked for speed V B and direction θ B, which requires some routine vector maths: 2.73 V B = (2.73)2 + (0.8) 2 = 2.84 m/s θ B 0.8 Check if Collision is Elastic: and KE before = 1 2 m52 = 12.5m KE after = 1 2 KE before KE after, so the collision is inelastic. 1 0.8 θ B = tan 2.73 = 16.3 ( ) 3 2 m 1 2 + 1 ( ) 3 2 2 m 2.84 2 = 6.8m