Exam Question 6/8 (HL/OL): Circular and Simple Harmonic Motion. February 1, Applied Mathematics: Lecture 7. Brendan Williamson.

Transcription

1 in a : Exam Question 6/8 (HL/OL): Circular and February 1, 2017

2 in a This lecture pertains to material relevant to question 6 of the paper, and question 8 of the Ordinary Level paper, commonly referred to as circular motion, or simple harmonic motion. Here we will learn how to describe and study the motion of planets and moons moving on orbits, and the vibrations of strings, among others. However we will start by defining the radian, an alternative unit of measurement for angles that we will use instead of degrees. :

3 Consider the angle θ below, inscribed in a circle of radius r: in a : We say that θ = s r radians. Notice that the same angle in a circle of larger/smaller radius would have a proportionally larger/smaller arc length, so that this definition is consistent. By this measurement, rather than having 360 degrees in a circle, we have 2π radians. Furthermore 1 radian is equal to 360 2π degrees. In general to convert from degrees to radians we multiply by 2π 360 = π 180.

4 in a : This gives us the following special angles measured in radians: 30 = π 6 radians 45 = π 4 radians 60 = π 3 radians 90 = π 2 radians 180 = π radians 360 = 2π radians. One of the major uses of radians are the following limits: sin x lim x 0 x cos x 1 = 1, lim = 0, x 0 x which are only true if x is an angle measured in radians. These are used to prove that d dx sin x = cos x, d dx cos x = sin x, which again are only true if x is measured in radians.

5 in a : We now direct our attention to an object that moves with constant speed on a circle. Although its speed is constant, if an object is moving around the circumference of a circle then its direction is changing. Therefore it is accelerating. What is the size and direction of this acceleration? It seems reasonable to believe that the direction of acceleration is towards the centre of the circle; after all three elementary examples of objects that travel in a circle are the object swinging on the end of a rope, the object rolling around the side of a circular arc (like a person on a waterslide) and a moon or planet in the orbit of a larger object. In all three cases the only significant force on the object points towards the centre of the circle; either tension from the rope, the normal reaction from and perpendicular to the arc, and from the larger celestial object.

6 in a : A sketch of the proof is as follows. If the speed of the object moving in a circle is constant and equal to v then its angular speed is also constant; after t seconds the arc length travelled is vt and so the angle that arc subtends is ω = vt r radians. Assume that at time 0 we are at the East point on the circle, moving anti-clockwise. Then v(0) = 0 i + v j. At time t we have rotated ω radians anti-clockwise, so v(t) = v sin(ω) i + v cos(ω) j ( vt ) = v sin i + v cos r a(t) = d ( vt v(t) = v2 dt r cos r = v 2 ) i v 2 r cos (ω) i v 2 ( vt r ) j ( vt r sin r r sin (ω) j It is easy to see that a(t) = v 2 r. Also, a(t) points towards the centre of the circle. ) j

7 Example in a : We therefore have the following relationship between constant speed v, angular velocity ω, acceleration magnitude a and the magnitude F of the force (called centripetal force) that causes the acceleration: v = ωr, a = v 2 r = ω 2 r, F = ma = mω 2 r. Example: A particle of mass 5 kg moves with constant speed 3 m/s around a horizontal circle or radius 2 m. Find its angular velocity, the time it takes to complete 1 revolution, and the centripetal force on the particle. As r = 2, v = 3, ω = 3 2 radians. As the speed of the object is 3 m/s and the circumference is 4π, the time it takes to complete 1 revolution is 4π 3 s. Alternatively we could divide a full revolution, 2π, by ω. The centripetal force if F = mω 2 r = 45 2 N.

8 Example in a : Example: A particle moving in a circle of radius 3 m completes 10 revolutions every second. Find its angular velocity, speed and acceleration. We know immediately that ω = 10(2π) = 20π. Therefore v = 60π, a = 1200π 2. Example: A particle of mass 4 kg is connected to a string of length 5 m, which in connected to the ceiling 3 metres above a horizontal table. The particle is swung so that is describes a horizontal circle on the table directly below the anchor point of the string, completing 1 revolution every 2 seconds. Find the tension in the string. This is a visual description of the situation:

9 Example From Pythagoras Theorem we have that the radius of the circle is 4 m. It follows that ω = 2π 2 = π radians, v = 4π, a = 4π2. To find the tension in the string we resolve it into its horizontal and vertical components. The horizontal component of the tension T is equal to 4 5T, and is equal to the centripetal force acting on the particle. So 4 5 T = 12π2, T = 15π 2 N. in a :

10 Example: Banking in a : Example: A car turns a circular horizontal corner of radius 5 m at a speed of 10 m/s. The road is rough and the coefficient of friction between the road and the car s tyres is µ. For which values of µ will the car not skid? In one sense it is the steering wheel that it causing the car to turn. From a physics point of view though it is the friction between the car and road that is stopping the car from skidding out, so it is the centripetal force. So F = µr = m v 2 r = 20m, where R is the normal reaction between the car and the road and we assume that µ is the smallest value it can be before the car slips. As R = mg, we get µ = g. So if µ g the car won t skid. Example: Now imagine the road above is smooth, but is also banked at an angle θ. At what speed should the car move so that it doesn t slide?

11 Example: Banking The following is a diagram of all the forces acting on the car. in a : Here the centripetal force will be the horizontal component of R. As the car doesn t slip there is no vertical motion and so Mg = R cos θ Mg cos θ = R. As we need the centripetal force to equal M v 2 r = M v 2 5, we get R sin θ = M v 2 5 Mg cos θ sin θ = M v 2 5 5g tan θ = v.

12 Example: Banking in a : Now imagine we have a motorbike instead of a car, and revert to the case where the road is flat and rough, now of radius r. As centripetal force acts to pull the bike inwards, and as this force comes from friction it acts on the bottom of the bike s wheels. If the bike was upright it would then cause the bike to topple. This is why bikes bank when they go around corners. So what angle do we bank at? If we bank at an angle θ to the road then the following forces are in place on the bike, where henceforth we assume that the mass of the bike is M and its centre of gravity is 1m up the bike.

13 Example: Banking in a : F is the centripetal force, and R is the normal reaction between the bike and the ground. In this case our two equations for force come from the fact that there is no vertical movement, which means that R = Mg, and that the bike is not rotating so that the moments around the centre of gravity of the bike is 0. Equating the clockwise and anti-clockwise moments around this point gives us F sin θ = R cos θ tan θ = R F. If the radius of our turn is r, then F = Mv 2 r, so that tan θ = gr v 2. This may seem like a counter-intuitive answer, that even though the road is rough we have an exact answer with no apparent wiggle room. What s actually happening is that if we change the banking angle or the velocity then the radius of the turn must change, so that the above equation will hold true. There is a small amount of wiggle room coming from the thickness of the bike tyres, but something must give or the bike will topple.

14 Example Example: A ring of mass 2 kg is threaded through a string of length 3 m. The two ends of the string are attached to two points p and q of a vertical wire 2 m apart. The ring describes a horizontal circle with centre q. Find its angular speed ω. The following is a visualisation of the system. in a : Because we know that the total length of the string is 3m, we know that 3 = r r 2 which gives us r = 5 6. To get the angular velocity we need to relate the tension in the string to the centripetal force.

15 Example First, although the string appears separated into two parts, the tension on the diagonal and horizontal sections are the same. So resolving the tension in the diagonal part of the string into its horizontal components, we get the following forces acting on the object, in a : where θ is the angle between the two components of the string. So cos θ = , sin θ = 13 and as there is no vertical movement T sin θ = 2g and so T = 13g 6. Therefore the centripetal force is T cos θ + T = 3g. As 3g = mω 2 9g r, ω = 5 = 4.2 radians per second.

16 in a in a : We will now consider objects moving in a vertical circle. The most common occurrences of this are objects moving down the inside or outside of a sphere and objects swinging on a string. We saw this before when studying momentum, and we used the principle of conservation of energy. We will do the same here. Example: An object, starting at the bottom of the inside of a sphere of radius r, is projected up the sphere with velocity u. The object falls off the wall of the sphere when the angle between the object and the centre of the sphere is 45 above the horizontal. Find u. When the line between the object and the centre of the sphere makes an angle θ with the horizontal the following forces are acting on the object.

17 Example in a : The centripetal force is therefore R + Mg sin θ. To see why we have the sign we do, notice that gravity detracts from the centripetal force when θ < 0 and contributes to it when θ > 0. If we have velocity v at this time, R + Mg sin θ = Mv 2. r Also, due to the principle of conservation of momentum, Mg(0) Mu2 = Mgh Mv 2 = Mg(r + r sin θ) Mv 2. At the point when the particle falls off the wall off the sphere, R = 0 and θ = 45, so that g = v 2 2 r, 1 2 Mu2 = Mg(r + u = r ) + Mgr = Mgr gr

18 Example in a : Now consider the inverse problem. We start at the top of the outside of the sphere, and roll down the side with initial speed u. We leave the sphere when θ = 60. Find u. In this case, following a similar diagram we have that when speed is v and the angle in question is equal to θ, that and Mg sin θ R = Mv 2 Mgr Mu2 = Mgr sin θ Mv 2. When the object leaves the sphere R = 0 and θ = 60, so that 3Mg = Mv 2, 2 r Mgr Mgr 2 Mu2 = Mgr u = g r

19 Hooke s Law in a : It is an experimental observation that if we apply a certain force to extend a spring or elastic string, that the length that the spring or string extends beyond its natural length is proportional to the magnitude of the force, with the constant of proportionality referred to as the elastic constant, measured in Newtons per metre. Mathematically this is often written as where l 0 is the natural length. F = k(l l 0 ), Example: A string of natural length 1m has one end tied to the ceiling and the other tied to a mass of 5kg, and with this setup its length is 3m. Find the elastic constant. The force applied is the weight of the object, 5g. Therefore 5g = 2k k = 24.5N/m.

20 Hooke s Law: Example in a An object with mass 2kg is attached to one end of an elastic string of natural length 1m and elastic constant 5 N/m. It is swung in a horizontal circle with angular speed 1 radians/sec. Find the length of the string. If the length of the string is r, the centripetal force is mω 2 r = 2r. This is the force exerted on the string, so that 2r = 5(r 1) r = 5 3. :

21 Hooke s Law: Example in a Two nails on a horizontal table are 5m apart, each with a string attached to it, of natural lengths and elastic constants of 1m and 2m, and 3N/m and 4N/m respectively. The other ends of both strings are attached to an object of mass M. At equilibirum how far will the object be from each nail? The following is a visualisation of what is going on; assume the object is infinitesimally small but blown up for the purposes of the diagram. : If the system is in equilibrium then the forces sum to 0, which gives us 3(x 1) = 4(5 x 2) x = So the object is 15 7 m from one nail and 20 7 m from the other one.

22 in a : Now consider the previous problem, but where both strings have identical natural lengths and elastic constants of, say, 2m and 5N/m. Then the system will be in equilibrium when the particle is 2.5m away from each nail. But if we moved the particle a distance 0.3m to the right of this equilibrium the particle will oscillate. Say at some time t we are x(t) away from the equilibrium point; the force applied to the object is 5(0.5 x) 5(0.5 + x) = 10x. As F = ma, if we have a mass of say 2.5kg then we have a = 4x. Also, as a = d2 x we have a differential equation for dt 2 x(t). To solve it we also need x (0), which equals 0 as we simply released the object. Solving a second order differential equation requires replacing d2 x dt 2 with v dv dx where v is velocity so that we have v dv dx = 4x v 2 = 4x 2 + C.

23 : Example in a : As v = 0 when x = 0.3 we get v 2 = 4(0.09 x 2 ) v = ± x 2 dx = ±2 dt 0.09 x ( 2 sin 1 x ) = ±2t + C 0.3 x = ±0.3 sin(2t + C). Notice that as sin(2t + π) = sin(2t) there is a redundancy in making a choice for the ± sign and for C, so if we let ± = + then 0.3 = 0.3 sin(c) C = π 2 x = 0.3 sin ( 2t + π ). 2

24 : General Formulae in a : In general if an object with displacement x and acceleration a satisfies a = ω 2 x then we say the object performs simple harmonic motion, or. The general solution is x = A sin(ωt + C), where A and C depend on the initial conditions. A is the maximum value of x, and so is referred to as the amplitude of the object. In our previous situation, if we had move our object a certain distance away from its equilibrium and projected it with a non-zero velocity then as long as the velocity was small enough so that x(t) 0.5 for all t (so that neither string goes slack) the motion would look similar to the specific situation we dealt with. A would be the maximum distance we would travel from the equilibrium point.

25 Finding A and C, and some general formulae in a : If an object starts away from the equilibrium with zero velocity (like our first situation), then C = ± π 2 and x(t) = A sin ( ωt ± π 2 ) = ±A cos(ωt), where we decide on the sign based on whether or not we start at the left or right extreme point. If the object starts at equilibrium with non-zero velocity then C = 0, π, so that x = ±A sin(ωt) by similar calculations. In this case the sign is the same as the sign of the initial velocity. With general x(0), x (0), then as x (t) = ωa cos(ωt + C) we get x(0) = A sin(c), x (0) = Aω cos(c) tan C = ω x(0) x (0). We can see from the signs of x(0), x (0) whether sin(c), cos(c) are positive or negative, which gives us the correct quadrant for C. We can then use other equations to find A.

26 Some General Formulae in a : Along with the formulae for x in the specific case of C = 0, π 2, we came across the formula v 2 = ω 2 (A 2 x 2 ). This is a useful way of finding the velocity of the object when it is a given distance from equilibrium without using the solution for x(t). From this we can see that velocity is maximised when x = 0, i.e. at equilibrium, and is equal in magnitude to ωa. Similarly from a = ω 2 x we can see that the magnitude of acceleration is maximised when x = ±A, i.e. at the extreme points, and this magnitude is equal to ω 2 A. The time it takes to complete one cycle of motion, the period or periodic time, is given by 2π ω. This is the amount of time it takes for an object to return to its extreme point, or how long it takes for an object to return to any other point while travelling in the same direction. Finally, the average speed over one cycle is also worth noting. In a cycle we cover a distance of 4A. It takes 2π ω to do this, and so average speed over a cycle is 2Aω π.

27 Example in a : An object performs with amplitude 3m and periodic time s. Find its maximum speed and its average speed over a cycle. π 2 In this case we know that A = 3, 2π ω = π 2, so that ω = 4. Therefore the maximum speed is Aω = 12m/s and its average speed is 24 π m/s. How long does it take the same object to move from the equilibrium point to 1m away? We can assume that at t = 0 we are at equilibrium. We can also assume v(0) > 0 so that we are finding the first positive time that x(t) = 3 sin(4t) = 1. This occurs when t = 1 4 sin 1 ( 1 3).

28 Example in a : An object performing is travelling at 1m/s when it is 2m from equilibrium and at 3m/s when it is 1m from equilibrium. What is its maximum velocity? Like before we want A and ω, which we will get from the equation v 2 = ω 2 (A 2 x 2 ). We have 1 = ω 2 (A 2 4), 9 = ω 2 (A 2 1). By dividing one equation by the other we get 1 9 = A2 4 A A = 8. Putting this into either of the other two equations gives us 8 ω = 3, so that the maximum velocity is ωa = 35 3 m/s.

29 Example in a : An object moves so that at time t its displacement from some fixed point is given by x = 3 sin(2t). Show that this particle performs, and find how long it takes for the object to travel between its extreme points. Showing that an objects performs amounts exactly to proving that d2 x = kx for some k. In this case dt 2 d 2 x = 12 sin(2t) = 4x. dt2 In this case we can proceed using all of our previous formulae with A = 3, ω = 2. As the periodic time for the object is 2π ω, the time it takes to travel between its extreme points is π ω = π 2.

30 Infitesimally Small Elastic String in a : Consider a string of natural length l 0 and elastic constant k that is attached on one end to a nail on a flat table and on the other end to an object of mass m. If the object is pulled a distance greater than l 0 from the table and released, if x(t) is the displacement of the object at time t then while x > l 0 we have ma = k(x l 0 ) a = k m (x l 0). If l 0 is very small then a k mx and so we can approximate the k motion of the object with our formulae with ω = m.

31 Object Hanging From a Ceiling in a : An object of mass 2kg is hanging from a string of natural length 3m and elastic constant 4.9N/m attached to the ceiling. Find the current length of the string. If x is the current length of the string, then we have 4.9(x 3) = 2g so that x = 7. If the object is pulled up 3 metres from its equilibrium position and then released, show that it performs and find its displacement from its equilibrium at time t. If we are at a height x(t) above the equilibrium at time t (where x may be negative), then the total force on the object is 4.9((7 x) 3) 2g = 2a, so that a = 2.45x, so that it performs with ω = Furthermore its amplitude is 3 and we will use cos as we start at a positive extreme point, so that x(t) = 3 cos( 2.45t).

32 Coordinates of displacement vector under circular motion in a : Consider an object moving in a horizontal circle with uniform speed. If you viewed this object side on it would appear to perform. More mathematically, each coordinate of its displacement vector satisfies d2 x = kx for some k. This can be dt 2 seen quite directly, if an object is moving in a circle or radius A then its displacement is s(t) = A cos θ i + A sin θ j where θ is the angle the line between the object and the centre of the circle makes with the x axis. If speed is constant then so is angular velocity, so that dθ dt = ω which implies that θ = ωt + C. Therefore the coordinates of s are A cos(ωt + C) and A sin(ωt + C), which both satisfy the differential equation d2 x = ω 2 x. dt 2

33 Example: Object in Water Example: A cylinder of radius r and height h with relative density s is submerged upright in water. If it is pushed down an additional distance A and released. Show that it will perform and find its height above its equilibrium at time t. The above is a very basic visualisation of the situation. in a :

34 Example: Object in Water in a : Here we will us a basic fact from the hydrostatics section of the syllabus, that the water applies an upward force on the object which is equal to the weight of the water displaced. So when the water level is at a height y up the cylinder, the volume of the water displaced is πr 2 y m, and as the density of the water is 1000 kg/m 3 the weight of the water displaced is πr 2 y(1000)g. There is also the weight of the cylinder itself which constitutes a downward force of πr 2 h(1000s)g. Therefore the cylinder is at equilibrium when 1000πr 2 hsg = 1000πr 2 yg hs = y. So when the cylinder is at a height x above equilibrium, the resultant force is πr 2 (hs x)(1000)g πr 2 h(1000s)g = 1000πr 2 gx. As F = ma = πr 2 h(1000s)a, we have 1000πr 2 hsa = 1000πr 2 gx a = g hs x.

35 g hs So we have with ω =. Our amplitude is also A, and we ( ) start when x = A so that x(t) = A cos g hs t is the height of the cylinder above its equilibrium at time t. in a :

36 Another important example of an object that performs is a pendulum swinging along an arc. Consider an object of mass m attached to an inelastic string of length L hanging from the ceiling, where the object is swinging in an arc. Then when the string makes an angle θ with the vertical the following diagram shows all the forces acting on the object. in a :

37 in a : As we can see the force in the direction of the motion of the object is mg sin θ, so that a = g sin θ. However we would like to relate acceleration and displacement x. To do this, first note that when θ is small and measured in radians, θ sin θ (even when θ = 1 4 radians the difference is about 1%). Furthermore θ = x L from the definition of radians, so that we have the approximate equation a = g L x, so that the object approximately performs with ω = g L. This tells us that the periodic time of this is 2π L g. So, as expected, periodic time increases as the length of the pendulum is increased, and decreases if we move to a lower gravity location. Interestingly the periodic time is independent of the amplitude of the pendulum swing (in reality the periodic time would be slightly increased but because we replaced sin θ with θ we don t see this) and the mass.

38 Example in a : Example: If we double the length of a pendulum, by what factor is its periodic time multiplied? Say when the length of the pendulum is L the periodic time is T. Then T = 2π L g. So when the length is 2L the new periodic time is 2π 2L g = 22π L g = 2T. So the periodic time is multiplied by a factor of 2. Example: If we move a pendulum to the moon (where acceleration due to gravity is g 6, what is the ratio of the old and new periodic times? Say on Earth the periodic time is T = 2π L g where L is the length of the pendulum. Our new periodic time is 2π Lg 6 = 62π L g = 6T. So the ratio between the old and new periodic times is 1 : 6.