3/18/2016 Basics v PA 199 Advanced Game Design Lecture 4 Physics for Game Design Dr. Fotis Liarokapis 16th March 2016 Units Some Basic SI Units The numbers that specify a point have units The spaces we will deal with are mostly regular kinds of space — i.e. Units of distance • i.e. meters and kilometres Almost all physics applications use metric units — Called SI (Systeme International) Quantity Base Unit Derived Units Distance Meter (m) 1 kilometer (km) - 1000 m 1 m = 100 centimetres (cm) Mass Kilogram (kg) 1 kg = 1000 grams (g) 1 g = 1000 milligrams (mg) Time Seconds (s) Temperature Kelvin (K) Prefixes Derived Units Original Language Celsius tempe rat u r arge, quantity of electricity Electric potential, potential difference, lectromotivi Energy, work, quantity of heat Frequency (of a periodic pher Power, radiant flu/ n Terms of Other SI 3/18/2016 Cartesian Coordinate System Polar Coordinate System In the polar coordinate system each point on a plane is determined by an angle and a distance Useful in situations where the relationship between two points is most easily expressed in terms of angles and distance - In the Cartesian coordinate system this can be found through trigonometry Each point is determined by two polar coordinates: - The radial coordinate - The angular coordinate 4D Coordinate Systems Motion • Well this is a bit hard to draw! - Even 3D is hard to draw • Mathematically though it is easy to add more dimensions - Just name points with more numbers • In 4D you need four numbers - The last dimension is usually called w - So points have coordinates (x, y, z, w) Displacement • A positive x value implies that the body is located x meters to the right of the origin • A negative x value implies that the body is located |x| meters to the left of the origin • Here, x is termed as the displacement of the body from the origin track origin displacement body + + + -H + Displacement Over Time x-l+t+ 2 4 2 3/18/2016 Velocity ID • Determine the body's instantaneous velocity as a function of time: — Velocity is the rate of change of displacement with time • This definition implies that: Ax u = —— At • where u is the body's velocity at time t, and Ax is the change in displacement of the body between times t and t + At Velocity General Equation Ax dx v = Jim —7— = — a*->u At dt • where dx/dt represents the derivative of x with respect to t • The above definition is particularly useful if we represent x(t) as an analytic function — It allows us to immediately evaluate the instantaneous velocity u(t) via the rules of calculus Velocity Graph V When u>0 the body is moving to the right When u<0 the body is moving to the left When u=0 the body is instantaneously at rest Velocity and Speed The terms velocity and speed are often confused with one another A velocity can be either positive or negative - Depending on the direction of motion The conventional definition of speed is that it is the magnitude of velocity - A body can never possess a negative speed Vector Velocity Consider a body moving in 3D, and we know the Cartesian coordinates, x, y, and z, of this body as time, t, progresses - How can we use this information to determine the body's instantaneous velocity as functions of time? The vector displacement of the body is given by: :{t) =[i(t), »(*),*(*)] Vector Velocity The body's vector velocity v = (ux, uy, uz) is simply the derivative of r with respect to t vtt) = Hm i(t+At)-r(t) _ di At di When written in component form, the above definition yields to: dx dy dt dz dt 3 3/18/2016 Acceleration ID • The definition of acceleration is as follows: — Acceleration is the rate of change of velocity with time • This definition implies that: A i a = —r-At • where a is the body's acceleration at time t, and Au is the change in velocity of the body between times t and t + At Acceleration General Equation • A general expression for instantaneous acceleration can be obtained by taking the limit of the acceleration equation as At approaches zero: Av dv d2z a — lim —- = — = —-At-a At dt dt' • which is valid irrespective of how rapidly or slowly the body's acceleration changes in time Acceleration Graph V Vector Acceleration When a is positive the "> body is accelerating to the right When a is negative the body is decelerating 10 The body's vector acceleration a = (ax, ay, az) is simply the derivative of v with respect to t afi) = lim v(t + At)-v(t) dv d2i At dt dt2 When written in component form, the above definition yields dvT d2x ~dt ~ ~dt* dvy _ tPy dt dt2 d2; 3z dt dt2 Motion with Constant Velocity ID v Simplest type of motion - Excluding the trivial case in which the body under investigation remains at rest Occurs in everyday life whenever an object slides over a horizontal, low friction surface Motion with Constant Velocity ID The graph consists of a straight-line which can be represented as: X = + V t Here, x0 is the displacement at time t=0 - This can be determined from the graph as the intercept of the straight-line with the -axis u = dx/dt is the constant velocity of the body - Can be determined from the graph Note that: a= y) = (rcosej), rsindp) Relative Velocity Simple Example • Suppose that the plane's velocity relative to the air is 300Km/h, at a compass bearing of 120°, and the air's velocity relative to the ground is 85Km/h, at a compass bearing of 225° • It follows that the components of va and u (measured in units of km/h) are: v _ (300 cos 120°, 300 sin 120°) = (-1.500 X 10!, 2.598 X ID3) u = (85 ma 225°, 85 sili225°) = (-6.010 X ID1, -6.01D X 101) 10 3/18/2016 Relative Velocity Simple Example . • According to (vg = va + u), the components of the plane's velocity vg relative to the ground are simply the algebraic sums of the corresponding components of va and u Vg = (-1.500 X 10z - 6.010 X W\ 2.598 X 10z - 6.010 X 101) = (-2.101 X 102, 1.997 X ID2) • The task is to reconstruct the magnitude and compass bearing of vector vg, given its components (vgx, vgy) based on Pythagoras' theorem vg= vV)1+ («„)' = ^/(-2.1D1 X 102)2 + (1.S97 x 102)2 = 2SB.9km/l Relative Velocity Simple Example .. • The compass bearing of vg is given by: • Because: = i>s oos^ and vsy = vs mnf> — Remember that: r = (x, y) = (rcoscfj, rsincfj) • Unfortunately, the above expression becomes a little difficult to interpret if is negative • An unambiguous pair of expressions for d> is: 4> = Um-1 (^j , v„.>0 $ = 180° - tan-1 (^) , < 0 Relative Velocity Simple Example • These expressions can be derived from simple trigonometry: 4> = 180° - tan _! (1.997 X 10z\ [2.IOI x 10z) • Thus, the plane's velocity relative to the ground is 289km/h at a compass bearing of 136.5° Example 4 - ID Relative Velocity A train travels at 60 m/s to the east with respect to the ground A businessman on the train runs at 5 m/s to the west with respect to the train Find the velocity of the man with respect to the ground Example 4 Solution • The velocity of the train with respect to the ground is: vTG = 60 m/s • The velocity of the man with respect to the train is: vMT= -5 m/s • Putting these together, we get: Example 5 - 2D Relative Velocity • An airplane flies at 250 m/s to the east with respect to the air • The air is moving at 35 m/s to the north with respect to the ground • Find the velocity of Air Force One with respect to the ground 11 3/18/2016 Example 5 Solution • Both vPA and vAG are two-dimensional vectors • You can find vPG by vector addition: -VPG=VPA+ ——y • From Pythagoras: v„ = 1/<250^)'+(35^)! = 252^ Example 5 Solution . • Can find the angle of using basic trigonometric functions: 9 = tan"1 —- = 8s Uso^J • Therefore, the velocity with respect to the ground is 252 m/s at an angle of 8° north of east Newtonian Physics V Newton's Laws of Motion Every body continues in its state of rest, or uniform motion in a straight line, unless compelled to change that state by forces impressed upon it The change of motion of an object is proportional to the force impressed upon it, and is made in the direction of the straight line in which the force is impressed To every action there is always opposed an equal reaction; or, the mutual actions of two bodies upon each other are always equal and directed to contrary parts Newton's First Law of Motion Newton's first law was actually discovered by Galileo and perfected by Descartes This law states that if the motion of a given body is not disturbed by external influences then that body moves with constant velocity In other words, the displacement r of the body as a function of time t can be written r = r„ + vi where r0 and v are constant vectors Newton's First Law of Motion The body's trajectory is a straight-line which passes through point r0 at time t = Oand runs parallel to v In the special case in which v = 0 the body simply remains at rest 12 3/18/2016 Newton's Second Law of Motion Newton used the word motion to mean what we nowadays call momentum The momentum p of a body is simply defined as the product of its mass m and its velocity: p = mv Newton's second law of motion is summed up in the equation: - where the vector f represents the net influence, or force, exerted on the object, whose motion is under investigation, by other objects Newton's Second Law of Motion . • For the case of a object with constant mass, the above law reduces to its more conventional form: { = TUB • The net force exerted on a given object by other objects equals the product of that object's mass and its acceleration — Note that this law is entirely devoid of content unless we have some independent means of quantifying the forces exerted between different objects Example 6 A railway engine pulls a wagon of mass 10 000 kg along a straight track at a steady speed. The pull force in the couplings between the engine and wagon is 1000 N - A) What is the force opposing the motion of the wagon? - B) If the pull force is increased to 1200 N and the resistance to movement of the wagon remains constant, what would be the acceleration of the wagon? Example 6 Solution (A) • When the speed is steady, by Newton's first law, the resultant force must be zero • The pull on the wagon must equal the resistance to motion • So the force resisting motion is 1000 N Example 6 Solution (B) Example 7 The resultant force on the wagon is: - Ftotal = 1200 - 1000 = 200 N From Newton's 2nd law of motion: F = ma 200 =100003 a = 0.02m/ s Find the acceleration of a 20 kg crate along a horizontal floor when it is pushed with a resultant force of 10 N parallel to the floor How far will the crate move in 5s - Starting from rest? 13 3/18/2016 Example 7 Solution Hooke's Law v The acceleration can be calculated using Newton's 2nd law of motion: F =ma 10 = 20a a - 0.5m/s2 Distance travelled is given by: s-0 + -0.5x5J 2 s = 6.25m Method of quantifying the force exerted on an object This law states that the force f exerted by a coiled spring is directly proportional to its extension Ax The extension of the spring is the difference between its actual length and its natural length The force acts parallel to the axis of the spring Hooke's law only holds if the extension of the spring is sufficiently small - If the extension becomes too large then the spring deforms permanently, or even breaks • Such behavior lies beyond the scope of Hooke's law Hooke's Law V Use Hooke's law to quantify the force we exert on a body of mass m when we pull on the handle of a spring attached to it The magnitude f of the force is proportional to the extension of the spring: - Twice the extension means twice the force The direction of the force is towards the spring, parallel to its axis - Assumingthat the extension is positive Hooke's Law The magnitude of the force can be quantified in terms of the critical extension required to impart a unit acceleration (i.e. lm/s2) to a body of unit mass (i.e. 1 Kg) According to f = ma the force corresponding to this extension is 1 newton Thus, if the critical extension corresponds to a force of IN then half the critical extension corresponds to a force of 0.5N, and so on Hooke's Law Example Suppose that we apply two forces, f1 and/2 The forces are acting in different directions, to a body of mass m by means of two springs • According to Newton's second law of motion, the body does not accelerate • It either remains at rest or moves with uniform velocity in a straight line Newton's Third Law of Motion Suppose, for the sake of argument, that there are only two bodies in the Universe Let us label these bodies a and b Suppose that body b exerts a force fab on body a According to to Newton's third law of motion, body a must exert an equal and opposite force fba = -fabort body fa Thus, if we label fab the 'action' then, in Newton's language, fba is the equal and opposed 'reaction' 14 3/18/2016 Newton's Third Law of Motion Suppose, now, that there are many objects in the Universe - As is, indeed, the case According to Newton's third law, if object j exerts a forceon object / then object / must exert an equal and opposite force■ = -fq on objectj It follows that all of the forces acting in the Universe can ultimately be grouped into equal and opposite action-reaction pairs Note, incidentally, that an action and its associated reaction always act on different bodies Newton's Third Law of Motion Why do we need Newton's third law? - Actually, it is almost a matter of common sense Suppose that bodies a and b constitute an isolated system \ffba * -fab then this system exerts a non-zero net force f=fab+fba on itself - Without the aid of any external agency - It will, therefore, accelerate forever under its own steam ■ -x b Mass, Weight and Strings Mass and Weight The terms mass and weight are often confused with one another but in physics their meanings are quite distinct A body's mass is a measure of its inertia - i.e., its reluctance to deviate from uniform straight-line motion under the influence of external forces According to Newton's second law (f = ma) if two objects of differing masses are acted upon by forces of the same magnitude then - The resulting acceleration of the larger mass is less than that of the smaller mass Mass and Weight Example Mass and Weight Example Imagine a block of granite resting on the surface of the Earth Earth He «1 ck Í The block experiences a downward force fg due to the gravitational attraction of the Earth of magnitude mg The block exerts a downward force fw, of magnitude mg, on the ground beneath it - We refer to this force as the weight of the block According to Newton's third law, the ground below the block exerts an upward reaction force fR on the block - This force is also of magnitude mg Thus, the net force acting on the block is: - fR + fg = 0 and thus the block remains stationary 15 3/18/2016 Calculation of Weight So far, we have established that the weight W of a body is the magnitude of the downward force it exerts on any object which supports it Thus, W = mg where m is the mass of the body and g is the local acceleration due to gravity - Note that since weight is a force it is measured in Newton A body's weight is location dependent and is not an intrinsic property of that body - For instance, a body weighing ION on the surface of the Earth will only weigh about 3.8N on the surface of Mars, due to the weaker surface gravity of Mars relative to the Earth Weight in an Elevator Consider a block of mass m resting on the floor of an elevator Suppose that the elevator is accelerating upwards with acceleration a How does this acceleration affect the weight of the block? w WS Weight in an Elevator V The block is subject to 2 forces: - A downward force mg due to gravity - An upward reaction force W Hence: W - mg = ma or W = m(g + a) The upward acceleration of the elevator has the effect of increasing the weight W of the block: - i.e. if the elevator accelerates upwards at g = 9.81 m/s2 then the weight of the block is doubled if the elevator accelerates downward (i.e. if becomes negative) then the weight of the block is reduced Strings Consider a block of mass m which is suspended from a fixed beam by means of a string |_ The string is assumed to be light (its mass is negligible compared to that of the block) and inextensible (its length increases by a negligible amount because of the weight of the block) H--string J Strings If we apply Newton's second law to the block, the mass of the block is m, and its acceleration is zero — Since the block is assumed to be in equilibrium The block is subject to two forces, a downward force mg due to gravity, and an upward force due to the tension of the string so: T - mg = 0 In other words, in equilibrium, the tension of the string equals the weight of the block Three Strings Example A slightly more complicated example in which a block of mass m is suspended by three strings is shown on the right What are the tensions, T, T-l and T2 in these strings, assuming that the block is in equilibrium? 16 3/18/2016 Three Strings Example Horizontal Components h A 60° T The horizontal component of tension T is zero, since this tension acts straight down The horizontal component of tension 11 is: TjCOs60° = V2 - Since this force subtends an angle of with respect to the horizontal Likewise, the horizontal component of tension T2 is T2cos30° = - V3 T2/2 Since the knot does not accelerate in the horizontal direction, we can equate the sum of these components to zero: Tj/2 = - V3 T2/2 Vertical Components • Consider the vertical components of the forces acting on the knot • Let components acting upward be positive, and vice versa • Since the knot does not accelerate in the vertical direction, we can equate the sum of these components to zero - mg + Tj/2 + V3 T2/2 = 0 Putting all Together • From the previous equations: V2 = -V3T2/2 - mg + V2 + V3 T2/2 = 0 • We can calculate Jx and T2 T1=V3 mg/2 T2=mg/2 Inclines, Pulleys and Friction Inclines Consider a block of mass m sliding down a smooth frictionless incline which subtends an angle 8 to the horizontal The weight mg of the block is directed vertically downwards cos 6 'ng cos 6 17 3/18/2016 Incline Weight Analysis Weight can be resolved into components: - mgcos6 acting perpendicular (or normal) to the incline - mgsin6 acting parallel to the incline Note that the reaction of the incline to the weight of the block acts normal to the incline, and only matches the normal component of the weight - It is of magnitude mgcos6 In general the reaction of any unyielding surface is always locally normal to that surface - Directed outwards and matches the normal component of any inward force applied to the surface Incline Weight Analysis Applying Newton's second law to this problem we obtain: df- ■■ Tn g sin 6 • which can be solved to give: x — xq +vut + —g sinť?ť2 Accelerating up a Slope V Accelerating up a Slope Suppose that the block, mass m= 5 kg, is subject to a horizontal force F = 27 N. What is the acceleration of the block up the (frictionless) slope? Only that component of the applied force which is parallel to the incline has any influence on the block's motion - The normal component of the applied force is canceled out by the normal reaction of the incline. The component of the applied force acting up the incline is Fcos255 The component of the block's weight acting down the incline is mgsin259 Accelerating up a Slope .. Hence, using Newton's second law to determine the acceleration ct of the block up the incline, we obtain: F cos 25° - m g sin 25° a —- m Since m = 5 kg and F = 27 N, we have: 27 X 0.9063 - 5 X 9.81 X 0.4226 . , a =---= 0.7483m/s' Breakout Application of Inclines Tip: - You could to add 3 inclines to the ground so that the ball hits the upper bricks of the well - Ball will perform a projectile motion 18 3/18/2016 Pulleys • Consider two masses, m1 and m2 connected by a light inextensible string • Suppose that the: - First mass slides over a smooth, frictionless, horizontal table - Second is suspended over the edge of the table by means of a light frictionless pulley Pulleys Weight Analysis • Since the pulley is light, we can neglect its rotational inertia in our analysis • No force is required to turn a frictionless pulley •Can assume that the tension T of the string is the same on either side of the pulley • Again apply Newton's second law of motion to each mass in turn Pulley - First Mass • The first mass is subject to a downward force rrijg due to gravity - However, this force is completely canceled out by the upward reaction force due to the table • The mass m1 is also subject to a horizontal force T due to the tension in the string, which causes it to move rightwards with acceleration T Or — - '7'1 Pulley - Second Mass • The second mass is subject to a downward force m2g due to gravity, plus an upward force T due to the tension in the string • These forces cause the mass to move downwards with acceleration: T Putting Everything Together The rightward acceleration of the first mass must match the downward acceleration of the second, since the string which connects them is inextensible Equating the previous two expressions: T= m1m2g/(m1+ m2) a = m2g/ (171!+ m2) Another Pulleys Example Consider two masses m1 and m2 connected by a light inextensible string which is suspended from a light frictionless pulley Again must apply Newton's second law to each mass in turn 19 3/18/2016 Another Pulleys Example. • Without being given the values of nij and m2 we cannot determine beforehand which mass is going to move upwards • Let us assume that mass m1 is going to move upwards: - Note that in this case we will obtain a negative acceleration for this mass Another Pulleys Example - First Mass • The first mass is subject to an upward force T due to the tension in the string, and a downward force rrijg due to gravity • These forces cause the mass to move upwards with acceleration: T a —--g jrii ^^^nother Pulleys Example - Second v Mass • The second mass is subject to a downward force m2g due to gravity, and an upward force T due to the tension in the string • These forces cause the mass to move downward with acceleration: T Putting Everything Together • The upward acceleration of the first mass must match the downward acceleration of the second, since they are connected by an inextensible string • Hence, equating the previous two expressions, we obtain: T = 2m1m2g / (m1 + m2) a = (m2-m1)g/(m1 + m2) Example 8 Example 8 Solution Consider the diagram The platform and the attached frictionless pulley weigh a total of 34N With what force F must the (light) rope be pulled to lift the platform at 3.2 m/s2 ? platform Let W be the weight of the platform, m = W/g the mass of the platform, and T the tension in the rope From Newton's third law, it is clear that T = F Let us apply Newton's second law to the upward motion of the platform The platform is subject to two vertical forces: - A downward force W due to its weight, and - An upward force 2T due to the tension in the rope • The force is 2T, rather than T, because both the leftmost and rightmost sections of the rope, emerging from the pulley, are in tension and exerting an upward force on the pulley 20 3/18/2016 Example 8 Solution . The upward acceleration a of the platform is: IT-W a — - 773 Since T = F and m = W/g, we obtain: W(a/g + l) F= -2- Finally, given that W = 34N and a = 3.2 m/s2, we have: ^ _ 34 (3,2/9,81 + 1) _ 22.55N Friction • When a body slides over a rough surface a frictional force generally develops which acts to impede the motion - Friction, when viewed at the microscopic level, is actually a very complicated phenomenon • The frictional force exerted on a body sliding over a rough surface is proportional to the normal reaction Rn at that surface, the constant of proportionality depending on the nature of the surface Friction Definition V Friction Example Definition: where u. is termed the coefficient of (dynamical) friction For ordinary surfaces u. is generally of order unity Consider a block of mass m being dragged over a horizontal surface, whose coefficient of friction is u. by a horizontal force F The weight W = mg of the block acts vertically downwards, giving rise to a reaction R = mg acting vertically upwards Friction Example The magnitude of the frictional force f which impedes the motion of the block, is simply times the normal reaction R = mg Hence f = u.mg The acceleration of the block is, therefore: Assuming that F > f Example 9 Consider the diagram where the mass of block A is 75kg and the mass of block B is 15 Kg The coefficient of static friction between the two blocks is u. = 0.45 and the horizontal surface is frictionless What minimum force F must be exerted on block A in order to prevent block B from falling? F-- A 21 3/18/2016 Example 9 Solution Suppose that block A exerts a rightward force R on block B By Newton's third law, block B exerts an equal and opposite force on block A Applying Newton's second law of motion to the rightward acceleration a of block B, we obtain: a = R / mB where mR is the mass of block B Example 9 Solution . • The normal reaction at the interface between the two blocks is R • Hence, the maximum frictional force that block A can exert on block B is u.R • In order to prevent block B from falling, this maximum frictional force (which acts upwards) must exceed the downward acting weight, mBg, of the block • Hence, we require: u.R>mBg or a>g/u. Example 9 Solution V Breakout Application of Friction Applying Newton's second law to the rightward acceleration a of both blocks we obtain: a=F/(mA + mB) You can consider the ground to have friction - What will be the effects in the ball? - How would you start the ball if it stops? where mA is the mass of block A It follows that: F > (mA + mB)g / u. Substituting we get: F > (75 +15)9.81 / 0.45 = 1.962*103 N Conservation of Energy Conservation of Energy The conservation of energy is undoubtedly the single most important idea in physics Although the basic idea of energy conservation was familiar to scientists from the time of Newton onwards, this crucial concept only moved to centre-stage in physics in about 1850 - When scientists first realized that heat was a form of energy 22 3/18/2016 Forms of Energy • Energy is the substance from which all things in the Universe are made up • Energy can take many different forms: - Potential energy - Kinetic energy - Electrical energy - Thermal energy - Chemical energy - Nuclear energy - Etc Energy Transformation • Everything that we observe in the world around us represents one of the multitudinous manifestations of energy • Various processes in the Universe transform energy from one form into another: - i.e. Mechanical (which are the focus of this course), thermal, electrical, nuclear, etc... • However, all of these processes leave the total amount of energy in the Universe invariant Energy in Closed Systems • Whenever, and however, energy is transformed from one form into another, it is always conserved • For a closed system the above law of universal energy conservation implies that the total energy of the system in question must remain constant in time - i.e. a system which does not exchange energy with the rest of the Universe Energy Conservation During Free-Fail • Consider a mass m which is falling vertically under the influence of gravity - Know how to analyze the motion of such a mass • Let us employ this knowledge to search for an expression for the conserved energy during this process - This is clearly an example of a closed system, involving only the mass and the gravitational field Energy Conservation During Free-Fail Energy Conservation During Free-Fail v The physics equations of free-fall under gravity is summarized by: s = u0t - Vi gt2 u = u0 - gt u2= u02-2gs • Let us examine the last of these equations: - u2 = u02 - 2gs • Suppose that the mass falls from height hx to h2 its initial velocity is u1 and its final velocity is u. • It follows that the net vertical displacement of the mass is: s = h2 - h1 23 3/18/2016 lergy Conservation During Free-Fall Moreover, u0 = u1 and u = u2 Hence, the previous expression can be rearranged to give: or mUj2 + 2mgh1 = mu22 + 2mgh2 Vi mUj2 + mghx = Vi mu22 + mgh2 !ls>^metic and Gravitational Energy of w Mass • The kinetic energy of the mass can be defined as: K = Vi mu2 • The gravitational potential energy of the mass can be defined as: U = mgh Mass . • Note that kinetic energy represents energy the mass possesses by virtue of its motion • Likewise, potential energy represents energy the mass possesses by virtue of its position • Thus: E = K + U = constant cl!, Rigid Body Rotation Analysis The instantaneous angular velocity of the body u)(t) is defined: , SA dtb ui — lim — = — 5*^0 St dt Note that if the body is indeed rotating rigidly, then the calculated value of u> should be the same for all possible points P lying within the body — Except for those points lying exactly on the axis of rotation, for which id is ill-defined Rigid Body Rotation Analysis The rotation speed u of point P is related to the angular velocity a) of the body via: u = au) — where a is the perpendicular distance from the axis of rotation to point P Angular acceleration a(t) of a rigidly rotating body is defined as the time derivative of the angular velocity: dui d2tj> a~ ~dt ~ ~d¥ — where is the angular coordinate of some arbitrarily chosen point reference within the body, measured with respect to the rotation axis Rigid Body Rotation Analysis • For a body rotating with constant angular velocity uj the angular acceleration is zero, and the rotation angle <|> increases linearly with time: 4>(t) = 4>0 + u>t - where c))0 = 4>(t = 0) • Likewise, for a body rotating with constant angular acceleration a: w(t) = uj0 + at and the rotation angle satisfies: 4>(t) = 4>0 + w0t + 1/2 at2 - Here w0 = w(t = 0) 28 3/18/2016 Rigid Body Rotation Vectors Rigid body rotating with angular velocity to with assumption: - The axis of rotation, which runs parallel to u> is assumed to pass through the origin O of our coordinate system Point P whose position vector is r represents a general point inside the body - What is the velocity of rotation u at point P? Rigid Body Rotation Vectors The magnitude of this velocity is simply: u = ooj = cor sin8 where o is the perpendicular distance of point P from the axis of rotation and 8 is the angle subtended between the directions of u and r The direction of the velocity is into the page Rigid Body Rotation Vectors .. Alternatively the direction of the velocity is mutually perpendicular to the directions of u and r in the sense indicated by the right-hand grip rule when u is rotated onto r - Through an angle less than 180 u = uj x r (cross product) Centre of Mass • The co-ordinates of the centre of mass (or centre of gravity) of an extended object are the mass weighted averages of the coordinates of the elements which make up that object • If the object has net mass M and is composed of N elements, such that the ith element has mass m, and position vector r, then the position vector of the centre of mass is given by: Centre of Mass References If the object is continuous: - where p(r) is the mass density of the object and V, is the volume occupied by the ith element - http://www.dummies.com/how- to/content/how-to-calculate-a-spring- constant-using-hookes-la.html http://formulas.tutorvista.com/physics/hooke -s-law-formula.html http://www.efm.leeds.ac.Uk/CIVE/CIVE1140/s ection02/mechanics sec02 full notes02.html 29 3/18/2016 Useful Links Useful Links http://www.staff.amu.edu.pl/~romangoc/M2-motion-two-three-dimensions.html http://media.wilev.com/product data/excerpt/19/04717580/04717 58019.pdf http://phvsics.bu.edu/~duffv/pyl05/RelativeV.html http://www.physnet.org/modules/pdf modules/m37.pdf http://ierome.iouvie.free.fr/QpenGI/Tutorials/Tutorial26 Advanced ■ php http://glprogramming.com/red/chapter03.html http://en.wikipedia.org/wiki/Newton's laws of motion http://www.phvsicsclassroom.com/Class/newtlaws/ http://teachertech.rice.edu/Participants/louviere/Newton/ http://cseplO.phvs.utk.edu/astrl61/lect/historv/newton3laws.html http://en.wikipedia.org/wiki/Mass versus weight http http http http http http http http http http //www.edinformatics.com/math science/mass weight.htm ://www. nyu.edu/pages/mathmol/textbook/weightvmass. html //en.wikipedia.org/wiki/Friction ://www.da rvill.clara.net/enforcemot/friction. htm ://www.school-for-champions.com/5cience/friction uses.htm ://www.pa.ukv.edu/~phy211/Friction book.html //en.wikipedia.org/wiki/Conservation of energy //cnx.org/content/ml4106/latest/ //library.thinkquest.org/2745/data/lawcel.htm //en.wikipedia.org/wiki/Work (physics) 30