GAME PHYSICS (INFOMGP) FINAL EXAM 15/JUN/ 016 LECTURER: AMIR VAXMAN Student name: Student number: This exam is 3 hours long and consists of 5 exercises. The points add up to 100. Answer only in English. Allowed: (offline) calculators and official formulas sheets currently delivered. Not allowed: phones, books, personal notes, and computers. The written answers are to be written in the corresponding boxes as concisely as possible. In case any additional paper is needed, state student number and name on it. An explanation must accompany every answer. Simple or direct answers (such as 4 and no ) will not be given any credit even if correct. Indicate the unit of every quantitative answer (points would be deduced otherwise). More than two decimal points are usually not necessary for an answer; an exception is the time discretization question, where you need to show accuracy. In case you do not have a calculator, the answer will be accepted only if you leave a correct and full formula, that is then straightforward to compute. Game Physics Exam - June 016 Amir Vaxman 1
EXERCISE 1: NEWTONIAN MECHANICS (5 POINTS) A. (7 points) A dimensionless character is standing on the ledge of a platform at height h = 0m, loses its standing and starts falling, without any initial velocity. The wind carries the character horizontally in an acceleration of a = 0.5 m sec,. What is the horizontal distance d acquired by the character upon reaching the ground? Let us first find the time of the fall. The vertical acceleration is g = 9.81 m sec,. The initial velocity is 0. Therefore, we get: And t =.0[sec] h = 1 gt, Consequently, the horizontal distance is d = 5, at, = 1.0[m] B. (8 points) Another dimensionless character of mass m = 75Kg is Bungee-jumping from a cliff of the same height h = 0m, using a spring-like string of rest length l 9 = 5m. What should be the stiffness coefficient K of the spring, at the least, so that the character never crashes unto the ground? In the worst admissible case, the character is exactly touching the ground, which means the current length of the spring is: l = l 9 + Δl = 0 m, Δl = 15 m The force that the spring exacts at that moment is then F? = KΔl, where gravity is G = mg. Since we have an equilibrium, we get: K = mg 75 9.81 = = 49.05 N Δl 15 m More stiffness = more force upwards, and therefore K should at least 49.05 N m Common mistake: using energy preservation. No preservation of energy was promised. This strings swallows the movement in Bungee jumping. Otherwise you crash with speed. Game Physics Exam - June 016 Amir Vaxman
C. (10 points) The wind suddenly returns with a force of F E = (350,0)N. The stiffness of the springs and the mass are as in the previous question (B). How high is the character above the ground in an equilibrium? What is the horizontal displacement? How strained (Δl spring length difference) is the spring in total? This question is canceled since there is a problem in the question: the rest length of the spring is not negligible in the force equation. Thus, should the spring stay in the rest length, but let the wind take it up (so rotate the spring), the forces are not zero, but there is no strain! This is because what the wind does here essentially, is give a torque to the spring. Since this question did not mention this and alluded to a dimensionless being, no otherwise answer is really correct. Thus, I had to cancel it and give complementary 10 points for everyone (since you did invest the time to try to solve). EXERCISE : ROTATIONAL & RIGID-BODY MOTION (5 POINTS) A. (1 points) A round carousel of mass m = 00 kg and radius r = 1 m is rotating around the z axis, where its center of mass is the origin (0,0,0). The angular speed is ω = 3 Rad sec. Suddenly, four cylindrical children, each with m M = 0 kg and radius r M = 0. m, and without prior velocities (angular or linear), attach themselves to the carousel at points 1,0, 0,1, 1,0, (0, 1). Assuming no friction or loss of energy/momentum, compute the new angular speed of the system. The individual moment of inertia for each element (carousel or cylindrical child) around their own COM is I = 5, mr,. Game Physics Exam - June 016 Amir Vaxman 3
The angular momentum stays the same, and therefore related to the original movement of the Carousel: L = Iω = 1 00 1, 3 = 300 N m sec We need to compute the new moment of inertia I. The moment of inertia per child is I = 5, mr, around its own center of mass; with the parallel axis theorem, the moment of inertia around the origin (the total center of mass) per child is: I M R = 1 mr, + md, = 1 0 0., + 0 = 0.4 kg m, Since we are now around a common center of mass, the total moment of inertia is added linearly : I R = I + 4 I M R = 100 + 81.6 = 181.6 kg m, And the new angular speed scales accordingly: ω R = L I = 1.65 Rad sec Another error on my behalf: I said that no energy is lost. This is a mistake! Non-elastic collisions (coefficient <1), like this one, lose energy. The momentum is preserved. As I have made the misleading here, I accepted answers which used energy preservation, but know that they were wrong. B. (13 points) Consider the following system in the xy plane: h L d r Game Physics Exam - June 016 Amir Vaxman 4
It describes a stationary rod of mass m = 10 kg, radius r = 5 m, and moment of inertia 5 5, mr, around the z axis relative to its COM. The rod is hinged at a point (red) which is distanced by d =.5 m from the COM. There is an initially stationary (zero speed) ball at height h = 7 m from the horizontal line of the hinge (assume the thickness of the rod is negligible), beginning to fall down by earth s gravity. Compute the necessary impulse L, so that the rod will meet the ball horizontally (the dashed configuration) upon impact. Assume no air friction, or otherwise loss of energy. The easiest thing to compute is the time of impact: a ball should fall h meters in an acceleration g. The time of impact is then: h = 1 gt, t = 1.19 sec The rod has to cover V Rad (quarter round) to hit the ball as we want. Since there is no friction, following, the impulse, we get a uniform angular speed, which should be: ω = π t = 1.3 Rad sec Since L = Iω, the only thing missing is I. We get it by the parallel axis theorem: And therefore the answer is I = 1 1 mr, + md, = 83.33 kg m, L = Iω = 6.04 1.3 = 110 N m sec Very common mistake: assuming we give a torque instead of an impulse. This question is about hitting the object, not applying a continuous force! EXERCISE 3: TIME AND SPACE DISCRETIZATION (0 POINTS) A. (8 points) Name an advantage and a disadvantage of using the linear elasticity approximation vs. the full Lagrangian strain tensor, and shortly describe a way to handle the problem (explicit formulas are not required). Advantage: the linear formulation is constant within each triangle, allowing for constant stiffness matrices which are then easy and cheap to compute and manipulate. Disadvantage: Only a good approximation of a small deformation, not good with medium- and largescale rotations. Possible solution: corotational elements factor out the rotation from the deformation, thus making the approximation better. Note: giving an answer like advantage: cheap, disadvantage: not accurate is not acceptable. This is just the dictionary definition of approximation. You needed to be specific. Game Physics Exam - June 016 Amir Vaxman 5
B. (1 points) Assuming an object is decelerated by a drag force of a t, v = v and at t = 0 sec, the velocity of the object is 40 m sec. What will be the velocity of the object after 0.5 sec? Calculate v(t + t) with Euler s method, the midpoint method, the improved Euler s method, and RK4 method. Then, compare the results with the ideal solution ( dv = v t dt v t = v(0)e Z[ ). Euler s method: v t + t = v t + t a t, v = 40 + 0.5 40 = 0 [m/sec] Midpoint method: v t + t t = v t + a t, v = 40 + 0.5 40 = 30[m/sec] v t + t = v t + t a t + t t, v t + Improved Euler s method: v 5 = v t + t a t, v = 0[m/sec] v, = v t + t a t + t, v 5 = 40 + 0.5 ( 0) = 30.0[m/sec] v t + t = b cdb e = 5[m/sec], RK4: v 5 = t a t, v t = 0.5 40 = 0[m/sec] = 40 + 0.5 30 = 5[m/sec] v, = t a t + t, v t + 1 v 5 = 0.5 40 1 0 = 15.0[m/sec] v f = t a t + t, v t + 1 v, = 0.5 40 1 15 = 16.5[m/sec] v g = t a t + t, v t + v f = 0.5 (40 16.5) = 11.875[m/sec] v t + t = v t + b cd,b e d,b h db i = 40 15.79 = 4.708 [m/sec] j Ground Truth: a = dv = v t dt dv = v t dt v t = v(0)ez[ v t + t = v 0 + 1.0 = 40e Z9.k 4.61[m/sec] Game Physics Exam - June 016 Amir Vaxman 6
EXERCISE 4: COLLISION RESOLUTION (15 POINTS) A. (5 points) A resting contact is when we need to assume that two objects "stick" together. Shortly explain why a resting contact is needed in a simulation, and how to model this in practice. A resting contact is needed since two objects can "jitter" back and forth around a slightly inexact collision. This is inaccurate, and also wastes computation resources on something that is basically noise or just unmoving. The way to handle it is to artificially reduce the coefficient of restitution, either relative to the velocity, or directly to zero. B. (10 Points) Two equally-sized balls collide at the origin 0,0. The first is with mass m 5 = 15 Kg, and moving in a velocity of v 5Z =,5, and the second ball has m, = 5 Kg, and moving in a velocity v,z =,5. Assume a constant of restitution C n = 0.8 and compute both resulting velocities v 5d, v,d. Due to symmetry, it is clear that the collision normal is just (1,0). Computing the collision impulses (j, is just the negative): j 5 = (1 + C n) v 5Z v,z n (1.8) 4,5 1,0 1 + 1 = = 7 Kg m 4 sec m 5 m 15, The resulting velocity changes: v 5d = v 5Z + j 5 m 5 n =,5 1.8,0 = (0.,5)[ m sec] v,d = v,z + j, m, n =,5 + 5.4,0 = (3.4,5)[ m sec] Note that the balls continue in the same horizontal direction. Common mistake: not applying the impulses with opposite signs. Game Physics Exam - June 016 Amir Vaxman 7
EXERCISE 5: SOFT-BODY PHYSICS (15 POINTS) A. (9 Points) Explain 3 out of 5 of these concepts concisely, but accurately: material derivative, convective acceleration, incompressible flow field, Young's modulus, viscosity. Material derivative: the change in velocity of a fluid element as it flows. Convective acceleration: the part of material derivative which is only dependent on the change in velocity due to moving along the flow (and not due to time derivative). Incompressible flow field: a field which does not change the volume of the flow. Differentially it means divergence-free. Young's modulus: the ratio between stress and strain along a stretched direction. Viscosity: the ability to resist shear deformation. Computed as the force needed to induce a difference between the velocity of different layers (orthogonal to the force). B. (6 points) Sir Gregor Clegane is celebrating his name day and is blowing candles on a layered cake with shear modulus of 5[Pa]. The blow is directed towards the top surface of the cake. The cake is a cylinder of radius r = 15 cm and the height of each layer is L = [cm]. The blow breaks the cake apart if the top moves beyond the center of the cylinder, with regards to the layer below. What minimal force does the mountain have to exert to break the cake? The area of a layer is: A = πr, = 0.071[m, ]. The needed shear is by moving the cake in the length of a radius, and thus: L L = r L = 7.5 (A modulus has no units!) Thus, the needed force F obeys the shear modulus law: 5 = F A = F L 0.535 [Pa] L And the necessary force is: F = 13.31[N] Common mistakes: not converting to meters, using diameter instead of radius. Game Physics Exam - June 016 Amir Vaxman 8