Ch01 Kinematics of particles.docx

KINEMATICS OF PARTICLES 11.28 a = ksin (a) v = -4572013208000169164013208000-4572013208000 v = -4572013970000 v = s = -13716011938000-13716011938000178308011938000 -13716049276000s = s = k v = vmax when i.e. when t = T -13716016510000105156016510000-13716016510000 -13716044704000vmax = 2 S/t=2T = = = 228600-292735S = 00S = = S/t=2T = vave = -457205080Vave = 00Vave = A small object is released from rest in a tank of oil. The downward acceleration of the object Is (g-kv) where g is the constant of the object is (g-kv) where g is the constant acceleration due to gravity, k is a constant which depends on the viscosity of the oil and the shape of the object and v is the downward velocity of the object. Derive expressions for the velocity, v, and the vertical drop, y, as function of the time, t, after release. a = (separation of variables) Let (g-kv) = u -kd0v=du ln 45720162560V = g 00V = g v = y = 4572071755Y= 00Y= -320040304165xb xo e xA xE xD xP 1 00xb xo e xA xE xD xP 1 1863090778510T 00T 11.42 Consider rope TCD: (assume rope inextensible) 685800000 XB = XP + 169164015557500 -28956043180A D B C XC XF II XE I XD III XA Xt E F XB 00A D B C XC XF II XE I XD III XA Xt E F XB Determine the relationship which governs the velocities of the four cylinders. Express all velocities as positive down. How many degrees of freedom are there? Assumption: Ropes I, II & III are inextensible Rope I: XC + 2XE = Constant(i) Rope II: (XC-/XE) + XC + 2/XF = Constant (ii) Rope III: (XB-XF) +XB +XA = Constant (iii) Substitute (i) in (ii) 2XC + 2XF += Constant (iv) Substitute (iv) in (iii) 2XB + XA + XC + or 4XA +8XB +4XC +XD = Constant (v) Differentiate (v) with respect to t 411480215904VA + 8VB + 4VC +VD = 0 004VA + 8VB + 4VC +VD = 0 OR Only 3 of XA, XB, XC and XD in (V) can be chosen system has 3 degrees of freedom A ball is dropped vertically onto a 20° incline at A; the direction of rebound forms an angle of 40° with the vertical. Knowing that the ball next strikes the incline at B, determine the velocity of rebound at A, (b) the time required for the ball to travel from A to B 50292068580Y 3m 40o A B X 20o 00Y 3m 40o A B X 20o 77724014097000 Motion in the x-direction Motion in the Y-direction Geometry: then becomes: -1.092 = V0sin50°. -1.092 = 3 tan 50° - 4.5 324612079375T = 0.975secs 00T = 0.975secs 77724079375V0 = 4.785 00V0 = 4.785 (b) Calculate the minimum possible magnitude of the muzzle velocity, u, which a projectile must have when fixed from point A to reach a target B on the same horizontal plane 12 km away. 190515049512 km A Y B X 0012 km A Y B X Motion in the X=direction 2331720-76200018745208382000 12000 = ucos.t (i) Motion in the Y-direction At maximum height, VY = 0 0 usin-gt1t1= (iii) Because of symmetry, time taken, t, to travel from A to is twice t1 (t = 2t1) From (ii) 12000 = ucos u2 = 3429000-7620(negative value for u not physically possible) 00(negative value for u not physically possible) 594360-7620Umin = 00Umin = 0000 The vertical slot moves to the right with a constant speed v0 in meters per second for an interval of motion and causes the pin P to move along the parabolic slot x = y2/3 where x and y are in meters. Calculate the radius of curvature to the path for the position where y = 2 m and calculate the tangential acceleration at of the pin when it passes this position. Ans. = 6.94 m, at = -0.169v02 m/s2 -64770192405t X = 4/3m Y V y = 1m x y = 2m x = v0 n an 00t X = 4/3m Y V y = 1m x y = 2m x = v0 n an For y = 2m, x = 4/3 m (since x = y2/3) 37090352223135 00 -457202122170 00 At Y = 2m, X = V2 = 19773903886203 4 5 003 4 5 an = 6.94 m/s2 m/s2 333756014605000 A nozzle discharges a stream of water in the direction shown with an initial velocity of 25 m/s. Determine the radius of curvature of the stream (a) as it leaves the nozzle, (b) at the maximum height of the stream. At A 156210163830g t at an 00g t at an 3063240155575 00 At B VB = (VA)X (Since ax = 0 and B is at the maximum) 379476021590an 00an Height of the stream VB = VAcos = 25 (4/5) = 20 m/s -4572027940 00 457205715000As the hydraulic cylinder rotates about 0 the length, l, of the piston rod P is controlled by the action of oil pressure in the cylinder. If the cylinder rotates at the constant rate = 60 deg/sec and is decreasing at the constant rate of 152.4 mm/sec, calculate the magnitudes of the velocity and acceleration of end B when l = 127 mag. 45720215900O 1 T 1 B T1 00 O 1 T 1 B T1 Solution r = 381 + 127 = 508 mm Vt = 45720157480 00 ar = a = 45720160020 00 388620012573000 The slotted arm is pivoted at O and carries the slider C. The position of C in the slot is governed by the cord which is fastened at D and remains taut. The arm turns counterclockwise with a constant angular rate = 4rad/sec during an interval of its motion. The length DBC of the cord equals R, which makes r = 0 when = 0. Determine the magnitude of the acceleration of the slider at the position for which = 30o. The distance R is 15 in. Ans. A = 489 in./sec2 132588023431500Solution 228600150495C B D O R r 1 t1 R 00C B D O R r 1 t1 R Geometry: From triangle OBD, Kinematics: ac = t1 + 1 r = 2Rsin ac = = 30° /2 = 15°, R = 15", r = 30 sin 15° = 7.76" -45720109220 00 04318000A police car P is traveling at velocity vP = 20j km/h when its radar gives the following data about car A : r = 100 m, = -60 km/h (r decreasing), 0, = 0.5 rad, = -10-3 rad/s ( decreasing), and 0. Determine the velocity vA. Solution: 55245131445 VP P r A 00 VP P r A Relationship between the unit vectors i, j & r1, 1 13525541910 r1 y, j x, i 1 00 r1 y, j x, i 1 2263140142875Note: 1 is positive in the direction of increasing 00Note: 1 is positive in the direction of increasing r1 = i sin + j cos 1 = I cos - j sin Velocity car A relative to the Policecar: Absolute velocity of car A: Geometry of a space curve 36195435610S O B A T(s) I s t1 n1 m1 T(s+s) 00S O B A T(s) I s t1 n1 m1 T(s+s) Consider the following space curve along which a particle moves from A to B116205011176000 Unit vector tangent to the curve at A is t1 1051560-100965 00 t1 = lim s 0 Unit vector perpendicular to t1 and pointing toward centre of curvature of the curve is the principal normal n1 Binotmal unit vector m1 = = t1 x n1 -Plane of t1 & n1 : OSCULATING PLANE -Plane of m1 & n1: NORMAL PLANE -Plane of m1 & t1: RECTIFYING PLANE Consider segment AB of curve is more detail 95250102870Osculating plane t1 (s+s) T1 (s) A t1 B t1 (s+s) 00Osculating plane t1 (s+s) T1 (s) A t1 B t1 (s+s) The orientation of n1 (i.e. principal normal) is determined from the following: n1 1 t1 (ii) n1 lies in the osculating plane (iii) 1n11 = 1 t1 .t1 = 1 t1 1 1783080205740 00 2240280186690Lies in the osculating plane since 00Lies in the osculating plane since The vector Lim s 0 t1 =t1 =t1 (S + S) -t1 (S) lies in the same plane. Therefore, the vector satisfies conditions (i) and (iii) above; hence 41148012573000411480125730003200404000500022860021717000 Osculating Plane 105156029210X Cx A X P R Z r Y 00X Cx A X P R Z r Y C: Centre of curvature 318325515875V 00V : radius of curvature P: Osculating plane Consider the space curvilinear motion of a particle along the path shown. The motion at position A may be considered to be taking place in the plane P which contains the path at this position. This plane, often called the OSCULATING PLANE, may be defined by point A and two adjacent points, one on either side of A, on the path curve. As these points are brought closer to A, the plane containing the three points approaches the limiting plane P. The osculating plane is, by definition, the plane containing the unit tangential (t1) and unit normal (n1) vectors. The path coordinates t and n, are used for solving most of the problems of planar motion. The osculating plane thus remains fixed in the plane of motion. Although it is possible to obtain the principal normal n and the direction of the osculating plane for spatial motion, it is generally easier to define the spatial motion of a particle using other coordinate systems, e.g. X,Y,Z or T,, Z.