the equations for the motion of the particle are written as

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1 Dynamics 4600:203 Homework 02 Due: ebruary 01, 2008 Name: Please denote your answers clearly, ie, box in, star, etc, and write neatly There are no points for small, messy, unreadable work please use lots of paper Problem 1: Hibbeler, The v t raph for the motion of a car as it moves alon a straiht road is shown Draw the a t raph and determine the maximum acceleration durin the 30s time interval The car starts from rest at x = 0 v ft/s 40 v = t + 30 v = 04 t t s The acceleration is the derivative of velocity, so that a = v Therefore the acceleration takes two forms, the first durin the interval 0 t 10s, for which 8 a ft/s 2 ẍt = vt = 08ft/s 3 t Durin the second interval the acceleration reduces to ẍt = vt = 100ft/s 2 The a t raph is shown to the riht inally, the maximum acceleration occurs at t = 10 s, for which ẍt = 800ft/s t s Problem 2: Hibbeler, Show that if a projectile is fired at an anle from the horizontal with an initial velocity v 0, the maximum rane the projectile can travel is iven by R max = v 2 0/, where is the acceleration of ravity What is the anle for this condition? The position of the ball is described by r P O = xt î+yt ĵ When subject to ravity, and initial velocity, v 0 xt, yt v P = ẋ0î + ẏ0ĵ = v 0 cos î + sin ĵ, the equations for the motion of the particle are written as 1

2 xt = v 0 cos t, yt = t2 2 + v 0 sin t Therefore, when y is written as a function of x, this reduces to yx = 2v0 2 cos2 x2 + sin cos x = v 2 0 x 1 + cos2 x v2 0 sin2 The rane occurs when y = 0, or solvin for x R = v2 0 sin2 The maximum rane thus occurs for 2 = 90, or = 45, R max = v2 0 Problem 3: Hibbeler, The fireman standin on the ladder directs the flow of water from his hose to the fire at B Determine the velocity of the water at A if it observed that the hose is held at = 20 A v 0 30 ft 60 ft B help!! The vertical displacement of the water y as a function of the horizontal displacement x is iven as yx = 2v0 2 cos2 x2 + sin cos x The initial anle is iven as = 20, while the final displacement is x f,y f = 60ft, 30ft Therefore, solvin for v 0 yields v 0 = 2 cos sin x f cos y f x f With the iven values, this reduces to v 0 = 8968ft/s 2

3 Problem 4: Hibbeler, The stones are thrown off the conveyor with a horizontal velocity of 10ft/s as shown Determine the speed at which the stones hit the round at B see textbook for fiure The position of a stone can be described with the vector r P0 = xî + y ĵ, so that usin xt = v 0 cos t, y can be written as a function of x as yx = 2v0 2 cos2 x2 + sin cos x or this system v 0 = 10ft/s, and = 0, which reduces the above to yx = 2v0 2 x 2 The round at the bottom of the conveyor can be described with the equation y r = 100ft x r 10 Therefore, the stones hit the round when their trajectory intersects the equation for the surface That is 2v0 2 x 2 f = y f = 100ft x f 10, 0161ft 1 x 2 f 010x f 100ft = 0 This equation is quadratic in x f, and may be solved to yield x f = 2523ft To find the speed at which the stones hit the round, we return to the equations for the velocity, which can be written as ẋt = v 0, ẏt = t = v 0 x inally, the speed of the stones at impact can be written as v P = 2 ẋ 2 + ẏ 2 = v0 2 + xf v 0 Usin the above value x f = 2523ft, we find that the stones hit the round with speed v P = 8187ft/s Problem 5: Hibbeler, A truck is travelin alon the horizontal circular curve of radius r = 60m with a speed of 20m/s which is increasin at 3m/s 2 Determine the truck s radial and transverse components of acceleration see textbook for fiure 3

4 Given the path of the truck, it is natural to describe its position in terms of polar coordinates, so that r P O = r, v P = ṙ + r ê, a P = r r 2 + r + 2ṙ ê ê P ĵ î With constant radius of the curve, ṙ = 0 and r = 0, so that the kinematics reduce to r P O r P O = r, v P = r ê, a P = r 2 + r ê O The speed of the truck, 20m/s, is iven in terms of the coordinates as v P = r = 20m/s, while its rate of chane is d v P = r dt = 3m/s 2 Notice that the rate of chane of the speed is different from the manitude of the acceleration rom these, we can determine and as = 1 3 rad/s, = 1 20 rad/s2 inally, with these values the acceleration of the truck can be written as a P = r 2 + r ê = 20 3 m/s2 + 3m/s 2 ê 4

5 Problem 6: Hibbeler, At the instant shown, the water sprinkler is rotatin with an anular speed = 2rad/s and an anular acceleration = 3rad/s 2 If the nozzle lies in the vertical plane and water is flowin throuh it at a constant rate of 3m/s see textbook for fiure a determine the manitudes of the velocity and acceleration of a water particle as it exits the open end, r = 02m; b this part is not in the textbook but builds upon this problem once it exits the nozzle, find how far this water particle travels before hittin the round Assume that the nozzle is at round level a The kinematics of a particle of water P can be described in terms of polar coordinates as r P O = r, v P = ṙ + r ê, a P = r r 2 + r + 2ṙ ê ê ĵ î rom the problem statement, the coordinates and their derivatives of P at the nozzle exit are iven as r = 02m, ṙ = 3m/s, r = 0, = 2rad/s, = 3rad/s 2 Notice that the coordinate is not iven, and does not influence the kinematics when written in terms of the radial and tanential directions certainly does affect the orientation of and ê relative to the round The velocity and acceleration can be written as v P = 3m/s + 04m/s ê, a P = 08m/s m/s 2 ê inally, the manitudes of these quantities are v P = 303m/s, a P = 126m/s 2 Notice that the velocity of the water is not simply in the direction Instead, the water velocity is directed at an anle of 759 off the direction ê v P 5

6 b The rane of the water can be determined from the equation R = v2 0 sin2ψ, where the water has an exit speed of v 0 and a velocity direction of ψ Usin the above values, we find that v 0 = 303m/s and ψ = + 013rad, so that R = 093m sin rad In contrast, if the nozzle is held stationary at an anle, the rane of the sprinkler is R = 092m sin2 Problem 7: Hibbeler, A cameraman standin at A is followin the movement of a race car, B, which is travelin alon a straiht track at a constant speed of 80ft/s Determine the anular rate at which he must turn in order to keep the camera directed on the car at the instant = 60 see textbook for fiure With the perpendicular distance between the track and O iven as d, the distance r between O and C is r CO = r = d sin Here the velocity of the car is naturally written in terms of Cartesian coordinates However, the response of the cameraman is determined in terms of polar coordinates In terms of the former, the velocity of the care is written as v 0 î C v C = v 0 î, while in terms of the polar coordinates r and with v C = ṙ + r ê, O r CO ê ĵ î = cos î + sin ĵ î = cos sin ê, ê = sin î + cos ĵ ĵ = sin + cos ê, Settin these two descriptions of v C equal to one another, we find that v 0 î = ṙ + r ê This vector equation has two unknowns, ṙ and We could write the directions and ê in terms on î and ĵ, which would lead to two scalar equations, coupled in the two 6

7 unknowns However, instead we write î in terms of and ê Doin so yields v 0 cos sin ê = ṙ + r ê Thus, solvin the equation in the ê for provides = v 0 sin r = v 0 sin 2 d Therefore, at = 60, we find that = 06rad/s 7