. Differential and integral calculus, an introductory course for colleges and engineering schools. far from the base of the fortress willthe projectile strike the water? 155. Problem 3. A particle of mud is thrown from the rim of a carriage wheel when thecarriage has a speed of m feet per second. Neglecting the resistance ofthe air, determine the motion of the particle of mud. Solution. Before being thrown off, the particle of mud is describing acycloid, and as it leaves the rim its velocity is in the direction of thetangent to the cycloid at that point, P, and is equal in magnitude to ds ^ .
. Differential and integral calculus, an introductory course for colleges and engineering schools. far from the base of the fortress willthe projectile strike the water? 155. Problem 3. A particle of mud is thrown from the rim of a carriage wheel when thecarriage has a speed of m feet per second. Neglecting the resistance ofthe air, determine the motion of the particle of mud. Solution. Before being thrown off, the particle of mud is describing acycloid, and as it leaves the rim its velocity is in the direction of thetangent to the cycloid at that point, P, and is equal in magnitude to ds ^ . , . dddt Vo = — = 2 a sindt (Art. 104, exercise 4.) The slope of the tangent to the cycloid is -Or —0) (Art. 93,problem 1). In the figure, angle APX = - (tt — 0). Since the carriage has a speed A of m feet per second, wre have ad = mt, 0 m dddt Hence the tangential velocity at P isVo = 2 7?z sin |0. The problem is nowreduced to that of determining the mo-tion of a particle launched from thepoint P with a velocity 2 m sin^0, and at an angle of elevation of - Or — 0), 0 being a Hence we get the motion by setting v0 = 2 m sin and = - (tt — 0) in the formulae of 222 INTEGRAL CALCULUS §155 Art. 153, problem 1. The axes of reference will be the horizontal andvertical lines through P. To find the highest point the particle of mudwill reach, we make the foregoing substitutions in the last of equations(5) of Art. 153. The result is - _ Vo* sin2 4> _ 4 m2 sin2 \ 0 cos2 \ 0 _ m2 sin2 0V 2g ~ 2g 2g This is the particles greatest height above PX. The greatest heightfrom the ground is , m2sin20 . M . h = — h a(l — cos 0). 2 <7 It is found without difficulty that h attains its maximum value when cos 0 = — -?. and 0 = cos_1( -n V This gives as the maximum value m2 V m1 oih maximum h = ^— f1- • 2gm? When m2 = - (*■ — 0) in equations (2) of Art. 153, z and are x = 2msin2^0-^, y = 2ms
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