Homework 11 Engineering 233 Dynamics Winter 2018 Due 04/16/2018 Name: 1. (20 poi

Question

Homework 11 Engineering 233 Dynamics Winter 2018 Due 04/16/2018 Name: 1. (20 points) A popular anusement park ride, often called the Teacup Ride, involves riders sitting in circular rotating teacups", that are themseles rotating around a central axis In many cases, the teacups can also translate relative to the center of the rotating platform. Consider the case of a rider on such a ride. The cester of the teacup rotates around a cesteal axis at an angular elocity ok, and, at a given instant is also moving radially outward along a track at a speed oa At this instant the distance between the ride center (point O) and the teacup center (pnt A) is Ros- The teacup is itself rotating counter-clockwise with a angular speed wA. The rider sits a distance Ran away from the center of the teacup and, at the instant under consideration, is in a direction o degrees clockwise relative to the I. Find the velocity and acceleration ofthe rider at this instant ife=90.. toa = 2 m/s. Roa 5 m. ila = 0.5 tada. Raa = 1 m. aas = 3 10

Explanation / Answer

1. angular velocity of the platform = wOA k ( where i , j and k are unit vectors a long x, y and z directions respectively)

let, i j be unit vectors along x, y axis

let r', t' be unit vectors along radial and tangential directions relative to the central platform

then let center of the cup be at a radial distance r, angle theta

dr/dt = vOA r'

r = Roa ( initial distance)

angular velocity of tea cup = wAB

distance from center of tea cup = Rab

angle = phi

hence position of B relative to A = Rab*cos(-wAB*t) i + Rab*sin(-wAB*t) j

position of A relative to O = (vOA*t + Roa)(cos(wOA*t) i + sin(wOA*t)j)

hence B relative to O is

rBO = rBA - rAO = Rab*cos(-wAB*t) i + Rab*sin(-wAB*t) j - (vOA*t + Roa)(cos(wOA*t) i + sin(wOA*t)j)

rBO = i(Rab*cos(wAB*t) - (vOA*t + Roa)cos(wOA*t)) - j(Rab*sin(wAB*t) + (vOA*t + Roa)sin(wOA*t))

d(rBO)/dt = i(-Rab*wAB*sin(wAB*t) - (vOA*cos(wOA*t)) + (vOA*t + Roa)wOA*sin(wOA*t)) - j(Rab*wAB*cos(wAB*t) + (vOA*sin(wOA*t)) + (vOA*t + Roa)wOA*cos(wOA*t))

Rab = 1 m

wAB = 3 rad/s

wAB*t = phi = 0 deg

vOA = 2 m/s

wOA = 0.5 rad/s

wOA*t = theta= 90 deg

Roa = 5 m

hence

v = i(-3*sin(0) - 2cos(90) + (0 + 5)0.5*sin(90)) - j(1*3*cos(0) + 2*sin(90) + (0 + 5)0.5*cos(90))

v = 2.5i - 5j m/s

similiarly

dv/dt = i(-Rab*wAB^2*cos(wAB*t) + vOA*wOA*sin(wOA*t) + (vOA*t + Roa)wOA^2*cos(wOA*t) + vOA*wOA*sin(wOA*t)) - j(-Rab*wAB^2*sin(wAB*t) + (vOA*wOA*cos(wOA*t)) - (vOA*t + Roa)wOA^2*sin(wOA*t) + (vOA)wOA*cos(wOA*t))

a = dv/dt = i(-1*3^2 + 2*0.5 + (5)0.5^2*cos(90) + 2*0.5*sin(90)) - j(-3^2*sin(0) + (2*0.5*cos(90)) - (5)0.5^2*sin(90) + (2)0.5*cos(90))

a = dv/dt = i(-1*3^2 + 2*0.5 + 1) + j( (5)0.5^2)

a = -7i + 1.25j m/s/s

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