The mine skip is being hauled to the surface over the curved track by the cable

Question

The mine skip is being hauled to the surface over the curved track by the cable wound around the 30-in. drum, which turns at the constant clockwise speed of 120 rev/min. The shape of the track is designed so that y = x2/27, where x and y are in feet. Calculate the magnitude of the total acceleration of the skip as it reaches a level of 3.3 ft below the top. Neglect the dimensions of the skip compared with those of the path. Recall that the radius of curvature is given by

The mine skip is being hauled to the surface over the curved track by the cable wound around the 30-in. drum, which turns at the constant clockwise speed of 120 rev/min. The shape of the track is designed so that y = x2/27, where x and y are in feet. Calculate the magnitude of the total acceleration of the skip as it reaches a level of 3.3 ft below the top. Neglect the dimensions of the skip compared with those of the path. Recall that the radius of curvature is given by

Explanation / Answer

Given:

y = x2 / 27

y = 3.3 ft

x2 = 27*3.3 = 89.1

x = 9.439 ft

Differentiating y = x2 / 27

dy/dx = 2x / 27 = 2(9.439) / 27 = 0.699

Again differentiating

d2y/ dx2 = 2/27 = 0.074

so, p = (1 + (dy/dx)2)3/2 / d2y/ dx2

p = (1 + 0.4886)3/2 / 0.074

p = 24.543 ft

v = rw

r = 30 inch = 2.5 ft

w = 120 rev/min = 12.56637 rad/sec

v = 2.5*12.56637

v = 31.415 ft/sec

so, a = v2 / p

a = 40.21 ft/sec2

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