The mine skip is being hauled to the surface over the curved track by the cable wound around the 36-in. drum, which turns at the constant clockwise speed of 114 rev/min. The shape of the track is designed so that y=x2/49, where x and y are in feet. Calculate the magnitude of the total acceleration of the skip as it reaches a level of 1.8 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 1+ L d²y dx2 36"

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The mine skip is being hauled to the surface over the curved track by the cable wound around the 36-in. drum, which turns at the
constant clockwise speed of 114 rev/min. The shape of the track is designed so that y = x²/49, where x and y are in feet. Calculate the
magnitude of the total acceleration of the skip as it reaches a level of 1.8 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
Answer: a = i
p=
1+
dy 23/2
dx
d²y
dx²
36"
ft/sec²
Transcribed Image Text:The mine skip is being hauled to the surface over the curved track by the cable wound around the 36-in. drum, which turns at the constant clockwise speed of 114 rev/min. The shape of the track is designed so that y = x²/49, where x and y are in feet. Calculate the magnitude of the total acceleration of the skip as it reaches a level of 1.8 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 Answer: a = i p= 1+ dy 23/2 dx d²y dx² 36" ft/sec²
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