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"
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb8d5d6cb-94d4-4c4c-87a7-4a343e295456%2F0be37fa7-0cc0-40c3-8156-d8a6f9386983%2Fyc1mpqb_processed.png&w=3840&q=75)
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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