The mine skip is being hauled to the surface over the curved track by the cable wound around the 38-in. drum, which turns at the constant clockwise speed of 96 rev/min. The shape of the track is designed so that y = x2/28, where x and y are in feet. Calculate the magnitude of the total acceleration of the skip as it reaches a level of 3.5 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
an = v2/ρ
The Attempt at a Solution
I assumed that the angular speed of the drum is the same as the speed of the mine skip which is constant, so tangential acceleration would be 0.
v = 96 rev/min = 15.92 ft/s.
ρ = 15.33 ft using the given formula.
an = 15.922/15.33 = 16.53 ft/s2.
since at = 0 => a = an = 16.53 ft/s2, but it said my answer is wrong. Is that because at is not 0 and how do I find it?