- #1

Zang

## Homework Statement

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*=

*x*/28, where

^{2}*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

## Homework Equations

a

_{n}= v

^{2}/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.

a

_{n}= 15.92

^{2}/15.33 = 16.53 ft/s

^{2}.

since a

_{t}= 0 => a = a

_{n}= 16.53 ft/s

^{2}, but it said my answer is wrong. Is that because a

_{t}is not 0 and how do I find it?