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 sahpe of the track is designed so that y=x^2/40, where x and y are in feet. Calculate the magnitude of the total acceleration of the skip as it reaches a level of 2 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 ---In the picture above^^. Please show all work. Thank you!

Explanation / Answer

a(total)=a(tangential)+a(radial) a(tangential)=0 as the drum is rotating at constant angular speed a(radial)=w^2*rho w=120 rev/min=[2*pi*120]/60=12.56 rad/s rho=26.29 ft a(total)=a(radial)=4147.34 ft/s^2

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