Question

A boy twirls a 15-lb bucket of water in a vertical circle. If the radius of curvature of the path is 4 ft, determine the minimum speed the bucket must have when it is overhead at A so no water spills out.

Answer 1

Answer:

11.35 m/s

Explanation:

Given,

Weight of the bucket = 15 lb

radius of curvature, r = 4 ft

Minimum speed of the bucket = ?

Using equation of centripetal acceleration

$W=\dfrac{mv^2}{R}$

$15=\dfrac{\dfrac{m}{g}\times v^2}{4}$

$15=\dfrac{\dfrac{15}{32.2}\times v^2}{4}$

$v^2 = 128.8$

$v = 11.35\ m/s$

Speed of the bucket  = 11.35 m/s