The pail is rotated at a constant rate so it has the minimum speed at all points along its circular path. the water has mass m. when the pail is at the bottom of the circle, what is the magnitude of the force exerted by the water on the bottom of the pail?

Explanation:

The pail is rotated at a constant rate in vertical circular path  so it has the minimum speed at all points along its circular path . That means at top position the velocity is almost zero. In that case the centripetal force at top position will be provided by its weight or

mg = mv² / r ( r is radius of  vertical circular path )

v = √ rg

At the bottom position its velocity will be increased due to loss of potential energy

so 1/2 m V² = 1/2 m v² + mg x 2r  

V =√ 5 gr

If R be the reaction force at the bottom by bottom of pail

R - mg = mV² / r

R = mg +mV² / r

= mg + m x 5gr / r

R = 6mg

This is the magnitude of the force exerted by the water on the bottom of the pail .

a straight line sloped

0.19 m/s and 0.45 m/s