# Centripetal Acceleration of rope and bucket

## Homework Statement

A bucket of mass 2.00 kg is whirled in a vertical circle of radius 1.10 m. At the lowest
point of its motion the tension in the rope supporting the bucket is 25.0 N. (a) Find the
speed of the bucket. (b) How fast must the bucket move at the top of the circle so that
the rope does not go slack?

Fc=mv^2/r

## The Attempt at a Solution

For part a), since it is at the lower point. The forces acting on the bucket are gravity acting downwards, and force of tension acting towards the center. Therefore Fc = Ft-Fg. We do Ft-Fg because force of tension is stronger (because the rope isnt snapping at this point).

We set up our equation to be: mv^2/r = Ft - mg

After rearraging this equation and solving for v I get 1.72 m/s.

For part b) since it is now at the highest point, the forces acting on it are both still tension and gravity, however tension is acting towards the center so both of these forces are acting downwards. For there to be just enough for no slack the force of gravity must be less than the centripetal force.

In this case centripetal force is equal to Ft + Fg.

We set up the equation to be: mv^2/r = Ft + mg

after rearranging this equation and solving for v I get 4.83m/s.

Am I correct, please I need help anyone.

rcgldr
Homework Helper
part a) looks ok to me.

For there to be just enough for no slack.
If there's just enough for no slack, then what is Ft?

part a) looks ok to me.

If there's just enough for no slack, then what is Ft?

hmmm. 0N right?

rcgldr
Homework Helper
hmmm. 0N right?
correct.