Centripetal Acceleration of rope and bucket

AI Thread Summary
The discussion revolves around calculating the centripetal acceleration of a bucket being whirled in a vertical circle. For part (a), the speed of the bucket at the lowest point is determined to be 1.72 m/s, using the equation that accounts for the tension in the rope and gravitational force. In part (b), the required speed at the top of the circle to prevent slack in the rope is calculated to be 4.83 m/s, factoring in both tension and gravity acting downward. Clarifications are made regarding the tension force at the top of the circle, confirming it can be zero if the bucket is moving at just the right speed. Overall, the calculations for both parts appear to be correct.
physicsnobrain
Messages
123
Reaction score
0

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?

Homework Equations


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 isn't 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.
 
Physics news on Phys.org
part a) looks ok to me.

physicsnobrain said:
For there to be just enough for no slack.
If there's just enough for no slack, then what is Ft?
 
rcgldr said:
part a) looks ok to me.

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

hmmm. 0N right?
 
physicsnobrain said:
hmmm. 0N right?
correct.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top