Centripetal Acceleration of rope and bucket

In summary, the bucket has a speed of 1.72 m/s at the lowest point of its motion and must move at a speed of 4.83 m/s at the top of the circle to prevent the rope from going slack. The force of tension at the top is 0N.
  • #1
physicsnobrain
123
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.
 
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  • #2
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?
 
  • #3
rcgldr said:
part a) looks ok to me.

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

hmmm. 0N right?
 
  • #4
physicsnobrain said:
hmmm. 0N right?
correct.
 
  • #5


I would say that your approach and calculations are correct. You have correctly identified the forces acting on the bucket at the lowest and highest points of the circle and have used the appropriate equations to solve for the speed at each point. Your calculations also seem to be in line with the given information and the units are consistent. Therefore, your solutions for the speed of the bucket at the lowest and highest points seem to be correct. Good job!
 

1. What is centripetal acceleration?

Centripetal acceleration is the acceleration towards the center of a circular path. It is caused by a force directed towards the center of the circle, known as the centripetal force.

2. How is centripetal acceleration related to the rope and bucket experiment?

In the rope and bucket experiment, the rope is attached to the bucket and swung in a circular motion. The centripetal acceleration is the acceleration that keeps the bucket moving in a circular path, caused by the tension in the rope pulling the bucket towards the center of the circle.

3. How do you calculate the centripetal acceleration in the rope and bucket experiment?

The centripetal acceleration can be calculated using the formula a = v²/r, where v is the velocity of the object and r is the radius of the circle. In the rope and bucket experiment, v is the linear speed of the bucket and r is the length of the rope.

4. Can the centripetal acceleration of the rope and bucket be increased or decreased?

Yes, the centripetal acceleration can be increased by either increasing the speed of the bucket or decreasing the length of the rope. It can also be decreased by decreasing the speed or increasing the length of the rope.

5. What are some real-life applications of centripetal acceleration?

Centripetal acceleration is used in many real-life applications, such as amusement park rides, bicycle wheels, and centrifuges in laboratories. It is also important in understanding the motion of objects in circular orbits, such as planets orbiting around the sun.

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