What Is the Minimum Speed Needed for a Whirling Ball to Taut a String?

  • Thread starter Thread starter Callen9
  • Start date Start date
AI Thread Summary
To determine the minimum speed needed for the lower string to be taut when a 2.0 kg ball is whirling, it is essential to analyze the forces acting on the ball, including gravity and the tensions in both strings. A free-body diagram should be drawn to visualize these forces and apply Newton's second law for both horizontal and vertical directions. The centripetal acceleration formula, v²/r, will be crucial in establishing the relationship between speed and tension. When the ball moves at a constant speed of 6 m/s, two equations can be derived to solve for the tensions in the strings. Ultimately, the tension in the lower string at the point it becomes taut can be used to find the minimum speed required.
Callen9
Messages
5
Reaction score
0

Homework Statement


A 2.0 kg ball is attached to a vertical post with two strings, one 2.0 m long and the other 1.0 m long as shown in the figure. If the ball is set whirling in a horizontal circle, what is the minimum speed necessary for the lower string to be taught? If the ball has a constant speed of 6 ms-1, find the tension on each string.

The picture would look something like this

|\
| \
| \ 2m
1.73m | \
| \
| ____O <--- 2kg
| 1m
|

Homework Equations


I know I have to use this equation at some point --> F=(mv^2)/r


The Attempt at a Solution


I've went throw a few sheets of paper already trying to find the answer. If someone could give me a push in the right direction would be great. This is what I know for sure. There are 3 forces at work here, Gravity, Tension of the upper line, Tension of the lower line. After that I get confused as to what to do. If I could get the first step on what to do and why to do it I think I could figure the rest out.

Any help and I would be extremely grateful!
 
Physics news on Phys.org
It screwed up my picture so i'll try this again.

|\
| \
| \
| \
| \
|___O
|
|
|
 
Try starting out by drawing a free-body diagram on the ball, labeling all forces, and writing out Newton's second law for both the horizontal and the vertical direction. (Remember that centripetal acceleration is v^2/r). Now you can solve the second problem, where the ball is moving at 6 m/s. You'll get 2 equations with the tensions as the 2 unknowns, so you can solve for the unknowns.

For the first problem, what's the tension in the bottom string when the string is barely taught? Use this value in the 2 equations you found before to solve for the speed.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...

Similar threads

Replies
3
Views
888
Replies
9
Views
2K
Replies
5
Views
2K
Replies
2
Views
2K
Replies
3
Views
4K
Replies
5
Views
15K
Replies
15
Views
6K
Back
Top