Angular Velocity Keeping Strings Taut

Click For Summary
The discussion revolves around a physics problem involving a ball attached to two strings rotating around a vertical rod at a constant angular velocity. For part A, the user attempts to derive the forces exerted by each string on the ball, using equations for centripetal force and tension, but struggles to incorporate angular velocity (ω) into their solution. Clarifications are provided that the net force in the x-direction is not zero due to the acceleration involved, and that the tensions must be expressed in terms of ω rather than linear velocity (v). For part B, the correct approach is to set the tension in the lower string (T2) to zero to find the minimum angular velocity (ω_min) required to keep it taut. The discussion emphasizes the need to accurately account for forces and accelerations in the analysis.
newtophysics2
Messages
4
Reaction score
0

Homework Statement


A ball of mass m is attached by two strings to a vertical rod. as shown in the diagram attached. The entire system rotates at constant angular velocity ω about the axis of the rod.

a)Assuming ω is large enough to keep both strings taut, find the force each string exerts on the ball in terms of ω, m, g, R, and θ.

b)Find the minimum angular velocity, θ_min for which the lower string barely remains taut.


Homework Equations


F_centripetal=mv2/r
F=ma


The Attempt at a Solution


A) To keep the strings taut, the net force in the y-axis and the x-axis have to both equal 0. I used forces and tension, but my answer didn't contain ω, but I feel like it should...
T1 tension is making θ angle with the vertical
T1cosθ along vertical upward
T1sinθ along horizental i.e towards the center of the circular path
applying ΣFy =0 ΣFx =0
T1cosθ=mg
T1=mg/cosθ
T1sinθ+T2=mv2/R
T2=mgsinθ+mv2/R

B) I would solve for v from the equations above, but the question doesn't say that I can use T (tension) in my answer...

What am I doing wrong in this problem? Thanks in advance!
 

Attachments

  • Diagram.png
    Diagram.png
    2 KB · Views: 1,613
Physics news on Phys.org
Velocity and angular velocity are related by,

v = ωR

Also,

b)Find the minimum angular velocity, θ_min for which the lower string barely remains taut.

should read,

b)Find the minimum angular velocity, ω_min for which the lower string barely remains taut.
 
Ok, thanks! So did I do part A correctly? And for part B, do I just set T2=0 and then substitute v=ωR to solve for ω?
 
newtophysics2 said:
A) To keep the strings taut, the net force in the y-axis and the x-axis have to both equal 0.
No. The complete equation is Fnet=ma. There is an acceleration, so the net force must provide that.
 
haruspex said:
No. The complete equation is Fnet=ma. There is an acceleration, so the net force must provide that.

How do I determine the acceleration? So is my solution wrong?
 
newtophysics2 said:
How do I determine the acceleration? So is my solution wrong?
You already correctly used it in your equations, here:
T1sinθ+T2=mv2/R
I was just pointing out that your comment that the net force in the x direction is 0 was wrong.
T1cosθ=mg
T1=mg/cosθ
T1sinθ+T2=mv2/R
T2=mgsinθ+mv2/R
I don't think that last equation follows from the ones before.
Also, part requires you to find the tensions in terms of ω, m, g, R, and θ, not v. So you need Spinnor's equation there..
For part B, yes, put T2 = 0 and find ω.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
27
Views
1K
  • · Replies 3 ·
Replies
3
Views
989
  • · Replies 16 ·
Replies
16
Views
6K
  • · Replies 15 ·
Replies
15
Views
6K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 32 ·
2
Replies
32
Views
3K