In summary: Some tips might be to try to simplify your problem, or to break it down into simpler parts. It sounds like you're still having a hard time getting started, so it might help to look at some solutions that others have posted. Thanks!
  • #1
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Homework Statement



A ball with the mass m is attached to a rod, suspended by two strings both with lengths L.
The rod is rotating with the angular velocity ω and the ball rotates with it in such a way that the strings are taut and the ball moves in a circular pattern. I tried to draw it on my computer farily accurate. (See attached image)

The first task is to provide a relationship between the tensions in the strings. (The upper tension over the lower tension.)
And the second is to determine the minimum angular velocity ##ω## for the lower string to be taut.

Homework Equations



I know that the centripetal acceleration ##a=ω^2r## and the centripetal force ##F_c=ma=mv^2/r##

The Attempt at a Solution



So I have to get the X- and Y components of the tension force for each string. The weight ##w=mg## of the ball.
But I'm having real troubles to getting started. Any good tips on stuff that I've left out and how to tackle this one?

Thanks
 

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  • #2
Quadrat said:

Homework Statement



A ball with the mass m is attached to a rod, suspended by two strings both with lengths L.
The rod is rotating with the angular velocity ω and the ball rotates with it in such a way that the strings are taut and the ball moves in a circular pattern. I tried to draw it on my computer farily accurate. (See attached image)

The first task is to provide a relationship between the tensions in the strings. (The upper tension over the lower tension.)
And the second is to determine the minimum angular velocity ##ω## for the lower string to be taut.

Homework Equations



I know that the centripetal acceleration ##a=ω^2r## and the centripetal force ##F_c=ma=mv^2/r##

The Attempt at a Solution



So I have to get the X- and Y components of the tension force for each string. The weight ##w=mg## of the ball.
But I'm having real troubles to getting started. Any good tips on stuff that I've left out and how to tackle this one?

Thanks

Couple of questions to get you started:

1) Which string is holding the ball up against gravity?

2) What does that tell you about the tension in that string?
 
  • #3
Thanks for chipping in!

The upper string is preventing the ball from accelerating towards the ground. So I'm guessing the vertical component for the tension is T_1y=mgsinα and the horizontal is T_1x=Lcosα or am I dead wrong here? It's very confusing.
 
  • #4
Quadrat said:
Thanks for chipping in!

The upper string is preventing the ball from accelerating towards the ground. So I'm guessing the vertical component for the tension is T_1y=mgsinα and the horizontal is T_1x=Lcosα or am I dead wrong here? It's very confusing.

You're correct on the first point. The upper string will have more tension in it than the lower string. I think you need to draw all the forces on the ball.
 
  • #5
When the ball is in a uniform circular motion as in the picture the tension from the lower string is zero, right? I'm considering the critical case where the lower string is precisely taut. If so the x-component from the upper string tension would be ##Tsinα = ma = mω^2r## and the y-component would be ##Tcosα-mg=0##?
 
  • #6
Quadrat said:
When the ball is in a uniform circular motion as in the picture the tension from the lower string is zero, right? I'm considering the critical case where the lower string is precisely taut. If so the x-component from the upper string tension would be ##Tsinα = ma = mω^2r## and the y-component would be ##Tcosα-mg=0##?

That's right for the case where tension in the lower string is zero. Can you see how to calculate ##\omega## from that?

For the first part, can you modify those equations to take account of a tension in the lower string?
 
  • #7
I'm guessing for the ##ω## part I could divide my x-equation with my y-equation to get rid of T, and solve for ω. ##tanα=(mω^2r/mg)## According to my drawing if I want to describe the distance ##d## from the highest point to the lowest point and the length from the top to the point where it's perpendicular to the CM of the ball would then be ##d/2##.

##tanα=L/(d/2)## --> ##ω=sqrt(2g/d)##

Does this make sense? The dimension of the square root seems fine at least.

And for the first part I have no idea how to do that!
 
  • #8
Quadrat said:
I'm guessing for the ##ω## part I could divide my x-equation with my y-equation to get rid of T, and solve for ω. ##tanα=(mω^2r/mg)## According to my drawing if I want to describe the distance ##d## from the highest point to the lowest point and the length from the top to the point where it's perpendicular to the CM of the ball would then be ##d/2##.

##tanα=L/(d/2)## --> ##ω=sqrt(2g/d)##

Does this make sense? The dimension of the square root seems fine at least.

And for the first part I have no idea how to do that!

I'm not sure why you've introduced ##d##. You just need gto eliminate T. ##\omega## is going to depend on ##g, L## and ##\alpha##

For the first part, what's stopping you introducing a second Tension (##T_2##) for the lower string? And adding its horizontal and vertical components to your equations?
 
  • #9
PeroK said:
I'm not sure why you've introduced ##d##. You just need gto eliminate T. ##\omega## is going to depend on ##g, L## and ##\alpha##

For the first part, what's stopping you introducing a second Tension (##T_2##) for the lower string? And adding its horizontal and vertical components to your equations?

Oh, well.. This is how confused I really am regarding this problem. I feel dumb as rocks.
 
  • #10
Quadrat said:
Oh, well.. This is how confused I really am regarding this problem. I feel dumb as rocks.

If you have ##T_1sin(\alpha)## as the horizontal component of the upper string, the you should be able to write down the horizontal component for the lower string.

It is a tricky problem, but maybe you have to start to trust techniques like force diagrams and maths, as it is difficult to see intuitively what's going on.
 
  • #11
PeroK said:
If you have ##T_1sin(\alpha)## as the horizontal component of the upper string, the you should be able to write down the horizontal component for the lower string.

It is a tricky problem, but maybe you have to start to trust techniques like force diagrams and maths, as it is difficult to see intuitively what's going on.

Yeah it's quiet unintuitive. I'll do some reading up on it and try again tomorrow!
 
  • #12
Edit: Brainfreeze
 
Last edited:
  • #13
I cannot get my head around how to draw the forces from the lower string acting on the ball.
 
  • #14
Okay, so on the first task I'm trying to do some kind of force equation for the two strings.
In the x direction: It should be T1sin(a)+T2sin(a)=mw^2r
And in the y direction: T1cos(a)=T2cos(a)+mg.

What would be the best way to solve for any of the tension forces? I need practice in equation systems, so any help is very much appreciated.
 
  • #15
And as for the latter task I find that omega shouldn't be lower than ##sqrt(g/Lcos(a))## for the lower string to be precicely taut.
 
  • #16
Quadrat said:
And as for the latter task I find that omega shouldn't be lower than ##sqrt(g/Lcos(a))## for the lower string to be precicely taut.
That's correct. The relationship in the tensions is just your second equation.
 
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Likes Quadrat
  • #17
Thank you PeroK for your patience and your help! Have a great day mate!
 

1. What is circular motion?

Circular motion is the movement of an object along a circular path. This means that the object maintains a constant distance from a fixed point while continuously changing direction.

2. How is tension related to circular motion?

Tension is the force that is applied to an object in order to keep it moving in a circular path. In circular motion, tension is always directed towards the center of the circle, as it is responsible for keeping the object on its circular path.

3. What is angular speed?

Angular speed is a measure of how quickly an object is rotating around a fixed point. It is usually measured in radians per second (rad/s) and is calculated by dividing the angle traveled by the time taken to travel that angle.

4. How does mass affect circular motion?

Mass has no direct effect on circular motion, as long as the object is not changing its speed or direction. However, a heavier object may require more force (tension) to maintain its circular motion compared to a lighter object.

5. Can an object in circular motion have a constant speed?

Yes, an object in circular motion can have a constant speed. This means that the object is moving at the same speed along its circular path and is not accelerating. However, the object is continuously changing its direction, which means that it is still accelerating and experiencing a force (tension) to maintain its circular motion.

Suggested for: Circular motion, tension and angular speed

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