Help on circular motion problem

[toesockshoe]

Homework Statement

A ball of mass m is attached to a rigid vertical rod by means of two massless strings each a length L. Both strings are attached on the rod a vertical distance D apart. The rod, strings, and ball are rotated at some speed and both strings are taught. If the tension in the upper string is given as FT, find the speed of the ball and the tension in the lower string.

Homework Equations

F=ma
a=v^2/r

The Attempt at a Solution

look at my pictures in the attachments... kinda hard to do physics problems online in my opnion. The problem is I feel like the teacher didnt make the question clear enough. can we assume that both strings are at the same angle? can we assume that the y-acceleration is 0?
also, I didnt even use the value D at all...so i feel like i MUST be going wrong somewhere. the first attachment is the entire problem, but if you can't read it, then I've attached 3 more close ups. THANKS!

[/B]

 

[BvU]

The two strings are taut and equal length. What does that say about $\theta$ and $\alpha$ ?

 

[haruspex]

After using BvU's tip, note that you invented those unknowns for the angles, so they cannot feature in your answer.

 

[toesockshoe]

The two strings are taut and equal length. What does that say about $\theta$ and $\alpha$ ?

i KNEW it had something to do with the problem, but i didnt know what it meant.
https://www.google.com/search?q=def...l4.1864j0j7&sourceid=chrome&es_sm=93&ie=UTF-8

it just tells me that taught is the past tense of teach... can you tell me the definition of taught in physics?

 

[BvU]

In physics taught is the past tense of teach.

A string isn't taught, a string can be wet, dry, curled up or wound up or it can be taut. Which google.

 

[toesockshoe]

The two strings are taut and equal length. What does that say about $\theta$ and $\alpha$ ?

oh so it means both angles are the same.

 

[BvU]

Makes life a lot easier doesn't it ! And you get a chance to use D

 

[toesockshoe]

Makes life a lot easier doesn't it ! And you get a chance to use D

if they are the same angle then the y acceleration would not be 0 right? the y components of the tension forces cancel out and the y acceleration would be 9.8?

 

[BvU]

Ah, just because they are both taut doesn't mean the tensions are equal ! (In fact they are asking for the tension in the lower string, so perhaps that's already Obvious to you).

You do have me wondering if there are more givens: the angular speed $\omega$ for instance.
[edit] but perhaps having $F_T$ is enough.

 

[haruspex]

You do have me wondering if there are more givens: the angular speed $\omega$ for instance.
[edit] but perhaps having $F_T$ is enough.

We're given the geometry, the mass, and the upper string tension. The lower string tension can be deduced from those. One can go on to find the rotation rate if desired.

 

[toesockshoe]

Ah, just because they are both taut doesn't mean the tensions are equal ! (In fact they are asking for the tension in the lower string, so perhaps that's already Obvious to you).

You do have me wondering if there are more givens: the angular speed $\omega$ for instance.
[edit] but perhaps having $F_T$ is enough.

oh yeah, so the y acceleration is not 0 right?

We're given the geometry, the mass, and the upper string tension. The lower string tension can be deduced from those. One can go on to find the rotation rate if desired.

 

[haruspex]

oh yeah, so the y acceleration is not 0 right?

No, if the strings remain taut then the geometry doesn't change, so there can be no vertical movement.
What three forces act on the mass? Which way is the mass accelerating?

 

[toesockshoe]

No, if the strings remain taut then the geometry doesn't change, so there can be no vertical movement.
What three forces act on the mass? Which way is the mass accelerating?

the 3 forces are the 2 tension forces and the gravitatioanl force. there would only be vertical acceleration if there is air resistance right? it is acceleration horizontally towards the center as of now.

 

[haruspex]

the 3 forces are the 2 tension forces and the gravitatioanl force. there would only be vertical acceleration if there is air resistance right? it is acceleration horizontally towards the center as of now.

Yes, except that I don't see how air resistance would lead to vertical acceleration.
Are you able to solve it now?

 

[toesockshoe]

Yes, except that I don't see how air resistance would lead to vertical acceleration.
Are you able to solve it now?

yes i think i can solve it ... ill post a pic later so you can check my work. its just that as the ball slows down (due to air resistance) would the ball not be in the air anymore? it would go down due to gravity and would lay along the pole. (wouldnt that be a small vertical acceleration) this is ofcourse if air resistance was introduced into the problem.

 

[haruspex]

as the ball slows down (due to air resistance) would the ball not be in the air anymore? it would go down due to gravity and would lay along the pole.

Sure, but you are asked to find the tension while the strings are both taut.

 

[toesockshoe]

Sure, but you are asked to find the tension while the strings are both taut.

hmm i got an answer but it looks really complex, and i don't know if i want to simplify it. do you mind checking it out to see if I made any glaring errors? Thanks!

glaring errors?

 

[haruspex]

do you mind checking it

Your expression for the tension is correct.
Not sure if your final answer for velocity is right, but it certainly doesn't need to be that complicated.
Go back to where you had $r\sin(\theta)$ terms. Express either r or $\sin(\theta)$ in terms of the other. What do you notice?

 

[toesockshoe]

Your expression for the tension is correct.
Not sure if your final answer for velocity is right, but it certainly doesn't need to be that complicated.
Go back to where you had $r\sin(\theta)$ terms. Express either r or $\sin(\theta)$ in terms of the other. What do you notice?

hmm, well i took theta to be the angle at the pole... not the ball and I think its correct. the angle at the ball is 90-tehta and cos(90-tehta) is the same as sin theta. i agree its really complicated, but i think its because of all the squeare roots. i do notice however, that I can factor out the sin(theta) in the square root and put it outside the radical, but is the logic itself wrong?

 

[haruspex]

hmm, well i took theta to be the angle at the pole... not the ball and I think its correct. the angle at the ball is 90-tehta and cos(90-tehta) is the same as sin theta. i agree its really complicated, but i think its because of all the squeare roots. i do notice however, that I can factor out the sin(theta) in the square root and put it outside the radical, but is the logic itself wrong?

Maybe you didn't understand my post. I'm suggesting you can make the final steps much easier (and get a simpler answer) if you first look at the relationship between r and theta. You can express r in terms of theta (or sin theta in terms of r) and the given dimensions D, L. Do that in your expression for v².

 

[toesockshoe]

Maybe you didn't understand my post. I'm suggesting you can make the final steps much easier (and get a simpler answer) if you first look at the relationship between r and theta. You can express r in terms of theta (or sin theta in terms of r) and the given dimensions D, L. Do that in your expression for v².

i thought theta can't be in my final answer cause it isn't given...

 

[haruspex]

i thought theta can't be in my final answer cause it isn't given...

That's right, it can't be, but I'm not suggesting it should.
What is the relationship between r, theta, and L?

 

[toesockshoe]

That's right, it can't be, but I'm not suggesting it should.
What is the relationship between r, theta, and L?

sin(theta)=r/L

or thetha = sin^-1(r/L)

 

[haruspex]

sin(theta)=r/L

Right. Use that to eliminate r from the line where you have "v² = ...". (This is one reason it's such a pain commenting on work posted as images. It's hard to point you to the part I'm commenting on.)

 

[toesockshoe]

Right. Use that to eliminate r from the line where you have "v² = ...". (This is one reason it's such a pain commenting on work posted as images. It's hard to point you to the part I'm commenting on.)

ohhh i see. I am an idiot... but hopefully my teacher won't be an a$$ and take off points for making it over complicated. also, i used latex on mathhelpforum, but haven't used it on this forum yet... (i know there are only very small technicality changes like the tags) and i don't want to show my work for a problem without latex... i think its just confusing.