Mass swinging in a horizontal circle on 2 strings

[applesoranges]

Homework Statement

The ball is spinning in horizontal plane attached with 2 massless strings to a rotating stick,
as shown in the picture. Find tension of the lower cord using the given value of tension in the upper cord and the mass of the ball. The length parameters are also given, so the angles can be calculated.
I don't want to copy the problem from the assignment in order not to violate any policies of my institution.

Relevant Equations

Fg=m*g, a=v^2/r

I have already solved this problem, just would like to double check something with you conceptually. I've got a negative result for the tension in the lower cord. Intuitively I think it is right, because the lower cord does not support the ball in its opposing the force of gravity. It actually adds up to the tension T1 ought to have. Please guide me in the understanding this problem. Also, for the summation of forces in the y-direction, if T2 is negative, the vector of T2x is in the same direction as m*g, that supports my result of T2 being negative. Am I on the right track?

 

[hmmm27]

If the contraption were spinning 10x as fast, would there still be no tension on T2 ?

 

[applesoranges]

If the contraption were spinning 10x as fast, would there still be no tension on T2 ?

Fair enough! I guess I'll have to model a situation where v is the way that T2 is 0 and see how it goes...

 

[haruspex]

I don't want to copy the problem from the assignment in order not to violate any policies of my institution.

Ok, but you could post a purely algebraic solution which we would be able to check.

 

[applesoranges]

Ok, but you could post a purely algebraic solution which we would be able to check.

What throws me off in the summation of forces in y-direction is T2y vector being opposite in sign to T1y. I get a negative number for T2y... I'm trying to comprehend what to make of it. I'll try to work on it more when I come back after my night shift.

 

[haruspex]

I get a negative number for T2y

If you define up as positive for all vertical vectors, of course the Y component of T2 will be negative.
I see nothing stopping you from posting your equations.

 

[applesoranges]

Ok, but you could post a purely algebraic solution which we would be able to check.

What throws me off in the summation of forces in y-direction is T2y vector being opposite in sign to T1y. I get a negative number for T2y... I'm trying to comprehend what to make of it. I'll try to work on it more when I come back after my night shift.

 

[applesoranges]

Given:
T₁=22.0N
m=0.500 kg
$\theta = 30.0^{\circ}$
Find:
T₂ = ?
What I have so far:
$\sum_{}^{}F_{x}^{}: T_{1}cos\theta +T_{2}cos\theta =ma_{c}$
$\sum_{}^{}F_{y}^{}: T_{1}sin\theta =T_{2}sin\theta + mg$
$T_{2}=\frac{T_{1}sin\theta-mg}{sin\theta }$

The result for T₂ is a positive number
T₂=12.2N
Question:
The big question is that I'm trying to visualize this situation and conceptualize what the positive value for T₂ means. Since it is a positive vector pointing in the direction of force of gravity, does that mean that the tension is actually cancelling T₁ in the y-direction, meaning that it is pulling on the ball down?

 

[haruspex]

Given:
T₁=22.0N
m=0.500 kg
$\theta = 30.0^{\circ}$
Find:
T₂ = ?
What I have so far:
$\sum_{}^{}F_{x}^{}: T_{1}cos\theta +T_{2}cos\theta =ma_{c}$
$\sum_{}^{}F_{y}^{}: T_{1}sin\theta =T_{2}sin\theta + mg$
$T_{2}=\frac{T_{1}sin\theta-mg}{sin\theta }$

The result for T₂ is a positive number
T₂=12.2N
Question:
The big question is that I'm trying to visualize this situation and conceptualize what the positive value for T₂ means. Since it is a positive vector pointing in the direction of force of gravity, does that mean that the tension is actually cancelling T₁ in the y-direction, meaning that it is pulling on the ball down?

Yes, the tension in the lower string exerts a downward force on the mass. Is that a problem?
What would happen if the lower string were cut?

 

[applesoranges]

Is that a problem?

Not a all, just a brain freeze.

What would happen if the lower string were cut?

Good question, because now I understand that T₂ can't really be negative. I mean, algebraically you could make T₂ in
$T_{2}=\frac{T_{1}sin\theta-mg}{sin\theta }$
less than zero. But that situation can't physically exist, right?
When the string is cut and T₂=0
$T_{1}sin\theta=mg$

 

[haruspex]

When the string is cut and T₂=0
$T_{1}sin\theta=mg$

No, I meant suppose it is spinning steadily with a nonzero tension, then the string is cut. What will happen to the mass and the other string?

 

[applesoranges]

No, I meant suppose it is spinning steadily with a nonzero tension, then the string is cut. What will happen to the mass and the other string?

The angle between T₁ and horizontal will have to increase to make sure ma_c equals T_1x.

 

[haruspex]

The angle between T₁ and horizontal will have to increase to make sure ma_c equals T_1x.

Since T₂ was pulling down on the mass, which way would you expect the mass to move when it is cut?

 

[applesoranges]

Since T₂ was pulling down on the mass, which way would you expect the mass to move when it is cut?

Upwards.

 

[haruspex]

Upwards.

Right.
After oscillating a bit, it would settle at a new equilibrium.
It's not obvious what that does to the tension in the upper string. By conservation of angular momentum, the rotation rate would be lower, and it would no longer have to fight the downward pull of the lower string. On the other hand, it would have to provide the whole centripetal force.