# Mass swinging in a horizontal circle on 2 strings

• applesoranges
In summary: Since T2 was pulling down on the mass, which way would you expect the mass to move when it is... since the tension in the lower string is now gone?I'm not sure, I'll have to think about that. right now I'm just trying to understand T2 and its meaning.I'm not sure, I'll have to think about that. right now I'm just trying to understand T2 and its meaning.
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?

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

hmmm27 said:
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...

applesoranges said:
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.

haruspex said:
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 said:
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.

haruspex said:
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.

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?

applesoranges said:
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
haruspex said:
Is that a problem?
Not a all, just a brain freeze.
haruspex said:
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##

applesoranges said:
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?

haruspex said:
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 mac equals T1x.

applesoranges said:
The angle between T1 and horizontal will have to increase to make sure mac equals T1x.
Since T2 was pulling down on the mass, which way would you expect the mass to move when it is cut?

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

applesoranges said:
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.

applesoranges

## What is mass swinging in a horizontal circle on 2 strings?

Mass swinging in a horizontal circle on 2 strings refers to the motion of an object attached to two strings, swinging in a circular path in a horizontal plane. This is a common experiment in physics to demonstrate the concept of centripetal force.

## What is the purpose of this experiment?

The purpose of this experiment is to demonstrate the relationship between centripetal force, mass, and velocity. It also helps to understand the concept of circular motion and how it differs from linear motion.

## What factors affect the motion of the mass in this experiment?

The motion of the mass in this experiment is affected by the length of the strings, the mass of the object, the speed at which it is being swung, and the tension in the strings. These factors determine the centripetal force acting on the mass and the resulting motion.

## How does the centripetal force change as the speed of the mass increases?

As the speed of the mass increases, the centripetal force also increases. This is because the force required to keep the mass in a circular path increases with the velocity. If the speed becomes too high, the strings may not be able to provide enough tension to keep the mass in a circular path, causing it to fly off.

## What is the formula for calculating the centripetal force in this experiment?

The formula for calculating the centripetal force in this experiment is F = mv²/r, where F is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of the circular path. This formula shows that the centripetal force is directly proportional to the mass and velocity, and inversely proportional to the radius of the circle.

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