What is the Maximum Speed at Point Q Without Breaking the String?

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To determine the maximum speed at point Q without breaking the string, the tension in the string (T) must be analyzed in relation to the gravitational force (mg) acting on the ball. The correct equation to derive Vmax incorporates the opposing directions of tension and weight, leading to the expression T - mg = mv^2/r. This reflects that the net force acting on the ball is the difference between the tension and the gravitational force. Clarification on the signs of the forces is crucial, as they cannot be treated as having the same direction. Understanding these dynamics is essential for accurately calculating the maximum speed without exceeding the string's tension limit.
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Homework Statement


The maximum tension the string can have without breaking is Tmax. Derive an expression for Vmax, the maximum speed the ball can have at point Q without breaking the string.


Homework Equations


F=ma
Vc=(mv^2)/r
T=mg+ma

The Attempt at a Solution


I thought I could do T+mg=mv^2/r because mv^2/r-mg would give you the max speed to keep the same tension and anything great would produce a greater tension that the string doesn't have causing it to break. So, I pulled out a common factor in m and got a common denominator giving me m((v^2-gr)/r)=T


IMG_3397.jpg
 
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Sorry for the size of the pic. If you click on it it will expand.
 
Careful, the equation you have is incorrect -- tension and weight force act in opposite directions at point Q, so this needs to be reflected in their signs.
 
tdreceiver17 said:
Sorry for the size of the pic. If you click on it it will expand.

Thats what I thought at first but wanted to try something new. So is it the same thing I have put down but with a plus sign?
 
What I mean is that you have this:

tdreceiver17 said:
T+mg=mv^2/r

but T and mg cannot have the same sign. There has to be a negative somewhere on the left.
 
jackarms said:
What I mean is that you have this:



but T and mg cannot have the same sign. There has to be a negative somewhere on the left.

-T+mg=mv^2/r ?
 
No, you're mixing up your signs. Start with this:

At Q, which way do both forces point, and which way does the acceleration point?
 
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