Tension in a String just before it breaks (Circular Motion)

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The discussion focuses on calculating the tension in a string just before it breaks while a ball is swung in a vertical circle. The initial velocity of the ball at the moment the string breaks is found to be 2.969 m/s using the equation Vf^2 - Vi^2 = 2aS. The user attempts to calculate the tension by considering the forces acting on the ball, including tension and weight, but initially miscalculates the acceleration. They are advised to trust their free body diagram and check for any mathematical errors or omissions in the problem statement. The conversation emphasizes the importance of correctly incorporating gravitational force into the tension calculation.
parkskier
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Homework Statement


A 120 g ball on a 60 cm long string is swung in a vertical circle about a point 200 cm above the floor. The string suddenly breaks when it is parallel to the ground and the ball is moving upward. The ball reaches a height of 650 cm above the floor. What was the tension in the string an instant before it broke?


Homework Equations


Vf^2-Vi^2=2aS


The Attempt at a Solution


I used the above equation to find the intial velocity when it breaks away from the string. Here's how I set that up:

(0)^2-(Vi)^2=2(-9.8)(.45)

This yielded: Vi= 2.969 m/s
Now my problem is how do I use this velocity to find the tension of the string just before it breaks?
 
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Draw the FBD and see what forces are acting on the ball and where the Tension fits in.

also you're missing one important equation:

\vec{a}=\frac{\vec{v}^2}{R}
 
Okay, so my FBD looks like this:

Tension<------
......|
......|
......V
.....mg

So...the only force in the x direction is Tension. The force of tension is F=ma, so using the equation you gave me I get my a to be 14.691 m/s^2, then multipling by the mass I get the Force of Tension to be 1.76, but this isn't right. Is there some way I need to incorporate the weight into the equation, I'm sure there must be.
 
Any help? Please?
 
parkskier said:
Any help? Please?
Trust your FBD. The only force acting in the radial centripetal direction is the tension force. Check math and round off errors, problem statement, etc.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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