Uniform Circular Motion: Solving for String Tension

[QCHabsQC]

Homework Statement

A ball of mass 4 kg on a string of length 1.5 m is being swung in a circle in the vertical plane at a constant speed of 10 m/s. Find the tension in the string at the bottom and top of the balls path

Homework Equations

ƩF = m(v^2/r)

The Attempt at a Solution

This is how I attempted to solve this problem, but my answer is not correct.
Is it correct that the tension in the string at the bottom is: T = m(v^2/r) + mg
And at the top: T = m(v^2/r) - mg

 

[Simon Bridge]

Please show your working - not just some answers.

Consider:
T = m(v^2/r) + mg

This is saying that, at the bottom of the path, the tension has to overcome gravity and still have enough left over to give a centripetal force... the net force points upwards.
Does that make sense?

 

[QCHabsQC]

Yes; so the tension at the bottom would be, T = (4*(10)^2)/1.5 + 9.8*4 = 217 N. Is this correct?

 

[Simon Bridge]

I don't check arithmetic.
But you appear to have answered your own question.

 

[Nickie]

I would like to clarify that the equations used are correct, but there may be a mistake in the application of the equations. The equation ƩF = m(v^2/r) is the centripetal force equation, which represents the net force acting on an object in uniform circular motion. In this case, the object is the ball and the force is provided by the tension in the string.

At the bottom of the ball's path, the tension in the string is equal to the centripetal force plus the weight of the ball, since the ball is at the bottom of the circle and is experiencing both the centripetal force and its weight. Therefore, the correct equation would be T = mv^2/r + mg.

At the top of the ball's path, the tension in the string is equal to the centripetal force minus the weight of the ball, since the ball is at the top of the circle and is only experiencing the centripetal force. Therefore, the correct equation would be T = mv^2/r - mg.

It is important to note that the tension in the string is not constant throughout the ball's path, as it varies based on the position of the ball in the circle. To accurately solve for the tension at any point in the circle, the centripetal force equation should be used and the appropriate forces (such as weight) should be included in the calculation.