Circular Motion, a swinging ball and tension in string

[Sdarcy]

http://www.mech.uq.edu.au/courses/mech2210/yat/q/swinging_ball.jpg

The mass of the ball is m, as given below in kg. It is released from rest. What is the tension in the string (in N) when the ball has fallen through 45o as shown.

m[kg] = 3.335

I've used the formulae:

v = sqrt[2gL(sin alpha - sin alpha0)]
T = m(3 sin alpha - 2 sin alpha0)

and assumed that alpha0 = 0 degrees

I don't get the answer that the system is looking for (and it doesn't tell me what the right answer IS)

Any ideas where I've gone wrong?

Cheers...

 

[Doc Al]

v = sqrt[2gL(sin alpha - sin alpha0)]
T = m(3 sin alpha - 2 sin alpha0)

Where do these formulas come from?

Unless you've derived them yourself, stick to basic principles: Conservation of energy and Newton's laws.

 

[baba89]

First, let's discuss the concept of circular motion. When an object moves in a circular path at a constant speed, it is undergoing circular motion. In this case, the swinging ball is moving in a circular path due to the tension in the string. This tension is acting as a centripetal force, keeping the ball in its circular path.

Now, let's look at the formulae you have used to calculate the tension in the string. The first formula, v = sqrt[2gL(sin alpha - sin alpha0)], is the equation for the velocity of the ball at any point along its path. Here, g is the acceleration due to gravity (9.8 m/s^2), L is the length of the string, alpha is the angle the string makes with the vertical when the ball is at a certain point, and alpha0 is the initial angle of the string (in this case, 0 degrees).

The second formula, T = m(3 sin alpha - 2 sin alpha0), is the equation for the tension in the string at any point along its path. Here, T represents tension, m is the mass of the ball, and alpha and alpha0 have the same meanings as in the first formula.

Now, let's apply these formulae to the given scenario. The ball is released from rest, so its initial velocity is 0. This means that alpha0 is also 0 degrees. The ball falls through 45 degrees, so alpha is 45 degrees. Plugging these values into the first formula gives us:

v = sqrt[2(9.8)(L)(sin 45 - sin 0)]
v = sqrt[19.6L(0.707)]
v = 4.14L

To find the tension in the string, we can now use the second formula:

T = m(3 sin 45 - 2 sin 0)
T = (3.335)(3)(0.707) - (3.335)(2)(0)
T = 7.46 N

Therefore, the tension in the string when the ball has fallen through 45 degrees is 7.46 N. It is possible that the system is looking for a slightly different answer due to rounding errors or different values for g and m, but this should give you a good estimate.

If you are still not getting the expected answer, double check your calculations and make sure you are using the correct values for