Posting this question cause I struggled forever and could not get an answer. You attach one end of a string of length L to a small ball of inertia m. You attach the string's other end to a pivot that allows free revolution.

(a) What is the tension as a function of the angle theta it swept through?
(b) What maximum tension should the string be able to sustain if you want it to not break through the balls entire motion?

ANSWER: (a) 3mgSin(theta) I have no idea why
(b) Maximum of sin(theta) -> 1. You get 3mg

To solve this problem, we need to consider the forces acting on the ball attached to a string that can freely rotate.

(a) Let's first analyze the forces acting on the ball. The two main forces are gravity (mg) acting downwards and the tension force (T) acting along the string. At any given angle theta, there are two components of tension force: one along the direction of gravity and the other perpendicular to it.

The component of tension along the direction of gravity helps balance out the gravitational force, while the perpendicular component provides the centripetal force that keeps the ball in a circular path.

Using trigonometry, we can write the vertical component of the tension force as T * cos(theta) and the horizontal component as T * sin(theta).

When the ball is at the bottommost point (theta = 0), the vertical component of tension force balances out the gravitational force. Therefore, we have T * cos(0) = mg, which simplifies to T = mg.

At any other angle theta, the vertical component of tension and the gravitational force will create a net force in the vertical direction. However, the horizontal component of tension only provides the required centripetal force for circular motion.

To find the net vertical force, we subtract the gravitational force from the vertical component of tension:

Net vertical force = T * cos(theta) - mg

Since the string is in equilibrium, the net vertical force must be zero. Therefore, we can set up the equation T * cos(theta) - mg = 0 and solve for T:

T * cos(theta) = mg
T = mg / cos(theta)

Cosine of an angle can be rewritten using the Pythagorean identity:

cos(theta) = 1 / sqrt(1 + tan^2(theta))

Substituting this into our equation, we get:

T = mg * sqrt(1 + tan^2(theta))

To simplify this further, we can use the trigonometric identity tan^2(theta) = sin^2(theta) / cos^2(theta):

T = mg * sqrt(1 + (sin^2(theta) / cos^2(theta)))
T = mg * sqrt((cos^2(theta) + sin^2(theta)) / cos^2(theta))
T = mg * sqrt(1 / cos^2(theta))
T = mg / cos(theta)

Since 1 / cos(theta) is equal to sec(theta), we can write the final expression as:

T = mg * sec(theta)
T = 3mg * sin(theta)

Therefore, the tension as a function of the angle theta is 3mg * sin(theta).

(b) To find the maximum tension the string can sustain without breaking, we need to consider the maximum value of sin(theta), which is 1.

Substituting sin(theta) = 1 into the tension equation T = 3mg * sin(theta), we get:

T = 3mg * 1
T = 3mg

So, the maximum tension the string should be able to sustain without breaking is 3mg.