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, let's analyze the forces acting on the ball at different angles theta.
(a) Tension as a function of the angle theta:
At any given angle theta, the ball experiences two forces: the force of gravity acting downward (mg) and the tension force acting along the direction of the string away from the pivot.
The tension force consists of two components:
- A vertical component (T * cos(theta)) that counterbalances the weight of the ball (mg * cos(theta)).
- A horizontal component (T * sin(theta)) that provides the necessary centripetal force for circular motion.
Since the ball is in rotational equilibrium, the vertical components of force must cancel out. Therefore, T * cos(theta) = mg * cos(theta).
To find the tension T in terms of theta, we need to solve for T:
T = mg / cos(theta).
Now, let's substitute this value of T into the horizontal component of the tension force:
T * sin(theta) = (mg / cos(theta)) * sin(theta).
Simplifying this expression, we get:
T = 3mg * sin(theta).
Therefore, the tension as a function of the angle theta is T = 3mg * sin(theta).
(b) Maximum tension the string can sustain without breaking:
To find the maximum tension the string can sustain without breaking, we need to consider the maximum value of sin(theta), which is 1. This occurs when theta = 90 degrees or (π/2) radians.
Substituting this maximum value into our previous expression for tension, we have:
T_max = 3mg * sin(π/2) = 3mg.
Therefore, the maximum tension the string can sustain without breaking is 3mg.