Ball on two strings, rotating around a rod

[chiurox]

Homework Statement

A 1.0kg ball is attached to a vertical rod by two strings. Each string is 1.0m long, and they are attached to the rod a distance of 1.0m apart. The rod-ball-strings setup is rotating about the axis of the rod. As it rotates the strings are taut and form an equilateral triangle. The tension in the upper string is 25N.
a. Describe a free-body diagram showing all the forces on the ball.
b. What is the tension in the lower string?
c. What is the net force on the ball when the ball is in the position as described in the question?
d. What is the speed of the ball

Homework Equations

Moment of inertia for thing rod about the center =(1/12)Ml^2
l=1m
m=1.0kg
so I = 1/12

The Attempt at a Solution

For part B:
The tension of the lower string would be less than the upper string, since the upper string has to cope with mg. So would T2 = T1-mg ?

Part C:
I'm not sure how to start this one...

 

[rohanprabhu]

I don't know if you need the moment of inertia in this problem or not, but i better solve your misunderstanding about it. See, the moment of inertia you've mentioned $\frac{1}{12} ML^2$ is for a rod of mass 'M' and length 'L' when it is rotating about it's center of mass. However, in this case, our system is different. The rod's mass is not given and since it is rotating around an axis parallel to itself... for a thin and light rod, we can neglect it's contribution to the moment of inertia of the system.

For now, you only have a ball of mass 1kg, rotating at a distance '1.0m' from the axis and having a mass 1kg. Use the formula: $I = MR^2$ to calculate the moment of inertia of this system.

PartB: Yes, the lower string will have a lesser tension, but it is not given by $T_2 = T_1 - mg$. You need to break it into vector components and then subtract/add the magnitude of the vectors which are parallel or anti-parallel to each other. This being an equilateral triangle, you have the angles with you. Start by making a free body diagram.

PartC: The ball is moving only in the x-z plane as the y-directioned forces are cancelled. Hence, the only forces acting on the ball are the components of Tension in the x-z plane.

PartD: Once you know the answer from PartC, use that to compute the angular velocity, using the formula for centripetal force. With that and the radius, you can calculate the speed of the ball.

 

[Moetasim]

Part D:
The speed of the ball can be calculated using the formula v=ωr, where ω is the angular velocity and r is the distance from the axis of rotation to the ball. In this case, r = 1m since the ball is rotating around the axis of the rod. To find the angular velocity, we can use the equation ω = v/r = 2πf, where f is the frequency of rotation. Since the ball is rotating in a circle, we can also use the equation a=v^2/r to find the centripetal acceleration. We know that the net force on the ball is equal to the centripetal force, so we can set the equation T = mv^2/r equal to the tension in the upper string and solve for v.