Inside a box(mass "M"), a light string "X" is atthached to it's bottom and the other end of the string is attached to a mass "m"(this makes a pendulum).
This system is attached to the point O(the center of gravity of the system), by a string " Y" and moved with constant speed of "v" in a vertical circle of radius "R"
Find the tensions of the strings X and Y,when the system is in the right hand side of the point O.(The right hand side end of the circular motion, let's call it point B)
At the point B,if we apply F=ma towards the center (point O) considering the system,
Let's call the tension of the string Y at the point B as Ty
Ty - (M+m)v^2/R = (M+m)*0
Ty=(M+m)v^2/R
Am I correct and how do we find the tension of the string X?
*Ty - (M+m)v^2/R + (M+m)g = (M+m)*0
Ty= (M+m)[(v^2/R) - g ]
Yes, you are correct in your calculation for the tension of string Y at point B.
To find the tension of string X, we can use the fact that at point B, the tension in string X must provide the necessary centripetal force to keep the mass m moving in a circular path. The centripetal force is given by the equation:
F = m * a,
where F is the centripetal force, m is the mass of the object (in this case, mass m), and a is the acceleration towards the center (point O).
Since the system is moving with a constant speed of v, we know that the acceleration a is given by:
a = v^2 / R,
where v is the speed of the system and R is the radius of the circular path.
Therefore, the tension in string X (Tx) can be calculated as:
Tx - m * a = m * a,
Tx = 2 * m * a,
Tx = 2 * m * (v^2 / R).
So, the tension in string X is given by 2 times the mass m times the acceleration (v^2 / R).