Question: A 3.45 kg object is attached to a vertical rod by two strings as in Figure P6.11. The object rotates in a horizontal circle at constant speed 5.65 m/s. What are the tensions of the two strings, individually?
m = 3.45kg
length of both strings (individually) = 2m
vertical distance between two strings = 3m
Sum(F) = m(v^2/r)
Sum(F) = m(a)
The Attempt at a Solution
I assumed that the ball was revolving around the middle of the rod, thus we have a triangle with hypotenuse of 2m and side 1.5m and by solving we find the radius:
r = sqrt(1.75) = 1.32m.
We should be able to find the sum of the forces by using m(v^2/r):
(3.45kg)((5.65m/s)^2/1.32m) = 83.25N.
This is the centripetal force, right?
a_c = v^2/r, thus a_c = (5.65m/s)^2/(√1.75m)=33.429m/s^2
I then used Sum(F) = ma to find the x-component of T (tension of string):
(3.45kg)(33.429m/s^2) = 115.33N
We have a triangle with legs 1.5m and √1.75m and hypotenuse 2m, thus the angle made by T and r is θ = sin^-1(1.5m/2m) = 48.59°.
Therefore, T = 115.33N/cos(48.59°) = 174.36N
Both strings share the force equally and both have this value, correct?
I just want to make sure I am doing this right...I have already lost points at wrong attempts already (online homework). Could someone look this over and make sure I did it correctly and notify me if there's anything I might have misunderstood? Thanks!