On part (d), can't you just use trig to solve this; namely that ?
I found this problem very recently on the forum but mine seems to have a 4th and fifth part to it
The diagram shows a small ball of mass m=0.280kg that is attached to a rotating vertical bar by means of two massless strings of equal lengths L=1.10m. The distance between the points where the strings are attached to the bar is D=1.50m. The rotational speed of the bar is such that both strings are taut and the ball moves in a horizontal circular path of radius R at constant speed v=6.66m/s.
(a) Determine the magnitude and direction of the net force that is acting on the ball when it is in the position shown in the diagram.
(b) Determine the magnitude of the tension T1 in the upper string and the magnitude of the tension T2 in the lower string.
(c) Determine the speed of the ball at which the tension T2 becomes zero.
(d) Determine the angle of the top string when the velocity is v = 1.48 m/s.
(e) Determine the tensions T1 and T2 as a function of velocity (2 seperate equations).
I've already got the answers to a, b, and c, just need d and e.
no because in the new part D of the problem, they are changing the velocity so that T2 becomes 0 and T1 gets closer to the pole thus Distance isnt exactly half and radius becomes unknown
thanks for the help though, anything is appreciated as im stumped and quite furious at this point
Since 1.48 cm/sec is less than 2.91 m/sec, then the r should be less than the "taut, maximum" r that is 0.804674 cm.
The theta of the upper string then is arctan(r/y)
r = radius
y = vertical distance of the ball to the top anchor point.
The lower string is slacked or loose.
At the FBD of the ball now:
Let T = T1
Summation horizontal forces is zero, so,
T(r/1.10) = (.28)(1.48^2)/r
T = 0.674643 /r^2
Summation of vertical forces is zero,so,
T(y/1.10) = (0.28)(9.8)
T = 3.0184 /y
T = T,
0.674643/r^2 = 3.0184/y
r^2 = (0.674643*y)/3.0184 = 0.22351*y
We have 1.10^2 = r^2 +y^2, so,
r^2 = 1.21 -y^2
1.21 -y^2 = 0.22351*y
y^2 +0.22351*y -1.21 = 0By the Quadratic Formula,
y = 0.9939 m
So, r = sqrt(1.21 -0.9939^2) = 0.471341 m
Therefore, theta = arctan(0.471341 /0.9939) = 25.37 degrees -----answer
Cut the strings at the points of attachment so that you have a FBD.
Call the upper attachment point as A; the lower, as B.
Summation of moments about B is zero.
[(T1)(0.804674 /1.10)](1.50) = [(0.28)(v^2)/0.804674](0.75) +[(0.28)(9.8)](0.804674)
(T1)(1.097283) = (0.260975)v^2 +2.208025
T1 = (0.237837)v^2 +2.012266 .....in joules ------------answer
For the T2, following the same procedure above, but now taking moments about point A,
T2 = (0.237837)v^2 -2.012266 ....in joules -------------answer.