Asked by Allison
                The 4.00-kg block in the figure is attached to a vertical rod by means of two strings. When the system rotates about the axis of the rod, the strings are extended towards the rod and each have a length of 1.25.  The length from the upper string and lower string on the rod is 2. So it forms a triangle with sides 2, 1.25, 1.25; with the block between the 1.25 sides. The tension in the upper string is 85.0 N . Tension of the lower cord is 36 N.
a) How many revolutions per minute does the system make?
b)Find the number of revolutions per minute at which the lower cord just goes slack.
            
        a) How many revolutions per minute does the system make?
b)Find the number of revolutions per minute at which the lower cord just goes slack.
Answers
                    Answered by
            Jose
            
    For part a)
T1 = Upper cord
T2 = Lower cord
theta = angle between vertical rod and upper cord = arcos(1/1.25)
R = radius of circular motion = sqrt((1.25^2) + 1^2)
M = mass of object
T1sin(theta)+T2sin(theta) = (MV^2)/R
Solve for V
Turn your V (should be in m/s) into revs/min.
rev = circumference = 2Rpi
m/s * 1 rev/(2Rpi)m * 60s/1min = rev/min
V in rev/min = (60V)/(2Rpi)
Haven't figured out part b) yet
    
T1 = Upper cord
T2 = Lower cord
theta = angle between vertical rod and upper cord = arcos(1/1.25)
R = radius of circular motion = sqrt((1.25^2) + 1^2)
M = mass of object
T1sin(theta)+T2sin(theta) = (MV^2)/R
Solve for V
Turn your V (should be in m/s) into revs/min.
rev = circumference = 2Rpi
m/s * 1 rev/(2Rpi)m * 60s/1min = rev/min
V in rev/min = (60V)/(2Rpi)
Haven't figured out part b) yet
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