The 4.00 kg block 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 as shown in the (Figure 1) and the tension in the upper string is 76.0 N. Find the number of revolutions per minute at which the lower cord just goes slack.

The number of revolutions per minute at which the lower cord just goes slack is 45.36 RPM.

The given parameters;

The following triangle can be formed from the given diagram;

     |

     | 1 m

     |---------x--------(θ)  4.00 kg

The hypotenuse of this triangle = 1.25 m

The height of the triangle = 1 m

The base of the triangle is calculated as;

Resolving the vertical component of the force;

Tension at the upper = Tension at the lower string

Resolving the horizontal component of the force;

Thus, the number of revolutions per minute at which the lower cord just goes slack is 45.36 RPM.

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Let theta be angle of upper string with rod.

This a 3-4-5 triangle where R is distance of block from rod. (.75 - 1 - 1.25)

Fc = m w^2 R    where Fc the centripetal force and w the angular velocity

Fc = T sin theta = T * .75 / 1.25 = .6 T = 45.6 N

w^2 = 45.6 / (4 * .75)

w = 3.90 /sec

RPM = 3.90 / sec * 60 sec / min / (2 * pi) = 37.2 RPM      (2 pi rad / rev)

i believe the answer is a hope it and is correct

not too sure thinking all accelerations are pointed down