When the 1.25 m strings are fully extended, they form a pair of "3-4-5" right triangles with the pole, such that the block is .75 m out from the rod and 1.00 m below the upper attachment. (m is meters, N is newtons, 9.81 m/s

The force exerted upward by the upper string is 80 N(1/1.25) = 64 N

The force exerted downward by the block and the lower string must total this:

64 N = T

(64 N - 39.24 N) = T

T

The total inward force is that exerted by the two strings

F

= .6(110.95 N) = 66.57 N

This centripetal force corresponds to a rotation rate (w radians/s) of

F

66.57 N = (4.0 kg)(w r/s)

w = sqrt[66.57/3] = 4.71

4.71 radians/s = (4.71 radians/s)(60 s/min)(1 rev/2Pi radians)

= 4.71(60)/(2Pi) = 44.98 rpm

When the lower string goes slack, the upward force in provided by the tension in the upper string, as is the inward force. The upward force balances the downward force due to the mass.

T

T

F

F

29.43 N = (4.0 kg)(w r/s)

w = sqrt[29.43/3] = 3.13

(3.13 radians/s)(60 s/min)(1 rev/2Pi radians) = 29.91 rpm

When the rotation rate is less than this, the block will be closer to the pole than the maximum .75 meters allowed by the string geometry.

^{2}is the gravitational constant we are using)The force exerted upward by the upper string is 80 N(1/1.25) = 64 N

The force exerted downward by the block and the lower string must total this:

64 N = T

_{lower}(1/1.25)+(4.0 kg)(9.81 m/s^{2}) = T_{lower}/1.25+39.24 N(64 N - 39.24 N) = T

_{lower}/1.25T

_{lower}= (1.25)(64 N - 39.24 N) = 1.25(24.76 N) = 30.95 NThe total inward force is that exerted by the two strings

F

_{in}= T_{upper}(.75/1.25) + T_{lower}(.75/1.25) = .6(80 N + 30.95 N)= .6(110.95 N) = 66.57 N

This centripetal force corresponds to a rotation rate (w radians/s) of

F

_{in}= mw^{2}R66.57 N = (4.0 kg)(w r/s)

^{2}(.75 m) = 3.0 w^{2}Nw = sqrt[66.57/3] = 4.71

4.71 radians/s = (4.71 radians/s)(60 s/min)(1 rev/2Pi radians)

= 4.71(60)/(2Pi) = 44.98 rpm

When the lower string goes slack, the upward force in provided by the tension in the upper string, as is the inward force. The upward force balances the downward force due to the mass.

T

_{upper}(1/1.25) = (4.0 kg)(9.81 m/s^{2}) = 39.24 NT

_{upper}= (1.25)(39.24 N) = 49.05 NF

_{in}= T_{upper}(.75/1.25) = .6(49.05 N) = 29.43 NF

_{in}= mw^{2}R29.43 N = (4.0 kg)(w r/s)

^{2}(.75 m) = 3.0 w^{2}Nw = sqrt[29.43/3] = 3.13

(3.13 radians/s)(60 s/min)(1 rev/2Pi radians) = 29.91 rpm

When the rotation rate is less than this, the block will be closer to the pole than the maximum .75 meters allowed by the string geometry.