4.00 kg block of Figure 6 is attached to a vertical rod with two laces. When the

Question

4.00 kg block of Figure 6 is attached to a vertical rod with two laces. When the system revolves around the axis of the rod, the cords extend as shown in the diagram, and the tension in the upper cord is 80.0 N. a) the tension in the lower cord? b) How many revolutions per minute (rpm) gives the system? c) Calculate the rpm with the lower cord loses all tension, d) Explain what happens if the number of rpm is less than in subsection c). El bloque de 4.00 kg de la figure 6 esta unido a una varilla vertical con dos cordones Cuando el sistema gira en torno al eje de la varilla, los cordones se extienden como se indica en el diagram, y la tension en el cordon superior es de 80.0 N. a) Que tension hay en el cordon inferior? b) Cuantas revoluciones por minuto (rpm) da el sistema. c) las rpm con las que el cordon inferior pierde toda tension, d) Explique que sucede si el numero de rpm es menor que en el inciso c).

Explanation / Answer

Angle made by both the string with rod (or we can say with the vertical direction)

Angle C = cos-1(1/1.25) = 37°

Cos C = 4/5

SecC = 5/4

Resolving all the forces along X and Y direction ( X is along horizontal and Y is vertical up in the figure)

Suppose tension in upper string is Tu an tension in lower string is Tl.

(a). Let's balancing all the force along Y direction

Tu cosC = mg + Tl cosC

Thus Tl = Tu - mg secC = 80 - (40×5/4)= 30 N. ....answer. Assuming g=10m/s^2

(b) . since block is moving on a circular path of radius R = 1.25sinC= 0.75 m

So net force must be along negative X direction (= centripetal force = mw2R)

Tu sinC +Tl sinC = 4 ×0.75 ×w2 = 3w2

We have calculated the value of both tension

So putting those value we get

110 ×3/5 = 3 w2

w = 22 = 4.7 rad per sec = 44.9 rpm

(C) when Lowe string loses all tension and tension in string be 80 N .Then angle made by string with vertical will change

Suppose angle made by string with vertical be B

Than 80 cosB = 40

B = 60°

And so

T sin 60° = 4 w2 (1.25 sin60°)

Or

w2 = 80/5= 16 or w= 4 rad/sec = 38.21 rpm

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