The m 365kg block in the figure below is attached to a vert

The m = 3.65-kg block in the figure below 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 diagram and the tension in the upper string is 75.5 N. (Let d₁ = 1.21 m, and d₂ = 1.80 m, and neglect the height of the block.)

(a) What is the tension in the lower cord? (The answer is 27.4 N, I just want to know how to get this answer)

(b) How many revolutions per minute does the system make? (The answer is 46.1 rpm, I just want to know how to get this answer)

(c) Find the number of revolutions per minute at which the lower cord just goes slack. (THE Answer???31.5 RPM, I????WANT??KNOW???TO???THIS???????

Solution

Tension in the upper wire = Tu

Tension in lower wire = Tv

W = mg downwards on the block

A = angle of wire with horizontal

using force balance in x and y direction

in y-direction

Total Fy = 0 = Tu*sin A - Tv*sin A - mg

Tv = (Tu*sin A - m*g)/sin A = Tu - m*g/sin A

sin A = 0.9/1.21

A = arcsin (0.9/1.21) = 48.05 deg

Tv = 75.5 - 3.65*9.81/sin 48.05 deg

Tv = 27.36 N = 27.4 N

B.

in x-direction

Total Fx = m*w^2*r

Tv*cos A + Tu*cos A = m*w^2*r

r = sqrt (1.21^2 - 0.9^2) = 0.8087 m

w = sqrt (cos A*(Tu + TV)/(m*r))

w = sqrt (cos 48.05 deg*(75.5 + 27.4)/(3.65*0.8087))

w = 4.827 rad/sec = 4.827*60/2pi = 46.09 rpm

w = 46.1 rpm

C.

Tv = 0 in this case

mow using force balance:

in x-direction

Tu*cos A = m*w^2*r

in y*direction

Tu*sin A = m*g

dividing both equation:

tan A = g/w^2*r

w = sqrt (g/r*tan A)

w = sqrt (9.81/(0.8087*tan 48.05 deg)) = 3.302 rad/sec

w = 3.302*60/2*pi = 31.5 rpm