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The system is released from rest at θ = 0° when a constant couple moment M = 100 N.m is applied. If the mass of the cable and links AB and BC can be neglected, and each pulley can be treated as a disk having a mass of 6 kg, determine the speed of the 10-kg block at the instant link AB has rotated θ = 90°. Note that point C moves along the vertical guide. Also, the cable does not slip on the pulleys.

The system is released from rest at θ 0 when a constant couple moment M 100 Nm is applied If the mass of the cable and links AB and BC can be neglected and each class=

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okpalawalter8

For a system released from rest at θ = 0° when a constant couple moment M = 100 N.m is applied,  the speed of the 10-kg block is mathematically given as

V=4.33m/s

What is the speed of the 10-kg block?

Generally, the equation for the workdone  is mathematically given as

W=T tehta<dtheta

W=100(90*\pi/180)

W=1.5707*100

W=1.57Nm

The change in potential energy across the pulley

dP=mgh

dp=10*9.81*111.8

dp=10.97J

For the thrid position, potential energy is

dP=mg(0.3)

dP=17.658J

dP'=17.658J-13.125

dP'=-4.532J

For 2nd position dP=0

The change in Kinectic energy across the pulley\

dK.E=0.5mv^2

For 1st

dk.E=0.5m(10)^2

2nd

dK.E=0.5Iw^2

dK.E=0.5(7.5*10^-3)(v^2/0.05)^2

3rd

2nd=3rd

In conclusion,

157.07=dKE+dP.E

157.07=5v^2+ (7.5*10^-3)(v^2/0.05)^2+10.97-4.53

V=4.33m/s

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