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boyakko [2]
2 years ago
10

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 li

nks 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.

Physics
1 answer:
Fantom [35]2 years ago
4 0

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

<h3>What is the speed of the 10-kg block?</h3>

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

Read more about Kinetic energy

brainly.com/question/999862

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Which of the following statements about this experiment is FALSE? Before each trial one should reshape the bob into something like a ball. You may assume the collision between the bob and the box is completely inelastic. The initial position for the box should be just touching the pendulum bob when it is hanging straight down. To make the box move, the pendulum bob should hit close to the bottom of the box during the collision is given below

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So, The bob is held at angle theta from initial position, P. Find the displacement from P.  

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