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A pulley of radius b=20 cm is fixed with a shaft of radus a=10 cm. Moment of inertia of shaft puley system is I=(33-800)kgm^(2) and the system is free to rotate about axis O of the shaft without friction. A block B of mass m_(2)=8 kg is resting over ann ideal spring of force costant. K=2048Nm^(-1) Lower end of the spring is fixed to the floor and the spring is vertical. Thread connected betwen shaft and block B is just taut. Another thead is connected between pulley and block A of mass m_(1)=4 kg. Initially this thread is loose. When block A is released first it falls freely through a height h=405/1024m then the thread becomes taut and block B is jerked into motion. calculate: a. Initial compressiion of the spring b. Velocity of block B when it is jerked into motion, c. Loss of energy during that jerk (g=10 ms^(-2))

Answer» <html><body><p></p>Solution :Since initially thread connected with block `A` is loose, therefoe weight block `B` is balanced by compressive force in the spring.<br/> `ky_(0)=mgimpliesy_(0)=(mg)/K80/2048m` <br/> `=y_(0)=<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.25/32m` or `125/32cm` <br/> Since block `A` first fall freely through <a href="https://interviewquestions.tuteehub.com/tag/height-1017806" style="font-weight:bold;" target="_blank" title="Click to know more about HEIGHT">HEIGHT</a> `h`, <a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>, it acquires a velocity `v_(0)=sqrt(2gh)` just before the thread becomes taut. Now the thread becomes taut and a impulse is developed in it. Due to htis impulse pulley is jerked into rotational <a href="https://interviewquestions.tuteehub.com/tag/motion-1104108" style="font-weight:bold;" target="_blank" title="Click to know more about MOTION">MOTION</a> and block `B` vertically upwards. Therfore wil not be impulse in spring as spring force is not an impulsive force. let impulses developed in threads connected with blocks `A` and `B` be `J` and `J'` respectively and let the pulley start to rotate clockwise with angular velocity `omega_(0)` then velocities of blocks `A` and `B` will be `bomega_(0)` (downwards) and a `omega_(0)` (upwards) respectively. <br/> Cosidering impulses and moment a as shown in the figure. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VOL2_C03_S01_024_S01.png" width="80%"/> <br/> For block `B, J'=m_(2)aomega_(0)` <br/> `m_(1)v_(0)-J=m_(1)bomega_(0)` <br/> For block `A, J.b-J'.a=Iomega_(0)` <br/> From i, ii and iii <br/> `J'=6.4Ns, J=4.8Ns` <br/> `omega_(0)=8rads^(-1)` <br/> Kinetic ennergy of the system. just after the thread becomes taut, <br/> `E=1/2m_(1)(bomega_(0))^(2)+1/2m_(2)(aomega_(0))^(2)=+1/2omega_(0)^(2)=9J` <br/> Loss of energy during the jerk `E_(0)-E=873/128J`</body></html>


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