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Question

A ring of mass m = 1 kg can slide over a smooth vertical rod. A light string attached to the ring passing over a smooth fixed pulley at a distance of L = 0.7 m from the rod as shown in Fig. 8.217.
At the other end of the string mass M = 5 kg is attached, lying over a smooth fixed inclined plane of inclination angle 370. The ring is held in level with the pulley and released. Determine the velocity of ring when the string makes an angle (α=370) with the horizontal. [sin 370 = 0.6].
980451_1f0d6d0542a540f2a3a7bc95c4961f26.png

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Solution

m=1kg
M=5kg

let, after reaching the final position
L increased by L, the ring descends by x
From diagram , tan37=x/L on x=3L4.
cos37=LL+ΔLΔL=L(sec371)



let, at final position, velocity of ring =Vγ so, from diagram velocity of M=Vγcos53
=3vr5

now change in potential Energy = change in kinetic energy
Mgsin37xΔ1+mgx=12mv2p+12 M(3vp5)2
0π5×10×38×29+10×3L4=12(v2γ+5×9v2r25)
on vγ=0

Ans- Velocity =0m/s

1992478_980451_ans_11be7ac89d87495fa64a7e64c783987a.png

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