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A linear impulse `int Fdt` acts at a point `C` of the smooth rod `AB`. The value of `x` is so that the end `A` remains stationary just after the impact is A. `l/4`B. `l/3`C. `l/6`D. `l/5` |
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Answer» Correct Answer - C Let `J` be the impulse acting on the rod `J=mmv_(cm), Jx=1/12 ml^(2) omega`. Since the end `A` is stationary `V_(A)=V_(cm)-1/2 omega=J/m-((12 Jx)/(ml^(2)))l/2=0` |
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