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A vertically oriented uniform rod of mass M and length l can rotate about its upper end. A horizontally flying bullet of mass m strikes the lower end of the rod and gets stuck in it as a result the rod swings through an angle alpha. Assuming that mlt ltM. find a. the velocity of the flying bullet: b. the moment change in the system bullet rod during the impact: what causes that change of momentum. |
Answer» SOLUTION :The impulse acting on the rod -bullet system during the impact is shown in figure. `J_(1)` is the impulse acting on the rod by the hinge. Here the role of `J_(1)` is a little intersecting force. `J` could be greater than zero `J_(1)gt0, F_(1)` could be EQUAL to zero, `J_(1)=0`, and `F_(1)` could be less than zero `J_(1)lt0`. And as a consequence the linear momentum of the system may increase, may remain constant or may decrease.  a. During the impact, the moment ofimpulse acting o the rod bullet system ahout the hinge is zero. Angular momentum about the hine will remain constant. Let `omega` be the angular velocity of the system about the hinge. `["Angular momentum of the system"]_("initial")=["Angular momentum of the system"]_("final") ` `implies mvxxl=I_(0)xxomega`......i Where `I_(0)` is the moment of inertia of the rod bullet system after COLLISION about `O` `I_(0)=I_("rod")+I_("bullet")=1/3Ml^(2)+ml^(2)`.....ii `mvl=(1/3Ml^(2)+ml^(2))omega`..............iii As the system swings about hinge, the mechanical energy of the system will remain constant `/_\k+/_\U=0` `[1/2(1/3Ml^(2)+ml^(2))omega^(2)-0]` `=mgl/2(1-cosalpha)+mg(1-cosalpha)` From eqn iii `omega=(3mv)/(l(M+3m))`............iv Substituting this value of `omega` in eqn`v` we get `1/2(1/3Ml^(2)+ml^(2))((3mv)/(l(M+3m)))^(2)` `implies 1/2xx1/2l^(2)(M+3m)(9m^(2)v^(2))/(l^(2)(M+3m)^(2))` `=gl(1-cosalpha)/2(M+2m)` `implies v^(2)=(2gl)/3((M+2m)(M+3m))/m^(2)SIN^(2)alpha/2` Now for `mltltm, (M+3m)(M+3m)~~M^(2)` Thus `v^2=2/3gl(M^(2))/(m^(2))sin^(2)alpha/2impliesv=M/msqrt(2/3gl)sinalpha/2` b. Change in linear momentum during the impact `implies /_\p=p_(f)-p_(i)` `={M(l/2omega)+m(lomega)}-mv` `=kl((M+2m)/2)omega-mv`.........v From eqn iv and i `=l((M+2m)/2)(3mv)/(l(M+3m))-mv` `=3/2mv-mv=1/2mv` `=1/2m M/msqrt(1/2)gl sinalpha/2` `=Msqrt(1/6)glsinalpha/2` what causes linear momentum to change? Obviously, impulse `J_(1)` at hinge. Here `/_\p` is positive, linear momentum increase so `J_(1)` must be towards the right. |
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