330_Physics ProblemsTechnical Physics

# 330_Physics ProblemsTechnical Physics - 332 P11.19 Angular...

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332 Angular Momentum P11.19 The vector from P to the falling ball is rr v a ri j j =+ + =++ F H G I K J ii tt gt 1 2 0 1 2 2 2 AA cos ± sin ±± θθ ej The velocity of the ball is vv a j =+=− i tg t 0 ± So Lr v m Li j j j Lk =+ + F H G I K J L N M O Q P ×− =− mg t g t t A cos ± sin ± cos ± θ eje j 0 1 2 2 m l P FIG. P11.19 P11.20 In the vertical section of the hose, the water has zero angular momentum about our origin (point O between the fireman’s feet). As it leaves the nozzle, a parcel of mass m has angular momentum: Lm m r v m = ° = = rv sin . . . . 90 0 1 30 12 5 16 3 m m s ms 2 af bg The torque on the hose is the rate of change in angular momentum. Thus, τ == = = dL dt dm dt 16 3 16 3 6 31 103 .. . k gs Nm 22 v i O 1.30 m v f FIG. P11.20 Section 11.3 Angular Momentum of a Rotating Rigid Object
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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