A thin, uniform 3.90-km bar, 60.0-cm long, has very small 2.50-kg balls glued on

Question

A thin, uniform 3.90-km bar, 60.0-cm long, has very small 2.50-kg balls glued on at either end (the figure (Figure 1) ).It is supported horizontally by a thin, horizontal, frictionless axle passing through its center and perpendicular to the bar. Suddenly the right-hand ball becomes detached and falls off, but the other ball remains glued to the bar.Find the angular acceleration of the bar just after the ball falls off in rad/s^2?Find the angular velocity of the bar just as it swings through its vertical position rad/s.

Explanation / Answer

let m be the mass of the bar l be its length I = ml2/12 when the ball falls off total inertia of the system is Is= mbr2 +I mb--> mass of the ball r-->l/2=0.3m--> distance of the ball from the centre I--> moment of inertia of the bar torque on the system is t= mbgr t= Is a a--> angular acceleration (0.117+0.225)a=0.342*a= 2.5*9.8*0.3 a=21.49 rad / s2 conservation of energy 0.5Is?2 = mbgr on solivn we get ?=6.55 rad/s

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