A 450.0-{\ m g} bird is flying horizontally at 2.50{\ m{ m/s}} , not paying much

Question

A 450.0-{ m g} bird is flying horizontally at 2.50{ m{ m/s}} , not paying much attention, when it suddenly flies into a stationary vertical bar, hitting it 25.0cm below the top (the figure (Figure 1) ). The bar is uniform, 0.710m long, has a mass of 1.70kg , and is hinged at its base. The collision stuns the bird so that it just drops to the ground afterward (but soon recovers to fly happily away). a)What is the angular velocity of the bar just after it is hit by the bird? b)What is the angular velocity of the bar just as it reaches the ground?

Explanation / Answer

angular momentum before (due to the bird) I*? = angular momentum after due to the bar

So m*r^2*v/r = 1/3*M*L^2*? (I of the rod is 1/3*M*L^2 = 1/3*1.70*0.71^2 = 0.2856kg-m^2)

Therefore ? = m*r*v/(1/3*M*L^2) = 0.450*(0.710 - 0.250)*2.50/(1/3*1.70*0.710^2) = 4.026rad/s

B) Now use conservation of energy

So (U + K)1 = K2 Note U is based on the height of the center of mass)

M*g*L/2 + 1/2*I*?1^2 = 1/2*I*?2^2

1.70*9.8*0.71/2 + 1/2*0.2856*4.026^2 = 1/2*0.2856*?2^2

or 0.1428*?2^2 = 8.229

So ? = sqrt(8.229/0.1428) = 7.59 rad/s

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