A 510.0 g bird is flying horizontally at 2.30 m/s , not paying much attention, w

Question

A 510.0 g bird is flying horizontally at 2.30 m/s , not paying much attention, when it suddenly flies into a stationary vertical bar, hitting it 25.0 cm below the top (the figure (Figure 1) ). The bar is uniform, 0.800 m long, has a mass of 1.80 kg , 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).

Part A

What is the angular velocity of the bar just after it is hit by the bird?

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Part B

What is the angular velocity of the bar just as it reaches the ground?

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Figure 1 of 1

Explanation / Answer

mass of the bird m1=510 g

bird speed v=2.3 m/sec

bird hitting at h=25 cm below the top

length of the bar l=0.8 m

mass of the bar m2=1.8 kg

A)

by using conservation of angular mometum,

L_bird=L_rod

m1*(l-h)*v=I*w_rod

m1*(l-h)*v=1/3*m2*l^2*w

0.51*(0.8-0.25)*2.3=1/3*1.8*0.8^2*W_rod

====> w_rod=1.68 rad/sec

angular velocity of the rod just after it is hit by the bird is,

W_rod=1.68 rad/sec

B)

by conservation of energy,

1/2*I*W_rod^2 + m2*g*(l/2) = 1/2*I*W'^2

1/2*(1/3*m2*l^2)*W_rod^2) + m2*g*l/2 = 1/2*(1/3*m2*l^2)*W'^2

(1/3*l*W_rod^2) + g = (1/3*l)*W'^2

(W_rod^2) + 3*g/l = W'^2

W'=sqrt((W_rod^2) + 3*g/l)

w'=sqrt(1.68^2+3*9.8/0.8)

w'=6.29 rad/sec

angular velocity of the bar just as it reaches the ground is,

w'=6.29 rad/sec

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