Problem 10.95 A 460.0 ?g bird is flying horizontally at 2.30 m/s , not paying mu
ID: 1463297 • Letter: P
Question
Problem 10.95
A 460.0 ?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, 0.700 m long, has a mass of 2.20 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?
Part B
What is the angular velocity of the bar just as it reaches the ground?
Explanation / Answer
first Lets write the given data Mass of the bird m1= 460 g or 0.46 kgs
bird speed v=2.3 m/sec
bird hitting at h=25 cm below the top
length of the bar l = 0.7 m
mass of the bar m2= 2.2 kg
A) using the law of conversation of angular momntum L1 = L2
i,e L_bird=L_rod
where each L1 = m V r ; L2 = mL^2/3
m is mass V is velocity and r is radius
so
m1*(l-h)*v=I*w_rod
m1*(l-h)*v=1/3*m2*l^2*w
0.46 *(0.7-0.25)*2.3=1/3*2.2 *0.7^2*W_rod
W_rod=1.54 rad/sec
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B) Again From Conservation principle
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(1.54^2+ 3*9.8/0.7)
w'=6.67 rad/sec