Problem 10.95 A 460.0 ?g bird is flying horizontally at 2.30 m/s , not paying mu

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

--------------------------------------------------------
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

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.