Question

10.51 A uniform, 4.5-kg, square, solid wooden gate 1.5 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.1-kg raven flying horizontally at 5.0 m/s flies into this door at its center and bounces back at 2.0 m/s in the oppo- site direction. (a) What is the angular speed of the gate just after it is struck by the unfortunate raven? (b)During the collision, why is the angular momentum conserved, but not the linear momentum?

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Answer 1

a)
here

,
I gate = (1/3) *4.5 *1.5^2 = 3.375Kg.m^2

conservation of angular momentum

3.375 * w = 1.1*(2 - ( - 5 ))*.75

>> w = 1.71 rad/ s

b) during the collision , there is external moment on raven dosr system about the pivot ,

but there is external force acting at the pivot ,therefore , angluar momentum is conserved , linear momentum is not

Answer 2

Conservation of angular momentum.
Before the collision the total angular momentum, about an axis thru the upper edge of the gate, is due to the raven, with velocity "v" and perpendicular distance from the axis "a/2" (a=length of gate). So
Li = mv(a/2)

After the collision the raven has velocity "- u" and the gate has angular velocity "w" . So the total angular momentum after is;
Lf = - mu(a/2) + Iw
= - mu(a/2) + (1/3)Ma^2w

Lf = Li
- mu(a/2) + (1/3)Ma^2w = mv(a/2)
w = (3/2)(m/M)a(u+v)

EDIT__________________________________...
Typo. That should be;
w = (3/2)(m/Ma)(u+v)

Answer 3

First I calculated the momentum for the bird using L=mvl
L = (1.1 kg)(5.0 m/s)(1.5m/2) = 4.125 kg m/s^2

Then the total momentum after it strikes the square
L(total) = raven + the square

I wasn't sure about the moment of interia equation for the square so I used the one a thin rectangular plate axis along edge = Mr^2/3

= (1.1kg)(-2.0 m/s)(1.5m/s) + (4.5 kg)(.75)^2/3(wf) = -1.65 kg m/s^2 + 0.84375(wf)

I equated this to the intial momentum of the bird
4.125 kg m/s^2 = = -1.65 kg m/s^2 + 0.84375(wf)

wf = 6.8 rad/s which is way off

Answer = 1.71 rad/s