Rotational Motion - Conservation angular momentum

In summary: I'd use energy conservation for part (b). You can use the CM translational velocity just as you did in part (a), since the "radius" is still L/2.ok. I'd use energy conservation for part (b). You can use the CM translational velocity just as you did in part (a), since the "radius" is still L/2.But the moment of inertia would be different. In part (a) is was just a rod rotating about one end. In part (b) it is a rod rotating about its center of mass. In both cases the hinge is at the other end of the rod.ok. I'd use energy conservation for part (
  • #1
HoodedFreak
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Homework Statement


A 500.0-g bird is flying horizontally at 2.25 m>s, not paying much attention, when it suddenly flies into a stationary vertical bar, hitting it 25.0 cm below the top(Fig. P10.85). The bar is uniform, 0.750 m long, has a mass of 1.50 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).What is the angular velocity of the bar (a) just after it is hit by the bird and (b) just as it reaches the ground?

Homework Equations


L = Iw
L = p x r
I = ML^2/3 for a rod

The Attempt at a Solution


a) The angular momentum before the collision:
L1 = Lbird + Lrod = p x r + 0 = mvr
Then afterwards:
L2 = Lbird + Lrod = 0 + Iw

we get the equation: mvr = Iw from conservation of momentum

Plug in the numbers: 0.5 * 2.25 * (0.75-0.25) = 1.5 * (0.75)^2 * w/3

Solve for w = 2 rads/s

b) I'm not really sure how to attack this one, the angular momentum won't be conserved because of the gravity enacting a torque on the object.

Any suggestions?
 

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  • #2
HoodedFreak said:

Homework Statement


A 500.0-g bird is flying horizontally at 2.25 m>s, not paying much attention, when it suddenly flies into a stationary vertical bar, hitting it 25.0 cm below the top(Fig. P10.85). The bar is uniform, 0.750 m long, has a mass of 1.50 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).What is the angular velocity of the bar (a) just after it is hit by the bird and (b) just as it reaches the ground?

Homework Equations


L = Iw
L = p x r
I = ML^2/3 for a rod

The Attempt at a Solution


a) The angular momentum before the collision:
L1 = Lbird + Lrod = p x r + 0 = mvr
Then afterwards:
L2 = Lbird + Lrod = 0 + Iw

we get the equation: mvr = Iw from conservation of momentum

Plug in the numbers: 0.5 * 2.25 * (0.75-0.25) = 1.5 * (0.75)^2 * w/3

Solve for w = 2 rads/s

b) I'm not really sure how to attack this one, the angular momentum won't be conserved because of the gravity enacting a torque on the object.

Any suggestions?
physics-problem-2-png.105273.png


You might consider using conservation of energy for part (b).
 
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  • #3
SammyS said:
physics-problem-2-png.105273.png


You might consider using conservation of energy for part (b).

Okay, so if I say 1/2Mv1^2 + MgL/2 = 1/2Mv2

where v = wR

Then, we get w1*L/2 + gL = w2*L/2

THen 2 + 2g = w2

SO we get 2 + 2g = w2

And finally we get w2 = 21.62

Which isn't the right answer, where did I go wrong?
 
  • #4
How do you justify the use of L/2 for the "radius" when calculating the initial KE?

Instead, use the angular form for the KE: ##\frac{1}{2} I \omega^2##
 
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  • #5
gneill said:
How do you justify the use of L/2 for the "radius" when calculating the initial KE?

Instead, use the angular form for the KE: ##\frac{1}{2} I \omega^2##

I was using the linear velocity of the center of mass. Isn't the angular kinetic energy for rotation about the center of mass, wouldn't this be zero in this case?
This is why I was so confused for this question.
 
  • #6
I realized I made a lot of mistakes on my previous post, so let me correct myself

1/2Mv12 + MgL/2 = 1/2Mv22

where v = wR, R = L/2 in this case

Then, we get 1/2 * (w1R)2 + gR = 1/2 * (w2R)2

So, 1/2 * 4 * R + g = 1/2 * w22*R

THen (2R + g)*2 / R = ww2

Solving for w, gives w = 7.5

Not the right answer
 
  • #7
HoodedFreak said:
I was using the linear velocity of the center of mass. Isn't the angular kinetic energy for rotation about the center of mass, wouldn't this be zero in this case?
This is why I was so confused for this question.
You can either use linear KE of the centre of mass and add angular KE about the mass centre (which is not zero, since it is rotating), or use angular kinetic energy about the hinge.

For part a), I'm not sure how to interpret "the bird drops to the ground". You seem to have read it as saying the bird loses all horizontal momentum, which would imply some degree of elastic collision. You might be reading too much into the wording. I think you should consider it a purely inelastic collision.
 
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  • #8
haruspex said:
You can either use linear KE of the centre of mass and add angular KE about the mass centre (which is not zero, since it is rotating), or use angular kinetic energy about the hinge.

For part a), I'm not sure how to interpret "the bird drops to the ground". You seem to have read it as saying the bird loses all horizontal momentum, which would imply some degree of elastic collision. You might be reading too much into the wording. I think you should consider it a purely inelastic collision.

Hmmm, well I got the answer right for part a) considering it as an elastic collision, if I considered it as an inelastic collision, then I would get:

L1 = 0.5*2.25*(0.75-0.25)
L2 = (ML^2/3 + 0.5*0.52)*w

Equating the two: 2.25/4 = (1.5*0.752/3 + 0.53) * w
which gives me an answer of w = 1.38 for the rod and bird, which is not the answer the book gives
 
  • #9
HoodedFreak said:
Hmmm, well I got the answer right for part a) considering it as an elastic collision, if I considered it as an inelastic collision, then I would get:

L1 = 0.5*2.25*(0.75-0.25)
L2 = (ML^2/3 + 0.5*0.52)*w

Equating the two: 2.25/4 = (1.5*0.752/3 + 0.53) * w
which gives me an answer of w = 1.38 for the rod and bird, which is not the answer the book gives
The wording of the problem is not too clear on this point, but I interpreted part (a) the way you did initially.
 
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  • #10
HoodedFreak said:
Hmmm, well I got the answer right for part a) considering it as an elastic collision, if I considered it as an inelastic collision, then I would get:

L1 = 0.5*2.25*(0.75-0.25)
L2 = (ML^2/3 + 0.5*0.52)*w

Equating the two: 2.25/4 = (1.5*0.752/3 + 0.53) * w
which gives me an answer of w = 1.38 for the rod and bird, which is not the answer the book gives
ok.
 

Related to Rotational Motion - Conservation angular momentum

1. What is conservation of angular momentum?

Conservation of angular momentum is a fundamental principle in physics which states that the total angular momentum of a system remains constant as long as there are no external torques acting on it.

2. How is angular momentum conserved in rotational motion?

In rotational motion, angular momentum is conserved when the net torque acting on a system is zero. This means that the initial angular momentum of a system will remain constant unless an external torque is applied.

3. What is the equation for calculating angular momentum?

The equation for calculating angular momentum is L = Iω, where L is angular momentum, I is the moment of inertia of the object, and ω is the angular velocity.

4. How does the moment of inertia affect conservation of angular momentum?

The moment of inertia is a measure of an object's resistance to rotation. Objects with a larger moment of inertia will require more torque to change their angular momentum, making it more difficult to conserve angular momentum.

5. Can angular momentum be transferred between objects?

Yes, angular momentum can be transferred between objects in a closed system. For example, when a spinning ice skater extends their arms, their moment of inertia decreases, causing an increase in their angular velocity and therefore their angular momentum.

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