Calculating angular velocity of a ratatable bar in inelastic collision

[mmoadi]

Homework Statement

1 m long rotatable bar with a mass of 2 kg is held on vertical stick that goes through its center of gravity. A bullet with mass 0.2 kg is shot with a speed 1 m/s horizontally at the edge of the bar. After the collision the bullet lodges into the bar. Calculate the angular velocity of the bar after the collision, if the bar was at rest at the beginning?

Homework Equations

Conservation of the angular moment

The Attempt at a Solution

mRv= (MR + mR + I) ω

ω= mRv / (MR² + mR² + MR²/12)
ω= mv / (MR + mR + MR/12)
ω= 0.17 m/s

Are my calculations correct?

 

[kuruman]

ω= mRv / (MR² + mR² + MR²/12)
ω= 0.17 m/s
Are my calculations correct?[/SUB]

Why are there three terms in the denominator? You are calculating angular momentum about the pivot which is also the CM of the rod. In the denominator you need only two terms, the angular momentum of the rod about the pivot (or its CM) and the angular momentum of the bullet about the pivot. Is "R" the length of the rod or is it the distance of the bullet from the pivot?

 

[mmoadi]

mRv= (mR + I) ω

<sub>ω= mRv / (mR² + MR²/12)
ω= 0.19 m/s[/color]

"R" is half the length of the rod, in other words the distance of the bullet from the pivot.

Are now my calculations correct?</sub>

 

[kuruman]

The moment of inertia of a rod of mass M and total length L about its mid point is

I=\frac{1}{12}ML^2

what should the formula look like if you used R instead of L?

 

[mmoadi]

The moment of inertia of a rod of mass M and total length L about its mid point is

I=\frac{1}{12}ML^2

what should the formula look like if you used R instead of L?

<sub>I think it should be like this ω= mRv / (mR² + M(2R)²/12)?

And the answer is now:

ω= 12 m/s[/color]</sub>

 

[kuruman]

I didn't put in the numbers, but the algebraic expression you have is correct.

 

[mmoadi]

_{Thank you very much!}