# Angle of Swing (for a rod)

1. Jun 23, 2012

### roam

1. The problem statement, all variables and given/known data

I need some help with the following problem:

http://desmond.imageshack.us/Himg62/scaled.php?server=62&filename=problem1p.jpg&res=landing [Broken]

3. The attempt at a solution

(i) I used the conservation of angular momentum (since there is no external torque):

$L_{before} = mv(\frac{L}{2})$

$L_{after} = m \left( \frac{v}{2} \right) \left( \frac{L}{2} \right) + \frac{mL^2}{3} \omega_o$

Lbefore = Lafter

$mv(\frac{L}{2}) = m \left( \frac{v}{2} \right) \left( \frac{L}{2} \right) + \frac{mL^2}{3} \omega_o$

Solving for ωo:

$\omega_o = \frac{3mv}{4ML}$

Is this correct?

(ii) I'm very confused by this part. How do I know what v must be in order for the rod to swing at 90°?

I know that the distance the rod has moved is equal to: s=L(90°). And here is the equation for torque:

$\tau = mgL \sin \ 90 = m \frac{d \omega}{dt} = m \frac{d \frac{3mv}{4ML}}{dt}$

However the RHS becomes 0 when I differentiate it with respect to time. So how can I solve this? How else can I express v in terms of m, M, L, and g?

Any help is really appreciated.

Last edited by a moderator: May 6, 2017
2. Jun 24, 2012

### rock.freak667

For this question, I think it is easier to use conservation of energy as your equations would imply that the bullet has angular momentum with respect to the point A.

If you use the energy method, you would just have KE of bullet before = KE of bullet after + KE of rod.

Similarly, for the second part, since you know the rod is being displaced exactly 90 degrees, this means that it does not go past 90 degrees. So what kind of energy will it possess?

3. Jun 24, 2012

### roam

So why can't we use the conservation of angular momentum? Does this mean the expression I've obtained for ωo is wrong?

Gravitational potential energy? So the equation would become:

1/2 m v2 = mgL + 1/2 m (v/2)2.

Is that it?

4. Jun 24, 2012

### rock.freak667

For you expression of the bullet's angular momentum before and after impact, you used mv(L/2) and m(v/2)(L/2). But the bullet isn't rotating about point A.

Yes.

5. Jun 25, 2012

### Yukoel

Hello roam,
The rod is rotating about A right?So the angular momentum of the bullet should be calculated about A (at the instant) .This implies that the angular momentum of the bullet is exactly double of the value you gave.It should reflect in your answer as well.Energy conservation gives twice the value of the angular velocity compared to your answer if my calculations are correct.For the second part try conserving energy of rod by converting rotational energy of the rod to potential energy.Or much better as rock points out use conservation of energy directly on both .Just another thing ,if i suppose A as my reference point of 0 zero P.E. the rod's center of mass lies midway ,right? So if you suppose 90 degree rotation the rod only rises by a height of L/2.So in your equation
1/2 m v^2 = mgL + 1/2 m (v/2)^2.
I think the modification should be as
1/2 m v^2 = mgL/2 + 1/2 m (v/2)^2.
Correct me if I am wrong.

rock.freak667
The bullet does have an angular momentum about the "point A" .As far as I know we define Angular momentum as L=m(rXv).At the instant of the collision r vector form A to the bullet and bullet's velocity vector make an angle of 90 degrees with none of them being zero.This gives rise to a non zero angular momentum.Angular momentum has been attempted to be conserved because the question regards the collision as very short ,as such the rod deflects by an infinitesimal angle so that gravity doesn't have a torque on rod or bullet about point A .And as for the later calculations I believe energy conservation is the only way out as the the system has the torque due to gravity(in later cases),isnt it? Please do tell
Correct me if I am wrong.
regards
Yukoel

6. Jun 26, 2012

### ehild

Energy conservation applies only on the rod after the bullet left it. It is not valid for the whole process, as some energy is used to make a hole in the rod.
Write up the kinetic energy of the road after the bullet just left it in terms of its moment of inertia and initial angular velocity. That becomes equal to the final potential energy: mgL/2, as Yukoel pointed out.

ehild