Angular momentum problem - am I right?

[bcjochim07]

Homework Statement

A dumbbell consists of two mass points of equal mass m at opposite ends of a rigid rod of length l and negligible mass. Calculate the angular momentum of this dumbbell as it rotates with constant angular velocity \omega about an axis that makes a constant angle \theta with the dumbbell and passes through its center of mass.

Homework Equations

In my calculation, I'll use r=l/2

The Attempt at a Solution

I'm not sure if this is correct, but here's what I did:

L = rXp

Since there are two masses:
L=\SigmaL_i

=rXp1 + rXp2 = m[(rXv1)+ (rXv2)]

=m[(rX(rsin\theta\omega)+rX(rsin\theta\omega]

=mrsin\theta[rX\omega + rX\omega]

||L||=mrsin\theta[r\omegasin\theta+r\omegasin\theta]

=2mr²sin²\theta\omega, r=l/2

L=(1/2)ml²sin²\theta \widehat{n} ;

where \widehat{n} is a unit vector perpendicular to both r and \omega

 

[Dr.D]

In cylindrical coordinates, your two particles are located at
r1 = r sin(theta) er + r cos(theta) k
r2 = -r sin(theta) er - r cos(theta) k
the ang velocity vector is
omega = omega k
L1 = mr1 x v1 = r1 x (omega x r1) = mr1 x (omega k x (r sin(theta) er + r cos(theta) k))
= mr1 x r omega (sin(theta) k x er + cos(theta) k x k)
= mr1 x r omega sin(theta) eth
= mr^2 omega (sin(theta) er + cos(theta) k) x sin(theta) eth
= mr^2 omega (sin(theta)^2 (er x eth)+cos(theta) sin(theta) (k x eth))
= mr^2 omega (sin(theta)^2 k - sin(theta) cos(theta) er)
It looks to me like this will be the contribution from one mass and the other will be developed similarly. I suggest that the algebra needs to be checked very carefully; I certainly could have made a mistake there, but this is the way I would approach it.

 

[bcjochim07]

I have not yet encountered cylindrical coordinates; do you think the way I have done it is correct?

 

[Doc Al]

The Attempt at a Solution

I'm not sure if this is correct, but here's what I did:

L = rXp

Since there are two masses:
L=\SigmaL_i

=rXp1 + rXp2 = m[(rXv1)+ (rXv2)]

=m[(rX(rsin\theta\omega)+rX(rsin\theta\omega]

This is OK. (Realize that r and v are perpendicular.)

=mrsin\theta[rX\omega + rX\omega]

||L||=mrsin\theta[r\omegasin\theta+r\omegasin\theta]

Since r and v are perpendicular, there's no second sinθ factor.

 

[bcjochim07]

Doesn't the angular velocity vector point either up or down? Also, the problem doesn't necessarily say that the dumbbell is perpendicular to the axis, so don't I need that sin\theta term?

 

[bcjochim07]

Oh... maybe omega does not point up or down. Does it point in the same direction as the tangential velocity? In that case, I would just divide my answer by sinθ, and it should be correct.

 

[Doc Al]

Doesn't the angular velocity vector point either up or down? Also, the problem doesn't necessarily say that the dumbbell is perpendicular to the axis, so don't I need that sin\theta term?

Let's say that the angular velocity vector points along the y axis. The dumbbell's center is at the origin and its axis is at an angle θ from the y axis.

Oh... maybe omega does not point up or down. Does it point in the same direction as the tangential velocity?

The tangential velocity is perpendicular to both r and ω, thus the magnitude of r X v is just rv.

In that case, I would just divide my answer by sinθ, and it should be correct.

Simply correct your error in your original answer. You have mixed up the angle that the dumbbell makes with ω, which is θ, with the angle that the v makes with r, which is 90°.

 

[bcjochim07]

Ok... follow up question.

Another way to approach this problem is L=Iω
so I=2m(rsinθ)^2 (since rsinθ is the distance to the axis)
Then L = 2mωr^2(sinθ)^2 = (1/2)mωl^2(sinθ)^2

Why does this turn out different?

 

[Doc Al]

Ok... follow up question.

Another way to approach this problem is L=Iω
so I=2m(rsinθ)^2 (since rsinθ is the distance to the axis)
Then L = 2mωr^2(sinθ)^2 = (1/2)mωl^2(sinθ)^2

Why does this turn out different?

Because L=Iω only gives you the component of the angular momentum parallel to the axis of rotation. The angular momentum vector is not parallel to that axis. (Note that the dumbbell is not rotating about an axis of symmetry.)