# Angular momentum problem - am I right?

• bcjochim07
In summary, the angular momentum of a dumbbell rotating with a constant angular velocity about an axis that makes a constant angle with the dumbbell and passes through its center of mass can be calculated using the equation L=Iω=mrsinθ(2rωsinθ). This is different from the approach of calculating the angular momentum using the equation L=Iω, which only gives the component of angular momentum parallel to the axis of rotation.
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=$$\Sigma$$Li

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

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

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

||L||=mrsin$$\theta$$[r$$\omega$$sin$$\theta$$+r$$\omega$$sin$$\theta$$]

=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 $$\omega$$

Last edited:
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.

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

bcjochim07 said:

## 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=$$\Sigma$$Li

=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$$\omega$$sin$$\theta$$+r$$\omega$$sin$$\theta$$]
Since r and v are perpendicular, there's no second sinθ factor.

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?

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.

bcjochim07 said:
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.

bcjochim07 said:
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°.

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?

bcjochim07 said:

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.)

## 1. What is angular momentum?

Angular momentum is a physical quantity that measures the amount of rotational motion of an object. It is a vector quantity, meaning it has both magnitude and direction.

## 2. How is angular momentum calculated?

Angular momentum is calculated as the product of an object's moment of inertia and its angular velocity. It can be represented mathematically as L = Iω, where L is angular momentum, I is moment of inertia, and ω is angular velocity.

## 3. What is the conservation of angular momentum?

The conservation of angular momentum states that in a closed system, the total angular momentum remains constant. This means that if no external torques act on a system, the initial angular momentum will be equal to the final angular momentum.

## 4. What is the Angular Momentum Problem?

The Angular Momentum Problem is a concept in physics that refers to the difficulty in explaining the observed rotation of galaxies, stars, and other celestial bodies. According to the laws of physics, these objects should not be able to rotate as fast as they do, leading to the need for additional theories or explanations.

## 5. Am I right in thinking that angular momentum is always conserved?

Yes, the conservation of angular momentum is a fundamental law of physics and holds true in most situations. However, there are some exceptions, such as when external torques are present or when dealing with quantum mechanical systems.

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