Why Is the Calculated Angular Momentum of the Pucks Zero?

AI Thread Summary
The calculated angular momentum of the system of three pucks is initially thought to be zero due to vector cancellation, but this is incorrect. Each puck contributes to the angular momentum, which is calculated using the formula Lθ = Iwθ, where I is the moment of inertia and w is the angular velocity. The direction of the angular momentum vectors is perpendicular to the plane of rotation, meaning they do not cancel out. The correct approach involves calculating the angular momentum for each puck individually and summing them to find the total. Thus, the magnitude of the angular momentum should be derived from the contributions of all three pucks.
Joshua A
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Homework Statement


Three small, identical 0.70-kg pucks are attached to identical 0.50-m strings, tied together at a common center as shown in (Figure 1) . Pucks are whirled in circular motion at angular speed 3.0 s-1

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What is the magnitude of the angular momentum of the system about the common center?

Homework Equations


I = mr2
Lθ = Iwθ

The Attempt at a Solution


Angular momentum is a vector, therefore the angular momentum of this system should be 0 kg m2/s as all the vectors will cancel out. The magnitude of that is also 0 kg m2/s.

Apparently this answer is wrong. Where am I going wrong? When they ask for the magnitude of the angular momentum, do they want me to find the magnitude of the angular momentum for each puck and then add the magnitudes together?
 

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Joshua A said:

The Attempt at a Solution


Angular momentum is a vector, therefore the angular momentum of this system should be 0 kg m2/s as all the vectors will cancel out. The magnitude of that is also 0 kg m2/s.

Apparently this answer is wrong. Where am I going wrong? When they ask for the magnitude of the angular momentum, do they want me to find the magnitude of the angular momentum for each puck and then add the magnitudes together?

What is the direction of the angular momentum vectors?
 
PeroK said:
What is the direction of the angular momentum vectors?

The direction of the angular momentum vectors should be the same as the angular velocity vectors. I had assumed that the direction of the angular velocity vector would be in the direction of the motion of the puck (i.e. along the t axis). Looking back through my textbook, they do not actually specify a direction for the angular velocity, just the magnitude of the angular velocity. From a Google search, it seems that the direction is actually perpendicular to the plane of rotation, therefore the vectors would not cancel out as they are all pointing the same direction.

I assume the answer should then be:
Lθ1 = (0.70kg)(0.50m)2(3.0s-1)

for puck 1, and the same for pucks 2 and 3. Then the angular momentum of the system would be Lθ = Lθ1 + Lθ2 + Lθ3
 
Joshua A said:
From a Google search, it seems that the direction is actually perpendicular to the plane of rotation, therefore the vectors would not cancel out as they are all pointing the same direction.

Yes. In 2D motion, angular velocity (and momentum) is often simplified to a signed scalar: anticlockwise is positive and clockwise is negative. For 3D motion, you have to consider the full vector nature.

Joshua A said:
I assume the answer should then be:
Lθ1 = (0.70kg)(0.50m)2(3.0s-1)

for puck 1, and the same for pucks 2 and 3. Then the angular momentum of the system would be Lθ = Lθ1 + Lθ2 + Lθ3

What is the SI unit of angular velocity?
 
PeroK said:
What is the SI unit of angular velocity?

Angular velocity? s-1 (or rad/s - normally, as far as I know, rad isn't specified)
Angular momentum is kg m2/s
 
Joshua A said:
normally, as far as I know, rad isn't specified
The SI unit is rad/s, but radians are generally considered dimensionless.
 
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