Angular Momentum: Problem w/Calculating MoI at 100cm

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The discussion revolves around a homework problem involving angular momentum and the calculation of moment of inertia (MoI) for a system consisting of a particle attached to a meterstick. The confusion arises regarding the use of 0.5 m as the radius for calculating MoI instead of 0.4 m, given that the particle is at the 100 cm mark. The correct interpretation is that when pivoting about the 50 cm mark, the effective radius for the particle is indeed 0.5 m from the pivot point. This is because the distance is measured from the pivot to the location of the mass, not just its position on the meterstick. Understanding this concept is crucial for accurately calculating the angular momentum of the system.
EngrStud
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I need some clarification on a homework problem related to angular momentum. I understand how to calculate the angular momentum by using L= IW but when calculating the moment of intertia for the particle i don't understand why to use .5m as the radius instead of .4m due to being placed at the 100 cm mark. From all the answers I've seen online and in the back of the book, the answer is calculated using .5m as the radius.

1. Homework Statement

A particle of mass 0.400 kg is attached to the 100-cm
mark of a meterstick of mass 0.100 kg. The meterstick
rotates on the surface of a frictionless, horizontal
table with an angular speed of 4.00 rad/s. Calculate
the angular momentum of the system when the stick
is pivoted about an axis

(a) perpendicular to the table
through the 50.0-cm mark and

(b) perpendicular to
the table through the 0-cm mark.
 
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EngrStud said:
I need some clarification on a homework problem related to angular momentum. I understand how to calculate the angular momentum by using L= IW but when calculating the moment of intertia for the particle i don't understand why to use .5m as the radius instead of .4m due to being placed at the 100 cm mark. From all the answers I've seen online and in the back of the book, the answer is calculated using .5m as the radius.

1. Homework Statement

A particle of mass 0.400 kg is attached to the 100-cm
mark of a meterstick of mass 0.100 kg. The meterstick
rotates on the surface of a frictionless, horizontal
table with an angular speed of 4.00 rad/s. Calculate
the angular momentum of the system when the stick
is pivoted about an axis

(a) perpendicular to the table
through the 50.0-cm mark and

(b) perpendicular to
the table through the 0-cm mark.

Where comes 0.4 m as radius come from? The meter stick is 100 cm long, and rotates about the axis going through its middle.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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