Rotational Inertia of a Rigid Body with Joined Rods

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The discussion focuses on calculating the rotational inertia of a system of four particles with given masses and coordinates about the x, y, and z axes. Participants clarify that the parallel axis theorem is not necessary for point masses, as the definition of moment of inertia should be applied instead. A specific example involving two joined rods is also presented, prompting questions about calculating rotational inertia about different axes. The conversation highlights confusion around applying the correct formulas for both point masses and rigid bodies. Overall, the thread emphasizes understanding the fundamental concepts of rotational inertia in different contexts.
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Homework Statement


The masses and coordinates of four particles are as follows: 27 g, x = 2.0 cm, y = 2.0 cm; 16 g, x = 0, y = 4.0 cm; 43 g, x = -3.0 cm, y = -3.0 cm; 61 g, x = -2.0 cm, y = 4.0 cm. What are the rotational inertias of this collection about the (a) x, (b) y, and (c) z axes?


Homework Equations


I'm not sure.. would I use parallel axis theorem?


The Attempt at a Solution


I don't know how to solve this problem.. help please =)
 
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Shatzkinator said:
I'm not sure.. would I use parallel axis theorem?

Hi Shatzkinator! :smile:

You only need the parallel axis theorem for a solid body, for which you already know the moment of inertia about its c.o.m.

These are just point masses, so just use the definition of moment of inertia (rotational inertia) :wink:
 


Two thin rods (each of mass 0.50 kg) are joined together to form a rigid body as shown in Fig. 10-66. One of the rods has length L1 = 0.30 m, and the other has length L2 = 0.60 m. What is the rotational inertia of this rigid body about (a) an axis that is perpendicular to the plane of the paper and passes through the center of the shorter rod and (b) an axis that is perpendicular to the plane of the paper and passes through the center of the longer rod?


Fig. 10-66
Problem 114.

also this question i can't get part B
 
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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