Homework Help: Calculating Moment of Inertia & Radius of Gyration

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The discussion focuses on calculating the moment of inertia and radius of gyration using the formulas r = √(I/m) and the parallel axis theorem. The user is initially uncertain about the moments calculated about the center of mass (CM) and the hip, questioning the incorporation of acceleration and distance to the toe. Clarifications are provided, confirming that the moment about the hip can be calculated as m*0.42^2 + m*0.24^2. The conversation also addresses the relationship between tangential acceleration and angular acceleration, concluding that if tangential acceleration is 18 m/s², the angular acceleration is also 18 radians/s². The final calculations are affirmed, leading to a resolution of the user's queries.
lupinpooter
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Homework Statement



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Homework Equations



radius of gyration:
r = root (I/m)
I = moment of inertia
m = mass

parallel axis theorem given above

The Attempt at a Solution



Okay, so I think the moment about CM is just m*0.24^2, but after that, I'm less sure.
Is the moment about the hip just m*0.42^2 + m*0.24^2?

If so, it's taking the acceleration and the distance to the toe into account that I'm having difficulty with. Is it just F = ma? If so, is the m taken from the moment of inertia at the hip, or the toe? I figure the distance to the toe must be significant, but I don't know how to account for it.

Any help?
 
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lupinpooter said:
Okay, so I think the moment about CM is just m*0.24^2, but after that, I'm less sure.
Is the moment about the hip just m*0.42^2 + m*0.24^2?
Yes, you've got it.

If so, it's taking the acceleration and the distance to the toe into account that I'm having difficulty with. Is it just F = ma? If so, is the m taken from the moment of inertia at the hip, or the toe?
How would you write the equivalent of F = ma for rotational motion?
I figure the distance to the toe must be significant, but I don't know how to account for it.
Given the tangential acceleration, how would you find the angular acceleration?
 
Doc Al said:
Yes, you've got it.


How would you write the equivalent of F = ma for rotational motion?

Given the tangential acceleration, how would you find the angular acceleration?

Okay, maybe it's best to do the last two bits in reverse order;
if the tangential acceleration is 18m/s^2, and the radius is 1m, is the angular acceleration also 18 radians/s^2?

Then is the moment: angular acceleration * the moment about the hip?
And that's the final answer?
 
lupinpooter said:
Okay, maybe it's best to do the last two bits in reverse order;
if the tangential acceleration is 18m/s^2, and the radius is 1m, is the angular acceleration also 18 radians/s^2?
Right!

Then is the moment: angular acceleration * the moment about the hip?
And that's the final answer?
Right again.
 
Awesome,
thanks man, I really appreciate it.
 
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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