Correct Use of the Parallel Axis Theorem for Moment of Inertia

AI Thread Summary
The discussion focuses on the correct application of the Parallel Axis Theorem for calculating the moment of inertia. The user initially calculates the moment of inertia for a rod in two cases but arrives at incorrect values, particularly for the second case with arms stretched. The error stems from not properly applying the theorem, specifically neglecting to use the moment of inertia about the center of mass. After clarification, the user acknowledges the mistake and realizes the need to incorporate the correct formula. Accurate application of the theorem is essential for obtaining the correct moment of inertia values.
simphys
Messages
327
Reaction score
46
Homework Statement
In order to increase her moment of inertia about a vertical axis, a spinning figure skater stretches out her arms horizontally; in order to reduce her
moment of inertia, she brings her arms down vertically along her sides. Calculate the change of moment of inertia between these two configurations of the arms. Assume that each arm is a thin, uniform rod of length 0.60 m and mass
2.8 kg hinged at the shoulder at a distance of 0.20 m from the axis of rotation.
Relevant Equations
xD
so calculated, the moment of inertia for a rod about an axis at the end of the rod is I = 1/3 * M * L^2
here for case 1: arms to the side
I is calculated to be ##I = 0.224##

for case 2: arms stretched
## I = 1 / 3 * M * L^2 + M * d^2 ## with L = 0.6 m (length of rod) and d = 0.2 (dinstance from 'hinge' to axis of rotation)
## I = 0.896## is my answer.
Yet the actual answer is ##I = 1.568##. I don't really understand what I am doing wrong here so can someone help me please?

Reasoning:
In case 2 I basically have the moment of inertia at the end (the 'hinge'/shoulder so to say) and need to add a dinstance d to get the moment of inertia about the rotation axis via the parallel axis theorem yet this isn't correct.

Thanks in advance
 
Physics news on Phys.org
simphys said:
for case 2: arms stretched
## I = 1 / 3 * M * L^2 + M * d^2 ## with L = 0.6 m (length of rod) and d = 0.2 (dinstance from 'hinge' to axis of rotation)
## I = 0.896## is my answer.
You have not used the parallel axis theorem correctly. $$I = I_{\rm cm} + md^2$$ What does ##I_{\rm cm}## mean?
What does ##d## mean?
 
  • Like
Likes simphys and hutchphd
TSny said:
You have not used the parallel axis theorem correctly. $$I = I_{\rm cm} + md^2$$ What does ##I_{\rm cm}## mean?
What does ##d## mean?
yeah apologies it should have been 2 times the symbolic equation, just forgot to type it out.
Omg... totally forgot that... thanks a lot... I needed to use the moment of inertia about the center of mass basically..,.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top