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 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top