Answer Moment of Inertia: Doubling Mass, Adding Weight to Stick

Click For Summary

Homework Help Overview

The discussion revolves around the concept of moment of inertia in physics, specifically how various changes to a system affect its moment of inertia. The original poster presents a scenario involving a small, heavy weight on a light stick and poses several questions regarding the effects of doubling the mass, adding weights at different positions, and considering the stick's own mass.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply the moment of inertia formula I = mr² to different scenarios, questioning the effects of mass changes and weight placements. Some participants provide feedback on the implications of adding mass at the center of the stick and the need to consider the stick's mass in calculations.

Discussion Status

Participants are actively engaging with the original poster's reasoning, confirming some points while prompting further exploration of others. There is a focus on clarifying the implications of mass distribution and the correct expressions for moment of inertia in more complex systems.

Contextual Notes

The discussion highlights the challenge of understanding moment of inertia, particularly when transitioning from point masses to extended bodies, and the need for careful consideration of mass placement relative to the axis of rotation.

twiztidmxcn
Messages
43
Reaction score
0
Moment of inertia time in physics, hard as hell concept, but here's the question:

You swing a small, heavy weight on the end of a light stick (the stick is light enough you can neglect its moment of inertia, and you can consider the weight as a point mass). Remember that the moment of inertia is a measure of how an object resists angular acceleration. How would the following actions effect the moment of inertia of what you are swinging

A) doubling mass of weight
B) adding an equal weight to the first at the center of the stick
C) adding a weight that is 4 times the original mass to the center
D) using a stick that isn't light, but weighs as much as the original weight on the end

so, I am using the equation I = mr^2, for A i got that if you double the mass, the moment of inertia is doubled as well. for B i believe that if you add a weight to the center of the stick, it shouldn't really affect it because it is not a distance away from the axis of rotation, same thing with C. For D i put that the moment of inertia increases because you get mass stick * r^2 + mass weight * r^2.

Is what I am saying here right? I think A and D are right but I am quite unsure of B and C, any help would be appreciated
 
Physics news on Phys.org
twiztidmxcn said:
Moment of inertia time in physics, hard as hell concept, but here's the question:

You swing a small, heavy weight on the end of a light stick (the stick is light enough you can neglect its moment of inertia, and you can consider the weight as a point mass). Remember that the moment of inertia is a measure of how an object resists angular acceleration. How would the following actions effect the moment of inertia of what you are swinging

A) doubling mass of weight
B) adding an equal weight to the first at the center of the stick
C) adding a weight that is 4 times the original mass to the center
D) using a stick that isn't light, but weighs as much as the original weight on the end

so, I am using the equation I = mr^2, for A i got that if you double the mass, the moment of inertia is doubled as well. for B i believe that if you add a weight to the center of the stick, it shouldn't really affect it because it is not a distance away from the axis of rotation, same thing with C. For D i put that the moment of inertia increases because you get mass stick * r^2 + mass weight * r^2.

Is what I am saying here right? I think A and D are right but I am quite unsure of B and C, any help would be appreciated

For A) your answer is correct.

For B) and C) : the implication from the question is that you're grabbing the stick by one end and swinging the mass from the other end. If you add the mass to the center of the stick, you're adding it midway between the original mass and the axis of rotation. Can you figure out the correct expressions now ?

For D) you need to know that for more complex masses than point masses situated at a certain distance from an axis, the expression for the moment of inertia is not simply [itex]mr^2[/itex]. For a uniform thin rod, the moment of inertia about the end is [itex]I = \frac{1}{3}mr^2[/itex], where r is the length of the rod (and the radius of the rotation). (To prove that you can use simple calculus). Now figure out the total moment of inertia of the rod-mass combination.
 
Just some extra insight on that last idea:

The same fundamental principle applies: you are still multiplying each point mass by the square of its distance from the axis. The difference for an object that is not itself a point mass is that now you have a collection of infinitely small point masses of infinite number. We express the moment of inertia (the infinite sum of all the moments of inertia of these theoretical point masses that make up the object) as a continuous distribution using the integration over the mass:

[tex]I=\sum_{i}{r_{i}^{2}m_{i}}=\int_{m}{r^{2}}\,dm[/tex]

This general expression is then used to develop the specific equations for moments of inertia of different objects.
 
thank you both for your help
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
3K
Replies
11
Views
4K
Replies
2
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 9 ·
Replies
9
Views
7K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
24
Views
4K
Replies
8
Views
2K