What Is the Moment of Inertia About a Point One-Fourth the Length from One End?

AI Thread Summary
The discussion centers on calculating the moment of inertia of a system of small blocks clamped to a rod, specifically about an axis located one-fourth the length from one end. Participants emphasize the importance of visualizing the problem by drawing a diagram to clarify the axis's position. The axis is described as being one-fourth of the way from one end, which is halfway between that end and the center of the rod. The original poster resolves their confusion before receiving further assistance. Clear communication and visual aids are highlighted as effective strategies for understanding physics problems.
K.QMUL
Messages
54
Reaction score
0

Homework Statement



Small blocks, each with mass m, are clamped at the ends and at the center of a rod of length L and negligible mass

Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through a point one-fourth of the length from one end.

Homework Equations



I = Mr2

What I am having difficulty with

I don't understand where this axis is, it would help if someone would explain or better draw a diagram to help me understand.
 
Physics news on Phys.org
K.QMUL said:
I don't understand where this axis is, it would help if someone would explain or better draw a diagram to help me understand.
Did you try drawing a diagram for yourself? The directions seem pretty straightforward.

Draw a line representing the rod. Now draw a line perpendicular to it. If it that perpendicular line were in the middle of the rod, it would be 1/2 the length from one end. Move it so that it's 1/4 the length from one end.
 
Do you agree that one-fourth of the rod length, as measured from one end, must be midway between that end and the center point?
 
Sorry, i figured it out before everyones reply, so I am all good. Thanks for the help though
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Correct Use of the Parallel Axis Theorem for Moment of Inertia
Replies
2
Views
2K
Moment of inertia of a cuboid parallel to one of its faces (repost with diagram)
Replies
25
Views
2K
Moment of inertia of a disk about an axis not passing through its CoM
Replies
11
Views
3K
What is the "r" in the moment of inertia?
Replies
13
Views
2K
Calculate the moment of inertia of this body
Replies
4
Views
1K
Calculating Moment of Inertia for a Curved Rod with Respect to a Specific Axis
Replies
1
Views
1K
Moment of Inertia in Planet System
Replies
9
Views
2K
Moment of inertia of 2 uniform thin rods
Replies
3
Views
1K
Quick question about moment of inertia
Replies
21
Views
2K
Moment of inertia of a system of masses
Replies
10
Views
2K

Hot Threads

Recent Insights

Back
Top