Inertia: blocks attached to rod of negligible mass

AI Thread Summary
The discussion centers on calculating the moment of inertia (I) for small blocks with mass m attached to a rod of negligible mass. The initial formula proposed is I = m(0.5L)^2 + m(0.5L)^2, leading to I = 0.5L^2 m. Participants clarify that the moment of inertia should not be expressed in meters and emphasize the importance of correctly identifying the mass distribution. A mistake in the placement of variables is noted, highlighting the need for careful notation to avoid errors. Ultimately, the calculation is confirmed to be correct after addressing these concerns.
cantgetno
Messages
23
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.
Find the formula for I


Homework Equations


I=MR^2


The Attempt at a Solution



I=m(0.5 L)^2 + m(0.5 L)^2
I=2(0.5L)^2 m
I=0.5L^2 m

am i right?
 
Physics news on Phys.org
Hi cantgetno! :smile:
cantgetno said:
Small blocks, each with mass m, are clamped at the ends and at the center of a rod of length L and negligible mass.
Find the formula for I

I=m(0.5 L)^2 + m(0.5 L)^2
I=2(0.5L)^2 m
I=0.5L^2 m

am i right?

mmm …

i] moment of inertia isn't measured in metres

ii] about which point is I being measured?

iii] what about the mass in the middle? :wink:
 


oh its around the middle
so that makes the mass in the middle mass x 0

metres? m=mass
 
cantgetno said:
oh its around the middle
so that makes the mass in the middle mass x 0

metres? m=mass

D'oh! m is mass … should have spotted that! :redface:

In that case, everything is fine. :biggrin:
 


oh crap
i put 1/2 l m^2 ... I am an idiot

thanks anyway
 
cantgetno said:
oh crap
i put 1/2 l m^2 ... I am an idiot

hee hee :biggrin:

Useful tip: keep your m's in the same place :wink:

for some reason, you shifted them from the left to the right …
I=m(0.5 L)^2 + m(0.5 L)^2
I=2(0.5L)^2 m
… which makes it much easier to make a mistake, and much less easy to spot a mistake. :cry:
 
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...
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...
Back
Top