Find the moment of inertia about the x axis

AI Thread Summary
The discussion focuses on calculating the moment of inertia about the x-axis for a system of particles connected by rigid rods. The formula I = sum of (r^2 * m) is used, but confusion arises regarding the use of negative distances from the axis of rotation. A participant initially miscalculates the moment of inertia, arriving at 140 instead of the correct value of 92. The conversation also touches on finding the total kinetic energy using the formula (sum of) mi vi^2/2. Clarification on these calculations is sought to resolve the issues faced.
integra2k20
Messages
35
Reaction score
0

Homework Statement



Rigid rods of negligible mass lying along the y-axis connect three particles (Fig. P10.20). The system rotates about the x-axis with an angular speed of 2.90 rad/s.

p10-23.gif

Figure P10.20

(a) Find the moment of inertia about the x axis.
______kgm2
Find the total rotational kinetic energy evaluated from 1/2 I2.
______J
(b) Find the tangential speed of each particle.
______m/s (4.00 kg particle)
______m/s (2.00 kg particle)
______m/s (3.00 kg particle)
Find the total kinetic energy evaluated from mi vi2/2.
______J

Homework Equations



I = sum of (r^2*m)

The Attempt at a Solution


My teacher told us to use I = sum of (r^2 * m) i am not sure how this works if some of the r's (distances from the axis of rotation) are negative. i tried subtracting the negative ones, or just plugging them in and squaring them, thus making them positive, to no avail. Any help would be greatly appreciated! the most difficult part of this for me is the moment of inertia.
 
Last edited:
Physics news on Phys.org
What are you struggling with?
 
finding the moment of inertia of the structure, after that i think i should be able to handle it
 
So you've done something like this;

I = 4(3)^2 + 2(-2)^2 + 3(-4)^2

Is that correct?
 
Hootenanny said:
So you've done something like this;

I = 4(3)^2 + 2(-2)^2 + 3(-4)^2

Is that correct?


yup, i did exactly that...didn't get it right...
 
oh wow, never mind, i made some type of calculation mistake. for some reason i got 140 when it shoudl have been 92. my last question is about the last part, it says to find the total kinetic energy evaluated from (sum of)mi vi^2/2.
 
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...
Thread 'A cylinder connected to a hanged 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