Rotational Dynamics of 2 Particles and 2 Rods

AI Thread Summary
The discussion focuses on calculating the rotational inertia and kinetic energy of a system consisting of two particles and two rods rotating around a common axis. The particles each have a mass of 2.30 kg, while the rods have a mass of 0.360 kg and a length of 0.670 m. To find the total rotational inertia, it is suggested to consider each rod and particle separately, applying the parallel axis theorem where necessary. The parallel axis theorem is particularly relevant for the second rod, which does not rotate about the axis. The conversation emphasizes the need to sum the contributions from all components to arrive at the correct values.
cd80187
Messages
38
Reaction score
0
In Fig. 10-35, two particles, each with mass m = 2.30 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 0.670 m and mass M = 0.360 kg. The combination rotates around the rotation axis with angular speed ω = 0.211 rad/s. Measured about O, what are the combination's (a) rotational inertia and (b) kinetic energy?
(I'm not sure how to post pictures on here, because the picture is on wileyplus, and to view it, you need a password)

But for this problem, I am not even sure where to start. I know that I need to find the total inertia, but how many parts should i split it into? Should I take each rod seperately and each particle seperately, and use the summation formula for inertia, or should i use the parallel axis theorem? I am just so lost on where to start.
 
Physics news on Phys.org
If the configuration is axis-rod-particle-rod-particle, then you will want to use the parallel axis theorem for the rods and add the contributions from the masses. The parallel axis theorm has already been used for the rod rotating at one end in most places where you look up moments of inertia. Since the second rod is not touching the axis, you will have to compute it yourself.
 
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