How Do You Calculate Angular Momentum for a Rotating Rigid Body?

AI Thread Summary
To calculate the angular momentum of a rotating rigid body, one must first determine its rotational inertia (I) using the appropriate formulas. The structure in question consists of a circular hoop and a square made of thin bars, with specific dimensions and mass provided. The initial calculations for rotational inertia were incorrect, indicating a need to apply the parallel axis theorem for accurate results. After finding the correct I for both the hoop and the square, these values can be summed to obtain the total rotational inertia. Once I is established, angular momentum (L) can be calculated using the formula L = Iω, where ω is the angular velocity derived from the period of rotation.
math_head7
Messages
4
Reaction score
0
Angular Momentum Question?

Homework Statement


Figure 11-45 (attached) shows a rigid structure consisting of a circular hoop of radius R and mass m, and a square made of four thin bars, each of length R and mass m. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of 2.5s. Assuming R=0.50m and m=
2.0kg, calculate (a) the structure's rotational inertia about the axis of rotation and (b) its angular momentum about that axis.

Homework Equations


I=\intr2dm
L=I\omega

The Attempt at a Solution


(a) I did mr2 = (10)(.5)2 = 2.5 (the answer is supposed to be 1.6
Then I tried msrs2 + mcrc2 = (8)(.25)2 + (2)(.5)2 = 1.5 (still incorrect)

(b) Due to not being able to solve "a" I could not began my attempt on "b"
 

Attachments

Physics news on Phys.org


math_head7 said:
(a) I did mr2 = (10)(.5)2 = 2.5 (the answer is supposed to be 1.6
Then I tried msrs2 + mcrc2 = (8)(.25)2 + (2)(.5)2 = 1.5 (still incorrect)
Not sure what you're doing here. The thin rods and the hoop are standard shapes--you can look up their rotational inertia. (You'll need to use the parallel axis theorem.)
 


first find I of hoop along any diameter using perpendicular axis theorem and then use parallel axis theorem to find I along axis of rotation
do same for square

then just simply add their I
 


Thanks. I got it.
 
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