Explain to me the parallel axis theorem

[jaded18]

hi there. can anyone please explain to me the parallel axis theorem? the parallel axis theorem states that I = I_cm + M(d^2) where d = distance from the center of mass axis to the parallel axis and M is the total mass of the object. The rotational inertia of a thin rod about the center is = (1/12)M(L^2) and the rotational inertia of a thin rod about the end is = (1/3)M(L^2). Here is a picture I need you to use to explain to me the parallel axis theorem. i just don't see how inertia of the bar is (1.33m_bar)(l^2)/12

http://session.masteringphysics.com/problemAsset/1003167/20/136675C.jpg

i will understand if no one will be able to help me.. thanks for reading :)

 

[PhanthomJay]

hi there. can anyone please explain to me the parallel axis theorem? the parallel axis theorem states that I = I_cm + M(d^2) where d = distance from the center of mass axis to the parallel axis and M is the total mass of the object. The rotational inertia of a thin rod about the center is = (1/12)M(L^2) and the rotational inertia of a thin rod about the end is = (1/3)M(L^2). Here is a picture I need you to use to explain to me the parallel axis theorem. i just don't see how inertia of the bar is (1.33m_bar)(l^2)/12

http://session.masteringphysics.com/problemAsset/1003167/20/136675C.jpg

i will understand if no one will be able to help me.. thanks for reading :)

I_cm for the bar is as you noted. To detremine the I of the bar about the pivot point, the 'd' to use in the parallel axis theorem is the distance from the cm of the bar to the pivot point. What is that distance? The result as you indicated is the I of the bar about the pivot, not the I of the bar and point mass system.

 

[jaded18]

I_cm = (1/12)(m_bar)(L^2) right?
M = m_bar right?
so what's this (d^2) thing. if i take the answer (which i don't understand how it was figured out) and subtract I_cm, then i get (m_bar)(l^2)/36, and I guess this is what that M(d^2) term is then ... but where the heck did this d=1/6 come from then??

i just want to know how they got the answer (1.33m_bar)(l^2)/12 and I know that this is the I of the bar about the pivot as this is different from the I of the bar and point mass system.

 

[PhanthomJay]

I_cm = (1/12)(m_bar)(L^2) right?
M = m_bar right?
so what's this (d^2) thing. if i take the answer (which i don't understand how it was figured out) and subtract I_cm, then i get (m_bar)(l^2)/36, and I guess this is what that M(d^2) term is then ... but where the heck did this d=1/6 come from then??

i just want to know how they got the answer (1.33m_bar)(l^2)/12 and I know that this is the I of the bar about the pivot as this is different from the I of the bar and point mass system.

assuming the length of the bar is L , the center of the bar is at L/2 from the left end, and the fulcrum pivot is L/3 from the left end. Thus the center is (L/2 -L/3) from the pivot, where L/2 - L/3 = L/6. that is the 'd'.

 

[jaded18]

thanks lots for clearing that up ..

 

[Vijay Bhatnagar]

Distance of the cm of the bar from the left end = L/2
Distance of the pivot from the left end = L/3
Distance between the cm and pivot = L/2 - L/3 = L/6

Now determine MI about the pivot.