Moment of Inertia about an axis at the center of mass

[JesseJC]

Homework Statement

3.) What is the moment of inertia of the object about an axis at the center of mass of the object? (Note: the center of mass can be calculated to be located at a point halfway between the center of the sphere and the left edge of the sphere.)

https://www.smartphysics.com/Content/smartPhysics/Media/Images/Mechanics/15/momentofinertia2new.png

mass of rod = 7.31kg
length of rod = 5.68m
radius of sphere = 1.42
mass of sphere = 36.55kg
1.) I about axis at left end of rod = 1960.4kg-m^2
2.) angular acceleration w/ 498N exerted perpendicular to center of rod = .72rad/s^2

Homework Equations

I = 2/5mr^2, I = 1/12ml^2, I = Icm + Md^2 (?)

The Attempt at a Solution

I've calculated the centre of mass, which ended up being about 6.4m, however, I am completely at a loss as to how I should approach the rest of the question. I'd really appreciate a bit of in depth help on this one! [/B]

 

[AlephNumbers]

Could you please restate the whole question, clearly? I'm not sure what's going on here.

 

[JesseJC]

Could you please restate the whole question, clearly? I'm not sure what's going on here.

An object is formed by attaching a uniform, thin rod with a mass of mr = 7.31 kg and length L = 5.68 m to a uniform sphere with mass ms = 36.55 kg and radius R = 1.42 m. Note ms = 5mr and L = 4R.

https://www.smartphysics.com/Content/smartPhysics/Media/Images/Mechanics/15/momentofinertia1new.png

1)

What is the moment of inertia of the object about an axis at the left end of the rod?

A: 1960.4kg-m^2

2)

If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 498 N is exerted perpendicular to the rod at the center of the rod?

A: .72rad/s^2

3)

What is the moment of inertia of the object about an axis at the center of mass of the object? (Note: the center of mass can be calculated to be located at a point halfway between the center of the sphere and the left edge of the sphere.)

https://www.smartphysics.com/Content/smartPhysics/Media/Images/Mechanics/15/momentofinertia2new.png

A: ?, this is where I'm beginning to get lost, I'm not sure how to approach this part of the question at all

I calculated the CM like so -> Xcm = [(mr)(L/2) + (ms)(L+R)]/(mr + ms) = 6.4m, but can't conjure up any ideas for the rest of the question, I've tried various applications of the parallel axis theorem and so on, but not confidently, I don't have a solid idea about how moment of inertia about the centre of mass, or a body such as this works. What do the arrows and dotted lines imply? I understand that MOI determines the amount of torque needed to angularly accelerate an object, but I just don't understand all of the theoretical workings behind it.

 

[haruspex]

I am completely at a loss as to how I should approach the rest of the question.

You have quoted the three equations you need.
To keep it readable, please work symbolically ( r = radius, l = rod length, etc.; use two different symbols for the masses).
In terms of those symbols, try to answer these questions in sequence:
What is the moment of inertia of the rod about its mass centre?
How far is the rod's mass centre from the rotation axis?
What is the moment of inertia of the rod about the rotation axis?
Then the same questions for the sphere.

 

[haruspex]

My post crossed with yours.

What do the arrows and dotted lines imply?

It isn't completely clear from the wording, but I believe the dashed line indicates the rotation axis. No idea what the arrow is for.

 

[JesseJC]

What is the moment of inertia of the rod about its mass centre? 1/12ml^2
How far is the rod's mass centre from the rotation axis? CM - l/2 or 6.4 - 2.84 = 3.56m
What is the moment of inertia of the rod about the rotation axis? 1/12m(CM - l/2)^2 ?

What is the moment of inertia of the sphere about its mass centre? 2/5Mr^2
How far is the sphere's mass centre from the rotation axis? r/2 ?
What is the moment of inertia of the sphere about the rotation axis? 2/5M(r/2)^2 ?

Let me know where I went wrong, I'm not confident in the answers followed by question marks.

 

[haruspex]

What is the moment of inertia of the rod about the rotation axis? 1/12m(CM - l/2)^2 ?

No. As you posted in the OP

I = Icm + md²

You have correctly calculated Icm here as 1/12ml².
d is the distance from the mass centre to the rotation axis.

 

[JesseJC]

ok, so the MOI with respect to the rod is 1/12ml^2 + md^2 ---> 1/12(7.31)(5.68)^2 + (7.31)(3.56)^2 --> 112.3kg-m^2

and for the sphere; 2/5mr^2 + md^2 ---> 2/5(36.55)(1.42)^2 + (36.55)(.71)^2 --> 47.9kg-m^2, correct?

 

[haruspex]

ok, so the MOI with respect to the rod is 1/12ml^2 + md^2 ---> 1/12(7.31)(5.68)^2 + (7.31)(3.56)^2 --> 112.3kg-m^2

and for the sphere; 2/5mr^2 + md^2 ---> 2/5(36.55)(1.42)^2 + (36.55)(.71)^2 --> 47.9kg-m^2, correct?

Yes.

 

[JesseJC]

Right, thanks, that clarified things for me. So would i approach finding the MOI @ this point the same way?

https://www.smartphysics.com/Content/smartPhysics/Media/Images/Mechanics/15/momentofinertia3.png

 

[haruspex]

Right, thanks, that clarified things for me. So would i approach finding the MOI @ this point the same way?

https://www.smartphysics.com/Content/smartPhysics/Media/Images/Mechanics/15/momentofinertia3.png

Yes.