Moment of Inertia of a rod and two spheres

[QuantumKnight]

Homework Statement

A barbell that consist of a long thin rod of mass M and length L is attached to two uniform spheres on each end. Both spheres have mass M and (1/3)L. The sphere on the left is hallow (spherical shell) and the sphere on the right is solid. What is the moment of inertia for the system as it rotates about an axis about the center of hallow sphere if M=1.0 kg and L = 1.0 m?

Homework Equations

I assume the parallel axis theorem since it's not rotating about the center of mass.

The Attempt at a Solution

What I've set up is: I = Inertia, cm = center of mass

I = I_cm + MD² = (1/12)ML² + MM((L/2))²

This is where I am stuck. This is for the rod but I am assuming I ignore the hallow sphere and add the moment of intertia for the solid sphere. If someone could explain what's going on in this problem or the basics of how to solve moment of inertia problems. Thanks

 

[SteamKing]

Your OP is not clear. What does (1/3)L represent?

 

[QuantumKnight]

My apologies. The spheres are 1/3rd the length of the bar. L is the length of the bar. So the length of the spheres are (1/3)L

 

[Vrbic]

I don't understand your attempt...it is for rod I guess...so MD^2 is not $M^2(\frac{L}{2})^2$ , if I understand in good way ( and (1/3)L is diameter of spheres): Just use your theorem for rod and ball on right hand side. Moment of inertia of hollow and solid ball is same. Than $J=J_{leftball}+J_{rod}+J_{rightball}=...$ . For instance: $J_{rod}=\frac{1}{12}ML^2+MD^2$ and your D is not nice $D=\frac{L}{2}+\frac{L}{3*2}$, when the latter term is radius of ball (axis of rotation).
The result what I tried is terrible number...try and write yours.
Hope I am right :)

 

[HalfLostAndSearching]

I would begin by stating that moment of inertia is a physical property of an object that describes its resistance to rotational motion. It is dependent on the mass and distribution of the object's mass relative to its axis of rotation.

In this scenario, we are given a barbell consisting of a rod and two spheres. To calculate the moment of inertia for this system, we must first consider the individual moments of inertia for each component. The rod can be treated as a thin rod, with its moment of inertia given by I = (1/12)ML^2, where M is the mass and L is the length.

Next, we must consider the two spheres. The hallow sphere can be treated as a thin spherical shell, with its moment of inertia given by I = (2/3)MR^2, where M is the mass and R is the radius. The solid sphere can be treated as a solid sphere, with its moment of inertia given by I = (2/5)MR^2.

To calculate the moment of inertia for the entire system, we must use the parallel axis theorem, which states that the moment of inertia about an axis parallel to the axis through the center of mass is equal to the moment of inertia about the center of mass plus the product of the mass and the square of the distance between the two axes.

In this case, the center of mass of the entire system is located at the center of the hallow sphere. Therefore, the moment of inertia for the system can be calculated as:

I = Irod + Ishell + Isolid = (1/12)ML^2 + (2/3)MR^2 + (2/5)MR^2 + M(L/2)^2

Substituting the given values for M and L, we get:

I = (1/12)(1.0 kg)(1.0 m)^2 + (2/3)(1.0 kg)(0.5 m)^2 + (2/5)(1.0 kg)(0.5 m)^2 + (1.0 kg)(0.5 m)^2

I = 0.083 kgm^2

Therefore, the moment of inertia for the system as it rotates about an axis through the center of the hallow sphere is 0.083 kgm^2.