Moment of Inertia of this combined system?

In summary, the extra ML^2 is added to the moment of inertia because the axis for the sphere is not in the plane of the rod.
  • #1
jaewonjung
9
0
Homework Statement
A rigid rod of mass M and length L has moment of inertia 1/12 ML^2 about its center of mass. A sphere of mass m and radius R has moment of inertia 2/5 MR^2 about its center of mass. A combined system is formed by centering the sphere at one end of the rod and placing an axis at the other.


What is the moment of inertia of the combined system?
Relevant Equations
SumI=I1+I2+I3...
I=MR^2
Moment of inertia for a rod rotated around one end is 1/3ML^2 using the parallel axis theorem
Moment of inertia for a sphere is 2/5 MR^2

Itotal=Isphere+Irod=2/5MR^2+1/3ML^2

However, the answer is 2/5MR^2+1/3ML^2+ML^2

Why is there an extra ML^2 added to the moment of inertia?
 
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  • #2
jaewonjung said:
Moment of inertia for a rod rotated around one end is 1/3ML^2 using the parallel axis theorem
Moment of inertia for a sphere is 2/5 MR^2

Itotal=Isphere+Irod=2/5MR^2+1/3ML^2

However, the answer is 2/5MR^2+1/3ML^2+ML^2

Why is there an extra ML^2 added to the moment of inertia?

So you need to use the parallel axis theorem properly here. When you have compound systems, you need to shift both of their axes to the same point. You have done so for the rod, but not for the sphere. That moment of inertia is still through the centre of the sphere.

Also, always best to shift both separately to the same point and then add, rather than vice versa.

Hope that helps.
 
  • #3
Master1022 said:
Just for my understanding, which is the direction of the axis? Just to confirm that it is out of the plane of the rod?
Sorry for not including a picture. The axis is perpendicular to the rod, and is placed at the end of the rod that doesn't have the sphere attached to it.
 
  • #4
Master1022 said:
So you need to use the parallel axis theorem properly here. When you have compound systems, you need to shift both of their axes to the same point. You have done so for the rod, but not for the sphere. That moment of inertia is still through the centre of the sphere.

Also, always best to shift both separately to the same point and then add, rather than vice versa.

Hope that helps.
I see. shifting the axis from the center of the sphere to the other end of the rod is ML^2.
Thanks!
 

Related to Moment of Inertia of this combined system?

1. What is "moment of inertia"?

Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is affected by an object's mass, distribution of mass, and its shape.

2. How is moment of inertia calculated?

The moment of inertia of a combined system can be calculated by adding the individual moments of inertia of each component using the parallel axis theorem. This involves multiplying the mass of each component by the square of its distance from the axis of rotation and summing all of these values together.

3. What factors affect the moment of inertia of a combined system?

The moment of inertia of a combined system is affected by the mass and shape of each individual component, as well as the distance of each component from the axis of rotation.

4. How does the moment of inertia affect the rotation of a system?

The moment of inertia affects the rotational acceleration of a system. A larger moment of inertia means that more torque is required to produce the same acceleration, while a smaller moment of inertia results in a faster acceleration.

5. How is moment of inertia important in practical applications?

Moment of inertia is an important concept in areas such as engineering and physics, as it helps determine the stability and rotational behavior of objects. It is also used in designing machines and structures that require precise control and manipulation of rotational motion.

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