Moment of inertia of thin rod with non-uniform mass

[shotgunshogun]

Homework Statement

A thin rod is exactly 1.9 meters long. The density in this rod varies in a peculiar manner. If we call the left-hand end of the rod x=0 and the right-hand end of the rod x=L , then the linear density can be expressed in units of kilograms per meter as http://coswebhost.rit.edu/webwork2_files/tmp/equations/6b/d06d55bce8c4f91789b960b87d21021.png

You grab the RIGHT-hand end of the rod and prepares to swing the rod this end. What is the moment of inertia of the rod around this end?

The Attempt at a Solution

I tried to intergrate from -L to 0 since your changing the axis of rotation and i got 1.16 kgm^2. You integrate the linear density times x^2 dx if you revolve from the left end , the bounds would be from 0 to L (1.85 kgm^2). The right side though... it doesn't work just the same, what needs to be changed to fufill the question, change of bounds, change of the density equation?

 

[LowlyPion]

Homework Statement

A thin rod is exactly 1.9 meters long. The density in this rod varies in a peculiar manner. If we call the left-hand end of the rod x=0 and the right-hand end of the rod x=L , then the linear density can be expressed in units of kilograms per meter as http://coswebhost.rit.edu/webwork2_files/tmp/equations/6b/d06d55bce8c4f91789b960b87d21021.png

You grab the RIGHT-hand end of the rod and prepares to swing the rod this end. What is the moment of inertia of the rod around this end?

The Attempt at a Solution

I tried to intergrate from -L to 0 since your changing the axis of rotation and i got 1.16 kgm^2. You integrate the linear density times x^2 dx if you revolve from the left end , the bounds would be from 0 to L (1.85 kgm^2). The right side though... it doesn't work just the same, what needs to be changed to fufill the question, change of bounds, change of the density equation?

What did you try to integrate? Can you show your equation?

 

[shotgunshogun]

I tried to integrate \lambda(x)x^2 from -L to 0

 

[LowlyPion]

I tried to integrate \lambda(x)x^2 from -L to 0

And not 0 to L ?

Edit: Oops I see they are going to hold it by the other end away from 0

 

[LowlyPion]

OK. You need to revise your treatment of the distance of the moment arm then don't you?

Instead of multiplying by x² and then integrating 0 to L if you were holding it at the origin, your moment arms now are (L - x)² aren't they?

Then you can integrate from 0 to L as before?

 

[shotgunshogun]

i worked it out and i don't believe it works because the inertia is greater than the first situation. The value should be lower because the mass increases as it approaches 1.9. So when you rotate it from the right side, it should be a lot easier to rotate the object and thus, a lesser value moment of inertia

 

[LowlyPion]

Choosing to go along the negative x-axis still puts your hands on the wrong end of the bat. Besides I'm not certain that λ is valid for -x.

Consider my suggestion a little more carefully.

For any mass element along the X axis you have a mass element of λ (x) a distance of L - x away. Is that not true?

 

[Dr.D]

The easy way to work this problem is to figure the MMOI with respect to the left end and then use the parallel axis theorem to move it to the right end.