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Moment of Inertia of rod on an axis

  • #1
105
3

Homework Statement


You are a project manager for a manufacturing company. One of the machine parts on the assembly line is a thin, uniform rod that is
60.0 cm long and has mass 0.700 kg .

1What is the moment of inertia of this rod for an axis at its center, perpendicular to the rod?
Express your answer with the appropriate units.

2One of your engineers has proposed to reduce the moment of inertia by bending the rod at its center into a V-shape, with a
60.0∘ angle at its vertex. What would be the moment of inertia of this bent rod about an axis perpendicular to the plane of the V at its vertex?


Homework Equations


I of rod = 1/12 * m *L2
I of rod aorund end 1/3 * m * L2

The Attempt at a Solution



mass= .7
length= 60 cm = .6 m

Ok, I got part 1 it is .021

for part 2

If the rod is bent into a v shape and rotated around its axis, it is 2 rods half the length of the original rotation around their ends so the moment of Inertia should be

1/3 * .7 * .3^2 + 1/3 * .7 * .3^2
= .2716

why is this wrong?
 
Last edited:

Answers and Replies

  • #2
Doc Al
Mentor
44,892
1,143
I of rod = 1/12 * m *r2
r is the length of the rod, not the radius. (Radius of what?)
 
  • #3
105
3
r is the length of the rod, not the radius. (Radius of what?)
I am use to doing moment of Inertia of discs and circle so I write radius out of habit, it is supposed to be L instead L instead of R, regardless when any point on the disc rotates it forms a disc.

but my mistake was diving by 2, it should be the whole length, though i do not understand why, the moment of inertia for a single point is mr^2 as it rotates on a disc
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
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1/3 * .7 * .3^2 + 1/3 * .7 * .3^2
So the whole mass is now 1.4?
Anyway, I don't see how you got .2716 from the above expression. I get .042.
 
  • #5
Doc Al
Mentor
44,892
1,143
1/3 * .7 * .3^2 + 1/3 * .7 * .3^2
= .2716

why is this wrong?
Sorry, I didn't see your edits. (Thankfully, haruspex is on the ball.) In addition to his comment about the calculation, in your formula you forgot to divide the mass in two.

Hint: How should the answers to each part relate?
 

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