# Help me to understand how my teacher calculated the moment of inertia/ distance to CM

#### Lisa...

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#### Doc Al

Mentor
Hint: Treat the U-shaped piece as consisting of three thin rods. For calculating the center of mass, treat them as three separates masses (located at the centers of the three rods).

#### Lisa...

I have tried treating the U-shaped piece as consisting of three rods but this gives me:

Itotal= 2 I10 cm + I 6 cm =

2 (1/3) (20*10*10-3) 0.12+ (1/3) (20*6*10-3) 0.62

A rod has a moment of inertia of 1/3 ML2.... but this gives me the wrong answer!

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#### Doc Al

Mentor
Lisa... said:
I have tried treating the U-shaped piece as consisting of three rods but this gives me:

Itotal= 2 I6 cm + I 10 cm =

2 (1/3) (20*6*10-3) 0.62+ (1/3) (20*10*10-3) 0.12

A rod has a moment of inertia of 1/3 ML2.... but this gives me the wrong answer!
Two problems:

(1) Your teacher is assuming that the "vertical" rods are 10cm long, and the horizontal "bottom" piece is 6cm long.

(2) While the two vertical rods are being rotated about one end, the horizontal rod is not. Treat that one as just a point mass (as far as rotation is concerned).

#### Lisa...

Okay great!!! Now I understand the whole thing :) Thank you VERY much Doc Al! The trick was the second point you've mentioned, I need to remember that :)

"Help me to understand how my teacher calculated the moment of inertia/ distance to CM"

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