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Moment of inertia of door rotation

  • Thread starter sp3sp2sp
  • Start date
  • #1
100
4

Homework Statement


A 23 kg solid door is 220 cm tall, 95 cm wide. What is the door's moment of inertia for rotation about a vertical axis inside the door, 17 cm from one edge?

Homework Equations


I = I_cm + MR^2

The Attempt at a Solution


I = I_cm + MR^2

I = (1/12)(23kg)(0.95m)^2 + (23kg)((.95m/2) -.17m)^2

I = 1.73 +2.14 = 3.87kg*m^2
If this is wrong then Im not sure Im using right value for R. I am using the distance from the center of mass for the door which i think is middle of door. So .17m from edge would be .95/2 - .17 = .305 = R

thanks for any help
 

Answers and Replies

  • #2
RPinPA
Science Advisor
Homework Helper
552
316
That looks completely correct to me.

I checked it with integration of mr^2 across the width of the door and got the same answer.
 

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