# Calculating moment of inertia about a door hinge.à

by Becca93
Tags: door, hingeà, inertia, moment
 P: 84 The problem statement, all variables and given/known data A solid door of mass 39.30 kg is 2.34 m high, 1.68 m wide, and 3.23 cm thick. What is the moment of inertia of the door about the axis through its hinges? If the edge of the door has a tangential speed of 76.5 cm/s, what is the rotational kinetic energy of the door? The attempt at a solution I don't really know where to start. Should I find the center of mass of the door? Because I = Ʃmr^2? I know how to solve for the second half (Ek = (1/2)Iω^2), but I'm not sure how to calculate I.
 P: 939 The moment of inertia of a thin rod of mass m and lenght L about an axis at one of its ends and perpendicular to the rod is given by I = $\frac{mL^{2}}{3}$ Now let us imagine the door to be a set of thin rods parallel to the axis passing through the hinges.
P: 84
 Quote by grzz The moment of inertia of a thin rod of mass m and lenght L about an axis at one of its ends and perpendicular to the rod is given by I = $\frac{mL^{2}}{3}$ Now let us imagine the door to be a set of thin rods parallel to the axis passing through the hinges.
I have the answer to the first question, but I'm not nearly as well off with the second question as I had assumed.

I got 37 kgm^2 for the first question.

I assumed you would take that value and take 76.5 cm/s (.765 m/s) as ω and plug it into
E = (1/2)Iω^2

I get 10.8 when I do that (I used m/s and I'm assuming the units are joules), but that's incorrect.

Any advice?

P: 939

## Calculating moment of inertia about a door hinge.à

$\omega$ is the ANGULAR speed i.e. measured in ANGLE per second i.e. rad/s and so you have the convert the TANGENTIAL speed given to ANGULAR speed.
P: 84
 Quote by grzz $\omega$ is the ANGULAR speed i.e. measured in ANGLE per second i.e. rad/s and so you have the convert the TANGENTIAL speed given to ANGULAR speed.
Oh! Okay. How do I calculate angular speed from tangential speed?
 P: 939 angle (in radians) rotated = distance along the arc / radius therefore distance along arc = radius x angle i.e. distance along arc/time = radius x (angle /time) v = r x w v in m/s because linear or tangential w in rad/s because angular
P: 84
 Quote by grzz angle (in radians) rotated = distance along the arc / radius therefore distance along arc = radius x angle i.e. distance along arc/time = radius x (angle /time) v = r x w v in m/s because linear or tangential w in rad/s because angular
So to get velocity, I take the linear velocity and multiply it by the width of the door to get ω, and then plug that value into the energy equation?

Edit: when I do that I get 30.6 J and that is incorrect.
P: 939
 Quote by Becca93 ... I take the linear velocity and multiply it by the width of the door to get ω ...
v = R$\omega$
therefore .... = $\omega$
P: 84
 Quote by grzz v = R$\omega$ therefore .... = $\omega$
Got it! Thanks!

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