Calculating moment of inertia about a door hinge.


by Becca93
Tags: door, hinge, inertia, moment
Becca93
Becca93 is offline
#1
Dec4-11, 02:34 PM
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.
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grzz
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#2
Dec4-11, 02:43 PM
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 = [itex]\frac{mL^{2}}{3}[/itex]

Now let us imagine the door to be a set of thin rods parallel to the axis passing through the hinges.
Becca93
Becca93 is offline
#3
Dec4-11, 03:02 PM
P: 84
Quote Quote by grzz View Post
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 = [itex]\frac{mL^{2}}{3}[/itex]

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?

grzz
grzz is offline
#4
Dec4-11, 03:21 PM
P: 939

Calculating moment of inertia about a door hinge.


[itex]\omega[/itex] 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.
Becca93
Becca93 is offline
#5
Dec4-11, 03:25 PM
P: 84
Quote Quote by grzz View Post
[itex]\omega[/itex] 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?
grzz
grzz is offline
#6
Dec4-11, 03:34 PM
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
Becca93
Becca93 is offline
#7
Dec4-11, 03:40 PM
P: 84
Quote Quote by grzz View Post
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.
grzz
grzz is offline
#8
Dec4-11, 03:48 PM
P: 939
Quote Quote by Becca93 View Post
... I take the linear velocity and multiply it by the width of the door to get ω ...
v = R[itex]\omega[/itex]
therefore .... = [itex]\omega[/itex]
Becca93
Becca93 is offline
#9
Dec4-11, 04:06 PM
P: 84
Quote Quote by grzz View Post
v = R[itex]\omega[/itex]
therefore .... = [itex]\omega[/itex]
Got it! Thanks!


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