
#1
Dec411, 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. 



#2
Dec411, 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. 



#3
Dec411, 03:02 PM

P: 84

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? 



#4
Dec411, 03:21 PM

P: 939

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




#5
Dec411, 03:25 PM

P: 84





#6
Dec411, 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 



#7
Dec411, 03:40 PM

P: 84

Edit: when I do that I get 30.6 J and that is incorrect. 



#8
Dec411, 03:48 PM

P: 939

therefore .... = [itex]\omega[/itex] 



#9
Dec411, 04:06 PM

P: 84




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