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Rotation and Inertia

1. The problem statement, all variables and given/known data

A 44.0-cm diameter wheel, consisting of a rim and six spokes, is constructed from a thin rigid plastic material having a linear mass density of 25.0 g/cm. This wheel is released from rest at the top of a hill 56.0  high.

How fast is it rolling when it reaches the bottom of the hill?How would your answer change if the linear mass density and the diameter of the wheel were each doubled?

2. Relevant equations

I of a hoop=mr^2
I of a rod=1/12 ML^23. The attempt at a solution
I set mgh=.5Iw^2 Inertia=m(rim)r^2+6(1/12 ML^2). I found the mass of the rim by finding the circumference and multiplied it by the density giving me 3.456 kg, plugged .22 m in for the radius.  Then I found the mass of a spoke by multiplying .44cm by 2.5 g/cm which gave me 1.1 kg. I plugged that in with L being .44. This gave me an inertia of .274. This made the energy equation (10.056)(9.8)(56)=.5(.274)w^2. Solving for w gave me 200.2 rad/s which I then multiplied by .22 to give me a velocity of 44.2 m/s. But that was wrong. Where did I go wrong? Thanks so much!  

    I believe you've confused diameter and radius. Each of the six spokes should have length 22 cm each.  

    Okay, well I tried that which changed inertia to be .1806 and ultimately gave me a velocity of 54.4 m/s, which was also wrong. Do you have any other ideas?  

    Hi Julie!                               Originally Posted by Julie323                   A 44.0-cm diameter wheel, consisting of a rim and six spokes,
#8230;
I set mgh=.5Iw^2 Inertia=m(rim)r^2+6(1/12 ML^2) #8230;                  
No, 1/12 ML2 is the moment of inertia of a rod about its centre #8230;

you need the moment of inertia about the centre of the wheel, ie the end of each spoke #8230;

so either use the formula for the end (which you should know, but if you don't, you can use the parallel axis theorem ),

or treat it as three spokes, of mass 2M and diameter 2L !   

    Hey!
So I figured out that I had the wrong equation a little earlier. I redid the problem, and it still came out wrong.  I will go through what I did, and maybe you can tell me what is wrong:
I started with the equation:
mgh=Iw^2.
I of rim=MR^2, R=.22m M=(.44m)(#960;)(2.5kg/m)=3.456kg
I of spoke=1/3ML^2, L=.22m M=(.22m)(2.5kg/m)=.55kg
I of rim+I of 6 spokes=.2205.

Returning to initial equation:
(3.456+6*.55)(9.8)(56)=.2205w^2
w=129.67
wr=v, r=.22m
v=28.5 m/s

Any idea where I went wrong? Thanks!  

    Hey Julie!

(have an omega:  and try using the X2 icon just above the Reply box )

I'm finding it rather difficult to read your equations, but I don't think you've included the linear KE.

(total KE = rotational KE about c.o.m. + linear KE of c.o.m.)
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