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Old Aug 16th 2017, 04:51 AM   #1
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Rotation of rigid bodies problem

Hello everyone,
I came across a problem recently which, in hindsight I believe I do not fully understand, and as a consequence of this I cannot seem to get the correct solution no matter how I tackle it. As I have gone through several different attempted solutions to the problem I will just list below the problem and my most recent attempted solution, and would appreciate it if someone could point me in the right direction as to how to solve the thing.

Problem
You hang a thin hoop with radius R over a nail at the rim of the hoop. You displace it to the side (within the plane of the hoop) through an angle of B from it's equilibrium position and let it go.

What is the angular speed when it returns to it's equilibrium position ?
[Hint : Use equation - I = I (centre of mass) + m d (squared)]

Now I started off with the above equation then I assumed that the thin hoop corresponds to the same shape as a thin-walled hollow cylinder. So this would mean that the moment of Inertia of the hoop's around it's centre of mass is

I (center of mass) = m R (squared)

Now since the pivot point of the new axis (the nail) is at the rim of the hoop then the value of 'd' (distance of new axis to centre of mass axis) in the original equation is equal to the radius of the hoop (R) .

If we then substitute these values into the equation above we get

I = m R (squared) + m R (squared) = 2 m R (squared)

I we then apply the energy equations to this system we have ..

K (i) + U (i) = K (f) + U (f) ....
where K (i), K (f) are initial and final kinetic energy values and
U (i), U (f) are initial and final potential energy values

Now K (i) and U(f) are both zero
and K (f) = 1/2 I w (squared) (w = angular speed)
and U (i) = m g y (g = acceleration (gravity), y= height displacement)

So we now have ..
0 + m g y = 1/2 I w (squared) + 0
m g y = 1/2 I w (squared)
Giving ... w (squared) = 2 m g y / 2 m R (squared)
so w = sqrt (g y / R (squared))

However the answer the book gives form this problem is ..

w = sqrt (g/r (1 - cos B))

I don't understand this answer and don't know where i have gone wrong, can anyone help ?

Regards,
Jackthehat
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Old Aug 16th 2017, 08:12 AM   #2
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When displaced an angle $\beta$ from equilibrium, the hoop's center of mass is raised a height $h=r-r\cos{\beta}=r(1-\cos{\beta})$ above its lowest position.

Using conservation of energy ...

$\dfrac{1}{2}I \omega^2 = mgh$

$\dfrac{1}{2} \cdot 2mr^2 \cdot \omega^2 = mgr(1-\cos{\beta})$

Now solve for $\omega$
Attached Thumbnails
Rotation of rigid bodies problem-hoop_disp.jpg  

Last edited by skeeter; Aug 16th 2017 at 08:49 AM.
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Old Aug 16th 2017, 09:22 AM   #3
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Originally Posted by jackthehat View Post
However the answer the book gives form this problem is ..

w = sqrt (g/r (1 - cos B))
I think this should be:

w = sqrt (g (1 - cos B)/r)

Skeeter's approach gives this correct result.
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Old Aug 17th 2017, 08:21 PM   #4
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Hi Skeeter,
Thanks for taking the time to help with this problem. I hadn't thought that the height raised in the displacement wasn't just given by straight line displacement up but a curve upwards and therefore involved an angle. I see that now. I just worry that I didn't work that out from the start, as it seems obvious now that you have shown me in your diagram.
Once again many thanks.
Regards,
Jackthehat
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Old Aug 17th 2017, 08:24 PM   #5
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Hi ChipB,
Hi Skeeter,
Thanks for taking the time to get involved with my problem. I hadn't thought that the height raised in the displacement wasn't just given by straight line displacement up but a curve upwards and therefore involved an angle. I see that now. I just worry that I didn't work that out from the start, as it seems obvious now that I have a look at Skeeter's diagram.
Once again many thanks.
Regards,
Jackthehat
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