# Need help in proving this Moment of Inertia equation

1. Jul 11, 2014

### null void

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

Suppose I have a rod with a know length, L, and 2 point-liked mass with know mass, m, each of them are fixed at both ends of the rod. And the rode has an axle connected at its midpoint where the axle is connect to a circular spring.

The rod(with the 2 mass) is rotated 180 degree or ∏ in radian, and is released to oscillate. How to find the period, T of the oscillation?

The equation I want to prove is:

T2 = 8m2r2/τ + T02

m is the mass of the point like mass,
r is the distance of the point like masses from the midpoint of the rode,
T0 is contributed from the Rod...(base on my guessing)

2. Relevant equations

3. The attempt at a solution
My attempt of solution is in the attachment. The final expression is pretty much different from the equation i want to prove...Can anyone help me to check where did i did wrongly?

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2. Jul 12, 2014

### Simon Bridge

Please do not post proprietary format attachments - not everyone can read them.
doc and docx formats can be particularly troublesome even if you have MSOffice or use google-docs.
you can post in pdf format, or use the LaTeX functions.

I think I've managed to open the docx though.

You write:

Total energy from potential energy of circular spring: $$\frac{1}{2}\tau\pi = \text{KE}+\text{PE} =\frac{1}{2}I\omega^2+\frac{1}{2}\tau\pi\left( 1-\frac{\theta}{\pi} \right)$$
$\tau\theta = I\omega^2$
$$\omega = \frac{d\theta}{dt}=\sqrt{\frac{\theta\tau}{I}}\\ \frac{d\theta}{\sqrt{\theta}}=\sqrt{\frac{\tau}{I}}dt$$

By integrating (I consider al of the parameters are constant except $\theta$ and $\omega^2$ are changing.)
$$2\sqrt{\theta} = t\sqrt{\frac{\tau}{I}}\\ t=2\sqrt{\frac{I\theta}{\tau}}$$
After 1/4 of the T time pass from the moment the rod is released $\theta = \pi$
$$T=8\sqrt{\frac{I\pi}{\tau}}$$
Squaring both sides an inserting the moment of point-like mass and rod into the equation:
$$T^2=64\frac{\pi}{\tau}\left( 2mr^2+\frac{m_{rod}L^2}{12} \right)$$

... is all that correct?

I take it that this system is basically a dumbell rotating back-and-forth about it's center and the spring is linear - so the restoring torque is proportional to the angle of deflection from equilibrium $\tau=-k\theta$?

You appear to have not yet finished - there is no "$r$" or "$\tau$" in the problem statement for example.
If that $\tau$ in your equation is, indeed, torque, then it is not a constant.

You should already know what sort of equation to expect for $\theta(t)$ in terms of the period T ... why not just plug that into Newton's second law?

Last edited: Jul 12, 2014
3. Jul 13, 2014

### null void

I hope i get it right...Since the oscillation oscillate maximally 180° degree or ∏ in radian, my equation of angular displacement is:
Θ(t) = ∏ sin(2∏ft)

then i differentiate the equation to get the angular velocity and acceleration:
ω(t) = 2∏2f cos(2∏ft)

α(t) = -4∏3f2 sin(2∏ft)

Then from Newton's 2nd law,
τ=Iα
τ=-4I∏3f2sin(2∏ft)

The maximum τ happens when the sin(...) = 1

Then I can be decomposed to mr2 and MR2/12, and the f is the inverse of T,

.....well i am not very sure if i can simply ignore the negative sign.....

T2=8mr23/τ + MrodLrod23/3τ

The problem still at the ∏...i must have done something wrong again...the equation is:

T2 = 8m∏2r2/τ + T02

The T02 should be something contribute by the rod.

Last edited by a moderator: Jul 13, 2014
4. Jul 13, 2014

### Simon Bridge

You still have a $\tau$ and an $r$ term. Why don't you express these in terms of $\theta$ and $L$? (Why did you make the equations so big?!)

Newtons second law for rotation is: $\tau = I\ddot\theta$
If the spring is linear with spring constant k, then $\tau = -k\theta$
(I think this second equation is the one you are missing.)

You suggested $\theta(t)=\pi\sin (2\pi f t)$
... what is the angular displacement at t=0 in this equation?
... what is the angular displacement at t=0 in the problem statement?

Since you want to find the period, T, then why not express the $\theta(t)$ equation in terms of T instead of f?

5. Jul 13, 2014

### null void

sorry I didn't notice i mixed-up the term τ-instantaneous and τ-maximum in the previous post.

I started to doubt if my concept is right or not. My current idea is :

the rod is rotated 180o before released to oscillate, so the maximum amplitude of the Θ(t) is ∏

Θ(t) = ∏sin(2∏t/T)
angular acceleration = α(t)= ( -4∏3/T2 )cos(2∏t/T)

....i guess the Θ with 2 dots means double differentiation of Θ...............

from the Newton's second Law,

I α(t) = -k Θ(t)
I ( ( -4∏3/T2 )sin(2∏t/T) ) = -k ( ∏sin(2∏t/T) )

[T2= 4 I ∏3/k∏

and the k∏ is the maximum torque force generated by the restoring spring.

T2 = 4 I ∏3 / τmax

is this correct?

Last edited by a moderator: Jul 13, 2014
6. Jul 13, 2014

### Simon Bridge

According to that equation, Θ(0)=0, but according to the problem statement, Θ(0)=∏
But it does not make any difference for the period calculation but it can cost you marks.

If you use: $\Theta(t)=\pi\cos 2\pi t/T$, then $$\ddot\Theta = -\frac{4\pi^3}{T^2}\cos 2\pi t/T = -\frac{4\pi^2}{T^2}\Theta(t)$$ Note: You are correct- $\ddot y \implies \frac{d^2}{dt^2}y$ while $y'' \implies \frac{d^2}{dx^2}y$ ... it just lets you write out the derivatives on one line. If y=f(x,t) then the implied derivatives are partials.

You end up with: $$T^2=\frac{4\pi^2}{k}I=\frac{4\pi^3}{\tau_{max}}I$$ ... well done.

The next step is to get the correct expression for the moment of inertia.
All you need to finish that is to find an expression for r in terms of L.

Some pointers:
∏ is the cap-pi and refers to a product.
The ratio of the circumference to the diameter is a lower-case pi: π
It looks a bit like an "n" I know. That's why I use LaTeX rather than the quick-symbols.

7. Jul 14, 2014

### null void

yeah the Torque and the period symbol almost look the same too......

In my lab manual, the period calculation is:

$T^2 = \frac{8m\pi^2r^2}{D} + T^2_0$

where the $T_0$ is the period of oscillation when the mass is removed from the rod.
And the $D$ is stated in my manual that it is the restoring torque, i suppose it means the maximum torque generated by the 'circular spring'.

But no matter how i think and try, i never get the equation with $\pi^2$, do u think there is a mistake in my manual? I think the $D$ should be the spring constant.

$r = \frac{1}{2}L$
so should I use the $2mr^2$ and $\frac{M_{rod} L^2_{rod}}{12}$ ?

if so i will end up getting this equation:
$T^2 = \frac{4\pi^3}{\tau}(2m(\frac{L_{rod}}{2})^2) + \frac{M_{rod}L^2_{rod}}{12})$

or if I keep the r,
$T^2 = \frac{4\pi^3}{\tau}(2m(r^2) + \frac{M_{rod}L^2_{rod}}{12})$

Last edited: Jul 14, 2014
8. Jul 14, 2014

### Simon Bridge

You need to check the definitions of the variables in the model equation.
I see now why you didn't want to lose the "r" ... so you have to put all lengths in terms of r.
What is this "r" as it is used in your book?

It may well be a typo of course.

9. Jul 14, 2014

### null void

$r$ is the distance of the point-like mass to the center of axis and the $L_{rod}$ is the length of the rode. The point like mass can be moved closer to the center, so the $r$ can be change, but the $L_{rod}$ always fixed at a length.

10. Jul 14, 2014

### Simon Bridge

Ah that makes sense - you left that off the description before.
The more I think about it the more it looks like a typo in your book.
Devil is in the details - go through the book carefully.
There are not many places to misplace a pi.

11. Jul 14, 2014

### null void

12. Jul 14, 2014

### Simon Bridge

For the mass on a spring:

$m\ddot x = -kx$
$x=A\cos 2\pi t/T$

... if we say that D = the max restoring force then D=kA

then: $$m\frac{4\pi^2}{T^2} = k \implies T^2=4\pi^2\frac{m}{k}= 4\pi^2A\frac{m}{D}$$
... you can see that this has to be correct.

...you can see that if $A=\pi$ then you get a $\pi^3$ term.
I think that D, for your version, has to be understood as a spring constant (dimensions of torque per radien)... especially considering the form of eq(1) in the leaflet.

13. Jul 26, 2014

### null void

Sorry for late reply, I think everything is fine now. Thanks you very much Simon, for guiding me to get the right equation.

14. Jul 27, 2014

Well done.