Min Period of oscillation of a rod

[Tanya Sharma]

Homework Statement

An iron rod of length L is hung at a common point with threads of length 'l' which are attached to the two ends of the rod. The rod is displaced a bit in the plane of the threads. What is the length of the threads if the period of the swinging of the rod is the least, and what is this period?

Homework Equations

The Attempt at a Solution

Moment of Inertia of the rod about O is $$ I = \frac{ML^2}{12}+My^2 $$

Write torque equation , $I\ddotθ = -Mgyθ$

$$ \ddotθ = -\frac{Mgy}{I}θ $$

$$ ω^2 = \frac{Mgy}{I} $$

$$ ω = \sqrt{\frac{Mgy}{I}} $$

$$ T = 2\pi \sqrt{\frac{I}{Mgy}} $$

$$ T = 2\pi \frac{(l^2-\frac{L^2}{6})^\frac{1}{2}}{g(l^2-\frac{L^2}{4})^\frac{1}{4}} $$

Should I minimize this ? This looks a little more complicated than usual .

Edit : Fixed the errors

 

[TSny]

Moment of Inertia of the rod about O is $$ I = \frac{ML^2}{3}+My^2 $$

Check this expression. What is the moment of inertia of a rod about its center of mass?

Otherwise, your work looks good to me.

You might find it easier to express the period in terms of the variable y (instead of $l$) and minimize the square of the period with respect to y.

 

[Tanya Sharma]

Check this expression. What is the moment of inertia of a rod about its center of mass?

OK...I have fixed the errors. Does post#1 look alright now?

You might find it easier to express the period in terms of the variable y (instead of $l$) and minimize the square of the period with respect to y.

$$ T = 2\pi \frac{(y^2+\frac{L^2}{12})^\frac{1}{2}}{g(y)^\frac{1}{4}} $$

$$ T^2 = 4\pi^2 \frac{(y^2+\frac{L^2}{12})}{g^2(y)^\frac{1}{2}} $$

Now I should minimize T² . Right ?

 

[TSny]

O

$$ T = 2\pi \frac{(y^2+\frac{L^2}{12})^\frac{1}{2}}{g(y)^\frac{1}{4}} $$

In the denominator, the 1/4 power on y is not correct.

Now I should minimize T² . Right ?

Right.

 

[Tanya Sharma]

The minimum value occurs at y=L/6 or at $l= \frac{\sqrt{10}}{6}L$

Is it correct ?

 

[TSny]

The minimum value occurs at y=L/6 or at $l= \frac{\sqrt{10}}{6}L$

Is it correct ?

That's not what I get.

 

[Tanya Sharma]

The minimum value occurs at $y=\frac{L}{\sqrt{6}}$ or at $l= \sqrt{\frac{5}{12}}L$

Does this make sense ?

 

[TSny]

Still not what I'm getting. What is your expression for T²?

 

[Tanya Sharma]

Another attempt

The minimum value occurs at $y=\frac{L}{\sqrt{12}}$ or at $l= \frac{L}{\sqrt{3}}$

 

[haruspex]

$$ T^2 = 4\pi^2 \frac{(y^2+\frac{L^2}{12})}{g^2(y)^\frac{1}{2}} $$

You should get into the habit of checking equations for dimensional consistency. The exponents of both g and y are wrong in the denominator.
I agree with TSny that you are still getting the wrong answer. Please post all your steps from the above.
Clearly the constant factor in the equation isn't going to affect the value of y, so you can dispense with that.

 

[TSny]

Another attempt

The minimum value occurs at $y=\frac{L}{\sqrt{12}}$ or at $l= \frac{L}{\sqrt{3}}$

Bingo.

 

[Tanya Sharma]

Thanks.

You might find it easier to express the period in terms of the variable y (instead of $l$) and minimize the square of the period with respect to y.

This is something I would like to understand .

This technique is used often where we need to max/min a function involving square root and instead we max/min the square of the function .How do we know that the maximum and minimum of both the function and its square occur at the same point ?

 

[TSny]

The square root function increases whenever its argument increases. It's a "monatonically increasing" function. So, the square root function will take on its maximum (minimum) value where its argument is maximum (minimum).