The period for small oscillations of a system

[zeralda21]

Homework Statement

See picture :

Homework Equations

$\sum M_{O}=I_{O}\ddot{\theta }$

The Attempt at a Solution

Consider the free-body diagram associated with an arbitrary positive angular displacement $\theta$; The moment about point $O$ is given by

$\sum M_{O}=-k\left ( b\sin\theta \right )\left ( b\cos\theta \right )-k\left ( 2b\sin\theta \right )\left ( 2b\cos\theta \right )-\underbrace{\left ( 3mgb\sin\theta \right )\left ( 3b\cos\theta \right )}_{\mathit{Why \ shouldn't \ this \ be \ included?}}$

Further, by the parallel axis theorem, $I_{0}=\overline{I}+md^2=0+m(3b)^2=9mb^2$ and for small oscillations $\sin\theta\simeq \theta \ \ \wedge \cos\theta\simeq 1$ and $\tau =\frac{2\pi}{\omega _{n}}=2\pi\sqrt{\frac{m}{k}}$. BUT why does not the mass of the sphere contribute to the moment?

 

[ehild]

The springs are stretched initially to keep the rod and mass horizontal, so the torque of gravity is balanced by the torque of the tensions in the springs. You only need to take into account the additional torques if you displace the mass from the equilibrium. The torque of gravity does not change if the displacement is small as it is proportional to the cosine of the angle.

ehild

 

[Tanya Sharma]

Hello ehild

The springs are stretched initially to keep the rod and mass horizontal, so the torque of gravity is balanced by the torque of the tensions in the springs.

Why should both the springs be stretched in the equilibrium position ? Either the upper one should be stretched or the lower one should be compressed or may be both the conditions hold simultaneously .

Am I right ?

Let the lower spring be compressed by x₁ and upper stretched by x₂ in equilibrium .Then,

For translational equilibrium , kx₁+kx₂ = mg

For rotational equilibrium kx₁b+kx₂(2b) = mg(3b)

Solving the above ,we get x₂ = 2mg/k and x₁ = -mg/k .

What does minus sign in x₁ = -mg/k signify ?

 

[ehild]

Hello ehild



Why should both the springs be stretched in the equilibrium position ? Either the upper one should be stretched or the lower one should be compressed or may be both the conditions hold simultaneously .

Am I right ?

Let the lower spring be compressed by x₁ and upper stretched by x₂ in equilibrium .

By "stretched" I meant that the springs are under tension (well, not necessarily both). You do not know the relaxed lengths. And it can happen that both are really stretched.

Then,

For translational equilibrium , kx₁+kx₂ = mg

That is not true. You ignored the force of the pivot.

For rotational equilibrium kx₁b+kx₂(2b) = mg(3b)

That is correct.

Solving the above ,we get x₂ = 2mg/k and x₁ = -mg/k .

What does minus sign in x₁ = -mg/k signify ?

As the first equation is not valid you can not determine x1 and x2.

 

[Tanya Sharma]

As always you are right .

Writing torque equation,

$$ M(3b)^2\ddotθ = -[k(x_1+bsinθ)(bcosθ)+k(x_2+2bsinθ)(2bcosθ) - mg(3b)] $$

Now $kx_1(bcosθ)+kx_2(2bcosθ) ≈ mg(3b)$

$$ 9Mb^2\ddotθ = -[k(bsinθ)(bcosθ)+k(2bsinθ)(2bcosθ)] $$

Applying approximations ,

$$ 9Mb^2\ddotθ = -[k(bsinθ)(bcosθ)+k(2bsinθ)(2bcosθ)] $$

$$ 9Mb^2\ddotθ = -5kb^2θ$$

$$ \ddotθ = -\frac{5k}{9M}θ$$

$$ \omega = \sqrt{\frac{5k}{9M}}$$

$$ T = 2\pi \sqrt{\frac{9M}{5k}}$$

Is this correct ?

 

[ehild]

It looks good

ehild

 

[Tanya Sharma]

Thanks .