Solid pivoted pendulum attached to a spring - oscillation period?

[goraemon]

Homework Statement

The figure shows a 200 g uniform rod pivoted at one end. The other end is attached to a horizontal spring. The spring is neither stretched nor compressed when the rod hangs straight down. What is the rod’s oscillation period? You can assume that the rod’s angle from vertical is always small.

Homework Equations

τ=2∏/ω
ω=√(k/M) for spring
ω=√(Mgl/I) for solid pendulum

The Attempt at a Solution

If not for the spring, this problem would simply be a solid pendulum, and the solution would simply be:
τ = \frac{2∏}{√(Mgl/I)}

And if this were just a particle mass attached to the spring, the solution would be:
τ = \frac{2∏}{√(k/M)}

I'm having trouble figuring out what happens to the equation in this case, though. One guess is as follows:

τ = \frac{2∏}{√(k/M) + √(Mgl/I)}

But I'm clueless as to whether this is right or how to derive a correct equation. It seems like I might have to figure out the restoring force and/or torque in this case, but isn't that just -kx and -MglΘ? If anyone could help me toward the right direction, I'd appreciate it.

 

[TSny]

Thinking about torques is a good idea.

When the rod has a small angular displacement θ from the vertical, what is the net torque acting on the rod (about the axle) expressed in terms of θ?

 

[goraemon]

Thinking about torques is a good idea.

When the rod has a small angular displacement θ from the vertical, what is the net torque acting on the rod (about the axle) expressed in terms of θ?

OK, the gravitational force is being imposed downwards on the rod's center of mass, so the torque is:

-M*g*l*sin θ (where l = the distance between the pivot and the rod's center of mass, and θ = the angle of the rod from vertical)

I think? And when θ is small, then the above can be approximated to:

-M*g*l*θ

Is that right? And where to go from here? I feel like I should combine the above somehow with ω = √(k/m) for the spring. But how?

 

[voko]

What does $l$ mean? Is this the rod's length, is it its half-length?

What about the torque from the spring?

 

[goraemon]

What does $l$ mean? Is this the rod's length, is it its half-length?

What about the torque from the spring?

$l$ = the distance between the pivot and the rod's center of mass, so it would be the half-length...right?

OK, about the torque from the spring...since the angle is small, isn't it just -kx*L, where x = the distance by which the spring is stretched or compressed, and L = the full length of the rod?

If the above is correct, how do I combine them into a single equation to find the oscillation period?

 

[voko]

You should express $x$ in terms of $\theta$. Then, having the sum of torques equals, what can you do?

 

[goraemon]

You should express $x$ in terms of $\theta$. Then, having the sum of torques equals, what can you do?

OK, so then the torque imposed by the spring = $-kL\theta$?
And the torque from gravitational force = $-Mg(L/2)\theta$?

Adding the torques together...

Net torque = $(-kL\theta)$ + $(-Mg(L/2)\theta)$ = $-(kL+MgL/2)\theta$

Recalling Newton's 2nd law for rotational motion...

$\alpha = τ/I$

Replacing net torque for $τ$...

$\alpha$ = \frac{-(kL+\frac{MgL}{2})\theta}{I}Am I on the right track or way off base or somewhere in between?

 

[voko]

You are on the right track. But: does $kL\theta$ have the dimension of torque? Compare this with the earlier earlier expression.

 

[goraemon]

You are on the right track. But: does $kL\theta$ have the dimension of torque? Compare this with the earlier earlier expression.

Oh I think I see what you're saying. $kL\theta$ has units of N, which isn't right for a torque. Since $x = Lsin \theta ≈ L\theta$, does it follow that $kL\theta$ should be replaced by $kL^{2}\theta$?

 

[goraemon]

And if so, does the equation for $\alpha$ (above in post #7) change to the following?

$\alpha = \frac{-(kL^2+\frac{MgL}{2})\theta}{I}$

 

[TSny]

And if so, does the equation for $\alpha$ (above in post #7) change to the following?

$\alpha = \frac{-(kL^2+\frac{MgL}{2})\theta}{I}$

Yes, I think that's right.

 

[goraemon]

Yes, I think that's right.

OK, then what to do next...comparing $\alpha$ with linear acceleration $a$...
Since $a = \frac{-k}{m}x$, and $\omega = \sqrt{\frac{k}{m}}$

analogously with torque...
$\alpha = \frac{-(kL^2+\frac{MgL}{2})}{I}\theta$, so $\omega = \sqrt{\frac{(kL^2+\frac{MgL}{2})}{I}}$

meaning that the period = $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{I}{(kL^2+\frac{MgL}{2})}}$

And since the moment of inertia for a solid thin rod pivoting about its edge is $I = \frac{1}{3}ML^2$, the equation for the period becomes:

$T = 2\pi\sqrt{\frac{\frac{1}{3}ML^2}{(kL^2+\frac{MgL}{2})}} = 2\pi\sqrt{\frac{ML}{3(kL+\frac{Mg}{2})}}$


Is that right? Now I just need to plug in the numbers?

 

[TSny]

$T = 2\pi\sqrt{\frac{\frac{1}{3}ML^2}{(kL^2+\frac{MgL}{2})}} = 2\pi\sqrt{\frac{ML}{3(kL+\frac{Mg}{2})}}$

Is that right? Now I just need to plug in the numbers?

Looks good!