Torsion pendulum in Cavendish experiment

[Yoonique]

Homework Statement

In the Cavendish experiment, the two small balls have mass m each and are connected by a light rigid rod with length L. The two large balls have mass M each and are separated by the same distance L. The torsion constant of the torsion wire is κ.
b) Put the large balls a small distance away from the small balls, when the system reaches the equilibrium, the rigid rod rotates an angle θ and the distance between the centers of balls is r, as shown in the above figure. Find the expression for the gravitational constant G.
c)If the small balls are perturbed with small angle from the equilibrium position in (b), will the oscillation be harmonic? If so, find the expression for the period of the oscillation.

Homework Equations

∑τ = Iα
T = 2π/ω

The Attempt at a Solution

I solved for part b. But I can't get the answer to part c.

Part b: G = 2π²Lr²θ/(MT²)

Part c:
Let the new angle be θ₁
∑τ = Iα
GMmL/(r-0.5θcosθ₁)² - κθ₁ = Iα
α = 2GM/(L(r-0.5θcosθ₁)²) - 2κθ₁/(mL²)
I can't make α=-ω²θ₁ to find for ω to solve for T.

 

[paisiello2]

Isn't "r" a function of L and θ?

 

[Yoonique]

Isn't "r" a function of L and θ?

Originally when it is in equilibrium, the distance between M and m is r. Then the small balls are perturbed with small angle from the equilibrium position, so the new distance between M and m is r-0.5θcosθ.

 

[paisiello2]

That's not how I interpret the diagram which clearly shows r depending on the angle θ.

The distance between M and m at equilibrium when θ=0° is r=L/√2.

 

[Yoonique]

That's not how I interpret the diagram which clearly shows r depending on the angle θ.

The distance between M and m at equilibrium when θ=0° is r=L/√2.

In part b it says it rotates at an angle θ when it reaches equilibrium. Then in part c it says the small balls are perturbed with small angle from the equilibrium position in part b. So in part c, the new angle isn't it θ+small angle, which we call it θ₁. I'll edit my θ in my workings to θ₁. I think it's confusing to use the same θ.

 

[paisiello2]

OK, I see that now.

This expression is not correct:
r-0.5θcosθ₁

For one thing the dimensions are wrong.

Since θ₁ is a small angle then a good approximation for the distance between M and m would be:
r' = r-Lθ₁

Now you are squaring this number r'. How much different will r² be from r'²?

 

[Yoonique]

OK, I see that now.

This expression is not correct:
r-0.5θcosθ₁

For one thing the dimensions are wrong.

Since θ₁ is a small angle then a good approximation for the distance between M and m would be:
r' = r-Lθ₁

Now you are squaring this number r'. How much different will r² be from r'²?

Why is it Lθ and not θL/2? Since the radius is L/2.

 

[paisiello2]

Yes, you're right, L/2.

 

[Yoonique]

Yes, you're right, L/2.

So the new torque would be:
2GMm/(r-0.5Lθ₁)₂(L/2) - κθ₁ = Iα
The answer involves a cosθ and it is not in terms of G. Means I got to use part b answer?

 

[paisiello2]

No, do you know how to solve differential equations?

 

[Yoonique]

No, do you know how to solve differential equations?

How do I make it into a differential equation? I think I know how to solve them if it is those basic ones.

Edit: Means I got to change α into d²θ/dt²? And solve it so that I will get an equation relating θ and T?

 

[paisiello2]

Yes, you got it. The only trick is to simplify the equation by assuming (r-0.5Lθ₁)² is approximately r² for small θ₁.

 

[Yoonique]

Yes, you got it. The only trick is to simplify the equation by assuming (r-0.5Lθ₁)² is approximately r² for small θ₁.

It will become 2(GMm/r² - κθ₁) = Lm(d²θ₁/dt²). How do you solve this? It is not linear and homogeneous.

 

[paisiello2]

I think it's linear. Regardless, you only need to come up with the period. So why not assume θ₁ is in the form of a simple harmonic equation?

 

[Yoonique]

I think it's linear. Regardless, you only need to come up with the period. So why not assume θ₁ is in the form of a simple harmonic equation?

I tried re-arranging the equation into the form of α = -ω²θ so that I can use T = 2π/ω. But I can't separate out θ₁ away from the other terms.

 

[paisiello2]

Can you solve the homogenous version of this equation?

 

[Yoonique]

Can you solve the homogenous version of this equation?

I'm not sure how though. Maybe I haven't learn how to solve this type of differential equations. Is differential equation the only way to solve it?

 

[paisiello2]

Afraid so.

 

[Yoonique]

Afraid so.

Can you guide me on using differential equation to solve it?
The equation I have now is 2(GMm/r² - κθ₁) = Lm(d²θ1/dt²)

 

[paisiello2]

First do what I said in post # 12.

 

[paisiello2]

OK, I see that you did that already.

 

[Yoonique]

OK, I see that you did that already.

Yeah, how do I continue from there? How do I alter the equation so I can integrate them?

 

[TSny]

I tried re-arranging the equation into the form of α = -ω²θ so that I can use T = 2π/ω. But I can't separate out θ₁ away from the other terms.

The net torque may be written $$\tau = \frac{A}{r^2} - \kappa \theta = \frac{A}{(r_0- b\, \delta \theta)^2} - \kappa (\theta_0 +\delta \theta) $$ where ##A## and ##b## are constants.

##r_0## and ##\theta_0## are the equilibrium values of ##r## and ##\theta##, and ##\delta \theta## represents displacement of ##\theta## from equilibrium.

For small ##\delta \theta## you can linearize this expression to get ##\tau = -B \, \delta \theta## for some constant ##B##.

 

[paisiello2]

You're right, we forgot the original Kθ term. That will simplify the equation down to:

GMm/r² + K(θ+θ₁) = Iθ₁''

(GMm/r² + Kθ) + Kθ₁ = Iθ₁''

Kθ₁ = Iθ₁''

 

[paisiello2]

If we have simple harmonic motion then we can assume that the solution to the differential equation will take the form:

θ₁= A⋅sin(ωt) where A and ω are constants to be determined.

 

[Yoonique]

You're right, we forgot the original Kθ term. That will simplify the equation down to:

GMm/r² + K(θ+θ₁) = Iθ₁''

(GMm/r² + Kθ) + Kθ₁ = Iθ₁''

Kθ₁ = Iθ₁''

Why did you ignore (GMm/r² + Kθ)?

 

[Yoonique]

The net torque may be written $$\tau = \frac{A}{r^2} - \kappa \theta = \frac{A}{(r_0- b\, \delta \theta)^2} - \kappa (\theta_0 +\delta \theta) $$ where ##A## and ##b## are constants.

##r_0## and ##\theta_0## are the equilibrium values of ##r## and ##\theta##, and ##\delta \theta## represents displacement of ##\theta## from equilibrium.

For small ##\delta \theta## you can linearize this expression to get ##\tau = -B \, \delta \theta## for some constant ##B##.

So B/I is ω²?

 

[Yoonique]

If we have simple harmonic motion then we can assume that the solution to the differential equation will take the form:

θ₁= A⋅sin(ωt) where A and ω are constants to be determined.

How do I determine the constants when I do not have any values to sub into the equation?

 

[TSny]

So B/I is ω²?

Yes.

Once you find B, you will see that there is an approximation you can make to simplify B.

 

[Yoonique]

Yes.

Once you find B, you will see that there is an approximation you can make to simplify B.

Okay, but I have a problem. I do not know how to linearize it for the fact that δθ is small. What is the trick here? I mean a lot of the post here tells me to 'simplify it' but I just don't know how to get started. I know (r₀ - bδθ)² ≈ r₀ right?

 

[TSny]

Okay, but I have a problem. I do not know how to linearize it for the fact that δθ is small. I know (r₀ - bδθ)² ≈ r₀ right?

That's going too far in the approximation.

For ##\frac{1}{(r_0-b \, \delta \theta)^2}##, use the binomial approximation: http://en.wikipedia.org/wiki/Binomial_approximation

 

[Yoonique]

That's going too far in the approximation.

For ##\frac{1}{(r_0-b \, \delta \theta)^2}##, use the binomial approximation: http://en.wikipedia.org/wiki/Binomial_approximation

So for this question I just got to approximate properly and I can solve for ω? I do not need differential equation?

 

[TSny]

So for this question I just got to approximate properly and I can solve for ω? I do not need differential equation?

That's right. To first order in ##\delta \theta## the differential equation would be the SHM equation ##I \ddot{\delta \theta} = -B \delta \theta##.

So, as you said, ##\omega^2 = \frac{B}{I}##.

 

[Yoonique]

That's right. To first order in ##\delta \theta## the differential equation would be the SHM equation ##I \ddot{\delta \theta} = -B \delta \theta##.

So, as you said, ##\omega^2 = \frac{B}{I}##.

Okay I got (GMmL/2)(r₀^-2 + Lδθr₀^-3) - κ(θ₀+δθ). How can I simplify even further?

 

[TSny]

Okay I got (GMmL/2)(r₀^-2 + Lδθr₀^-3) - κ(θ₀+δθ). How can I simplify even further?

I don't get the 2 in the numerator of the first factor.

To simplify further, think about the condition that holds at equilbrium.