Torsion pendulum in Cavendish experiment

[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.

 

[Yoonique]

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

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

Isn't the 2 factor because the radius of the circle is L/2?
At equilibrium, GMmL/(2r₀²) = κθ₀. So I sub it in. I'll get -δθ(κ - κθ₀/r₀) = Iα
So ω² = 2κ(r₀ - θ₀)/(mL²r₀).
Therefore T = 2π√(mL²r₀)/2κ(r₀ - Lθ₀)

But my given answer is T = 2π√(mL²r₀)/2κ(r₀ - Lθ₀cosθ₀). Why is there an extra cosθ₀ ...

 

[Yoonique]

If I work backwards, when it is θ = θ₀ + δθ the distance between M and m is (L/2)δθcosθ₀. I thought since δθ is small, I can approximate the distance between M and m as (L/2)δθ?

 

[TSny]

Isn't the 2 factor because the radius of the circle is L/2?

Did you take into account that there are 2 gravitational forces acting on the rod?

At equilibrium, GMmL/(2r₀²) = κθ₀.

OK, except for that factor of 2. (Hope I'm not the one who's dropping a factor of 2 somewhere.)

So I sub it in. I'll get -δθ(κ - κθ₀/r₀) ...

Note that the second term in parentheses does not have the correct dimensions.

 

[Yoonique]

Did you take into account that there are 2 gravitational forces acting on the rod?OK, except for that factor of 2. (Hope I'm not the one who's dropping a factor of 2 somewhere.)Note that the second term in parentheses does not have the correct dimensions.

Ops, is a typo, I forgot about the L inside. It should be δθ(κ - Lκθ₀/r₀). My final answer is T = 2π√(mL²r₀)/2κ(r₀ - Lθ₀). However the given answer is T = 2π√(mL²r₀)/2κ(r₀ - Lθ₀cosθ₀). Why is there an extra cosθ₀? If I work backwards, when it is θ = θ₀ + δθ the distance between M and m is (L/2)δθcosθ₀. I thought since δθ is small, I can approximate the distance between M and m as (L/2)δθ?

 

[paisiello2]

cosθ₀≅1 for small θ₀

 

[Yoonique]

cosθ₀≅1 for small θ₀

But is θ₀ small? Because the question did not state that though. They only state that δθ is small. θ₀ is the equilibrium point though..

 

[haruspex]

cosθ₀≅1 for small θ₀

If cosθ₀ figures in the given answer then clearly θ₀ is not considered small enough to use that approximation.

 

[Yoonique]

If cosθ₀ figures in the given answer then clearly θ₀ is not considered small enough to use that approximation.

Then why is there an extra cosθ₀ in the given answer? Do you know?

 

[haruspex]

Then why is there an extra cosθ₀ in the given answer? Do you know?

No. I haven't tried to follow all the algebraic developments in the thread. I attempted the problem from scratch and got the given answer except that in place of the cosθ₀ I get $\cos(\phi)$, where $r_0 = L \sin(\phi)$. θ₀cosθ₀ feels wrong. I don't see how cosθ₀ could come into it. So I suspect a typo in the given answer.

 

[Yoonique]

No. I haven't tried to follow all the algebraic developments in the thread. I attempted the problem from scratch and got the given answer except that in place of the cosθ₀ I get $\cos(\phi)$, where $r_0 = L \sin(\phi)$. θ₀cosθ₀ feels wrong. I don't see how cosθ₀ could come into it. So I suspect a typo in the given answer.

So the distance between M and m is r₀ - Lδθcos(θ₀/2)/2, where $\cos(\phi)$ = cos(θ₀/2) because it is the angle at the circumference.
But why is the distance m moved is Lδθcos(θ₀/2)/2 not Lδθ/2 since δθ is small.

 

[haruspex]

So the distance between M and m is r₀ - Lθ₀cos(θ₀/2), where $\cos(\phi)$ = cos(θ₀/2) because it is the angle at the circumference.

Whoa, what do you think θ₀ stands for? Isn't it the angle the torsion fibre had turned through to reach the r₀ position? Why would that relate to the angle that m and M subtend at the rod's centre in the equilibrium position?

 

[Yoonique]

Whoa, what do you think θ₀ stands for? Isn't it the angle the torsion fibre had turned through to reach the r₀ position? Why would that relate to the angle that m and M subtend at the rod's centre in the equilibrium position?

Oh I see what you meant. So when m moves an angle of δθ, the distance between M and m is r₀ - δθL/2 right? Which means LsinΦ - δθL/2? But this will give me the answer without the cosΦ. How do you get the cosΦ in the answer? If I worked backwards from the answer, why when m moves an angle of δθ, m moves cosΦ⋅δθL/2?

 

[TSny]

How do you get the cosΦ

The small mass can be thought of as moving along a tangent to the circle, whereas r cuts across the circle. So the motion of the mass and r are not in the same direction.

I agree with haruspex's answer.

 

[Yoonique]

The small mass can be thought of as moving along a tangent to the circle, whereas r cuts across the circle. So the motion of the mass and r are not in the same direction.

I agree with haruspex's answer.

Okay, so the length of the tangent is δθL/2. But I can't visualise why cosΦ⋅δθL/2 will be in the same direction as r.

 

[TSny]

Okay, so the length of the tangent is δθL/2. But I can't visualise why cosΦ⋅δθL/2 will be in the same direction as r.

See figure.

 

[Yoonique]

See figure.

Oh I was drawing the wrong triangle that's why I can't see it. Thanks I got it now. Thanks everyone for helping me in solving this question! So the given answer was wrong.