Derive the Period of an Oscillating Sphere

[SpringPhysics]

Homework Statement

Derive the period of a sphere oscillating in a hemispherical concave surface at small angles.

Homework Equations

Net Torque = I x alpha
Delta E_m = 0

The Attempt at a Solution

The thing is, tried using the conservation of mechanical energy to get an expression and isolate for omega (7/10 M r^2 omega^2 = 0). Then I differentiated in order to get omega and alpha, and substituted alpha from the net torque equation (in this case, gravity = - Mg (R - r) sin theta) with the moment of inertia of the sphere. But then it gets really messy and I can't solve for omega without a theta. The bottom line is I don't think I quite understand whether circular motion plays a role here, and whether for the kinetic energy, the velocity in translational kinetic energy is (R-r) x omega (due to circular motion) or just r x omega, and whether the omega in rotational kinetic energy is v/(R-r) or just v/r.

Thanks

 

[tiny-tim]

Hi SpringPhysics!

(have an alpha: α and an omega: ω and a theta: θ and try using the X² tag just above the Reply box )
Yes, conservation of mechanical energy should give you θ'² + f(θ) = constant …

the question says "at small angles", so now replace cos or sin by polynomials, and integrate.

(btw, it isn't (R-r)sinθ, it's (R-r)(1-cosθ))

 

[SpringPhysics]

Oh, thanks.

So torque is completely irrelevant in this question? I attempted to solve using the conservation law, but I am unsure about the tangential velocity of the sphere - is it based on the radius of the bowl or just the sphere? Also, if I use the small-angle approximation, the gravity portion of the mechanical energy completely disappears. And why would we integrate? Shouldn't we differentiate in order to get an equation in terms of α and θ, in order to relate it to the SHM equation?

Thanks.

 

[tiny-tim]

Hi SpringPhysics!

So torque is completely irrelevant in this question?

It's irrelevant if you use conservation of energy, it's very relevant if you dont!

I attempted to solve using the conservation law, but I am unsure about the tangential velocity of the sphere - is it based on the radius of the bowl or just the sphere?

The speed of the point of contact is zero.

Also, if I use the small-angle approximation, the gravity portion of the mechanical energy completely disappears.

No it doesn't.

And why would we integrate? Shouldn't we differentiate in order to get an equation in terms of α and θ, in order to relate it to the SHM equation?

Yeah, ok, differentiate.

 

[SpringPhysics]

Sorry, I'm still very confused.

Suppose that I use the conservation of energy. Then if I set the zero of potential at the base of the bowl, the change in potential energy would be
mg[r - (R-r)(1 - cos \theta].
If I use the small angle approximation, then cos \theta becomes 1 and the change in potential energy becomes just mgr. Then the kinetic energy is the sum of translational energy and rotational energy
1/2 mv² + 1/2 I\omega².
Then the above becomes
1/2 m(r\omega)² + 1/2 (2/5 mr²)\omega²
Even without solving for omega, there is no R variable here!

Suppose I use torque. Then
\alpha = -mg(R-r)sin\theta / (2/5 mr²) = -5/2 g(R-r)/r².
When I integrate this from a small angle \theta to 0, I get
-5/2 g(R-r)/a² (cos \theta - 1).
If I use the small angle approximation, cos \theta = 1 and then the entire thing becomes zero.

What am I doing wrong?

 

[tiny-tim]

Hi SpringPhysics!

(what happened to that θ ω and α i gave you? )

… If I use the small angle approximation, then cos \theta becomes 1 …

No, cosθ becomes 1 - θ²/2.

Then the kinetic energy is the sum of translational energy and rotational energy
1/2 mv² + 1/2 I\omega².
Then the above becomes
1/2 m(r\omega)² …

No, v depends on R also.

 

[SpringPhysics]

Sorry, I couldn't make it work (more like, I don't know how to make it work).

According to my textbook though, cos theta = 1 It doesn't mention the second derivative approximation (is there a way to derive it?).

I'll try it out first, thanks.

 

[tiny-tim]

According to my textbook though, cos theta = 1 It doesn't mention the second derivative approximation (is there a way to derive it?).

Yes, cosθ = √(1 - sin²θ) ~ √(1 - θ²) ~ 1 - θ²/2.

Alternatively:
cosθ = 1 - θ²/2 + θ⁴/24 - … = ∑ (-1)ⁿθ²ⁿ/(2n)!
sinθ = θ - θ³/6 + θ⁵/120 - … = ∑ (-1)ⁿθ²ⁿ⁺¹/(2n+1)!

 

[SpringPhysics]

Thanks.

So I used the conservation of mechanical energy.

PE_i = mg[R-(R-r)cosθ]
PE_f = mgr

KE_i = 0
KE_f = 1/2 mv² + 1/2 I ω²

Then by equating the negative change in PE to the change in KE, I eventually end up with

g(R-r)(1-cosθ) = 1/2[(R-r)ω₁]² + 1/5 r² ω₂²

I know that somehow I have to relate the two terms for kinetic energy, but I do not know how.

EDIT: I finally got something better

1 - cos θ= 7/10 (R-r)/g ω²

Now do I differentiate? I have to somehow get

T = 2 pi [7/5 (R-r)/g]^1/2

 

[tiny-tim]

EDIT: I finally got something better

1 - cos θ= 7/10 (R-r)/g ω²

Now do I differentiate? I have to somehow get

T = 2 pi [7/5 (R-r)/g]^1/2

(just got up :zzz: …)

Now put 1 - cos θ ~ θ²/2.

hmm … that gives θ' proportional to θ, which isn't SHM …

ah, I think you're missing a constant, you've got ω= 0 when θ = 0, which can't be right.

 

[SpringPhysics]

Good morning

A constant? If I differentiate, I get $$\alpha$$ = 5g/7(R-r) sin $$\theta$$. Using the small angle approximation, I get $$\alpha$$ = 5g/7(R-r) $$\theta$$, which mirrors the SHM equation right?

How can there be a constant? (Although I agree that it can't be true that when $$\theta$$ is zero, $$\omega$$ is zero as well.)

Can someone please help me?

 

[tiny-tim]

… Using the small angle approximation, I get $$\alpha$$ = 5g/7(R-r) $$\theta$$, which mirrors the SHM equation right?

Good afternooooon.

No, you get α = a positive constant times θ, which is no good (it's exponential, ie positive feedback) …

you need α = a negative constant times θ for SHM (so it's negative feedback, or retarding).

I'm not sure, but I think you must have got your PE negative when you did KE + PE = constant.

(btw, it didn't occur to me before, but you can also use one of the standard trigonometric identities before approximating … 1 - cosθ = 2sin²(θ/2) )

(now I'm off to watch sean the sheep :tongue2:)​

 

[SpringPhysics]

I actually didn't start with KE + PE = constant, I started with the change in KE + change in PE = 0. So then the change in KE is just the negative of the change in PE.

EDIT:

It worked when I placed the zero of potential at the top of the concave surface (as opposed to at the bottom). But I don't know why it doesn't work if it's at the bottom.

 

[SpringPhysics]

Sorry for the double post, but I just found out that I used a relation incorrectly.

So I have that the displacement of the circumference of the ball is equal to $$\phi$$ times r, and also equal to R times $$\theta$$. So I differentiate and get a relation between the angular velocities, and isolate for the angular velocity for the ball. I substitute that into the PE-KE equation, and then I'm stuck. What do I do next?

 

[tiny-tim]

Hi SpringPhysics!

… I substitute that into the PE-KE equation, and then I'm stuck. What do I do next?

Sorry, you've lost me …

what is the PE-KE equation you got stuck with?

 

[comicnabster]

Hey, I have to derive the exact same equation for my lab too O_O

 

[SpringPhysics]

=D

I placed the zero of gravity at the top of the concave surface at the centre.
PE_i = -mg(R-r)cos_$$\theta$$
PE_f = -mg(R-r)
KE_i = 0
KE_f = 1/5 mv² + 1/2 I_$$\omega$$²

So I = mr²
and _$$\omega$$² = (R-r)/r _$$\omega$$
and v = (R-r)_$$\omega$$

When you substitute them into $$\Delta$$PE + $$\Delta$$KE = 0, you get g(1-cos_$$\theta$$) = 7/10 (R-r)_$$\omega$$²

When you differentiate with respect to time, you get gsin_$$\theta$$ = _$$\alpha$$ 7/5 (R-r)

Obviously, alpha is not a negative number, so what am I doing wrong?

 

[comicnabster]

I just used what was shown here:

http://www.egglescliffe.org.uk/physics/gravitation/balls/aproxg2.html

Enjoy :)

 

[SpringPhysics]

Thanks, but I would still like to know what went wrong with my solution.

 

[comicnabster]

You were trying to use trigonometry and torques, which overcomplicated matters and thus got you confused.

It's ok, I've had my fair share of experiences where I try to apply unneeded concepts to a problem and make things harder for myself :(

 

[SpringPhysics]

Torque o.0...

Nevertheless, I finally got the answer lol.

Thanks for your help guys