# Came up with a problem that I can't solve

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1. May 11, 2016

### guv

Imagine a hoop with mass M and radius R that will only roll without slipping on the floor. Place a point object with mass m on top of the hoop and then the system starts from at rest. Question: where does m leave M?

If one fixes the hoop or let the hoop slide, solutions can be found using high school mechanics. I ended up with 3 complicated equations with 4 unknowns. Can the given problem be solved using high school physics?

Thanks

2. May 11, 2016

### ProfuselyQuarky

Why not post what you've come up with? I'm trying to wrap my head around this problem.

3. May 11, 2016

### guv

The 3 equations I have are based on the following principles:

1. Conservation of energy (static friction between hoop and floor does no mechanical work):
mgh = 1/2 m v_{bg}^2 + 1/2 M v_{hg}^2 + 1/2 I \omega ^2
h = R(1-cos(\theta))

\theta is the angle where the point mass m (ball) leaves the hoop.

Note, one cannot use conservation of momentum in either horizontal and vertical direction and since the static friction and normal force acting on the hoop is not linearly related, there is not necessarily any simple relationship between horizontal and vertical impulse that allows the use of state function variables...

2. Escape condition Normal force = 0:
mg sin(\theta) = m \frac{v_{bh}^2}{R}

3. Relative motion ball relative to ground v_{bg}, hoop relative to ground v_{hg}, and finally ball relative to hoop form a triangle which allows the use of law of cosine.
\vec v_{bh} = \vec v_{bg} - \vec v_{hg}

Looking at the set of equations, one has the following unknowns:
\theta (where the ball leaves the hoop), 3 speeds v_{bg}, v_{bh}, and v_{gh}

4. May 11, 2016

### wrobel

This system has two degrees of freedom: $\psi$ is the angle of hoop's rotation and the angle $\phi$ which determines the place of point m till it does not slide out from the hoop. I expect that the Lagrangian of this system $L$ does not depend on $\psi$ and thus in addition to the law of total energy conservation you have the law of conservation of generalaized impuls $p=\frac{\partial L}{\partial \dot\psi}$ If you could obtain this by means of the high-school methods it would be useful to solve the problem

Last edited: May 11, 2016
5. May 12, 2016

### wrobel

Thanks a lot, very beautiful problem. We (me and students) solved it today on our classes

6. May 12, 2016

### guv

wrobel, do you mind posting the solution you used in your class? Thanks!

7. May 12, 2016

### guv

huh, found the 4th equation through the horizontal change of momentum, and the angular impulse-momentum relationship:

$\int f_s R dt = \Delta L = I \omega$
$\int f_s dt = \Delta p = m v_{bg,x} - M v_{hg}$

8. May 12, 2016

### wrobel

here it is

#### Attached Files:

• ###### hoopoint.pdf
File size:
33 KB
Views:
384
9. May 12, 2016

### guv

wrobel, thanks! I work with high school students interested in physics olympiad. I don't have the luxury to teach them lagrangians and hamiltonians. I do use the ideas and principles but I can't really use the detailed calculations.

what type of students are you working with? Do you think high school students should know lagrangians to have a shot in national/international physics olympiad?

10. May 12, 2016

### wrobel

I gave this problem to students which study the baccalaureate degree program in mathematics. I believe that this problem is suitable for theoretical mechanics course in university not in high school. I think this problem is too complicated for high school it is too complicated even for olympiad in high school

11. May 12, 2016

### wrobel

I have found a misprint in the pdf file above. It is written $\boldsymbol{OA}=(x+R\sin\phi)\boldsymbol e_x+R\cos\phi\boldsymbol e_y.$
Must be $\boldsymbol{OA}=(x+R\sin\phi)\boldsymbol e_x+(R+R\cos\phi)\boldsymbol e_y.$
Fortunately this does not change other formulas.

12. May 17, 2016

### Neandethal00

If the hoop doesn't have linear velocity, the angle will be the same as m falling from an inclined plane, related only to μs, coeff of static friction between m and M. If the hoop gets a velocity, the result is a quadratic equation in Cos or Sin of the angle. Quite solvable.

13. May 17, 2016

### wrobel

The answer is as follows. The angle of leaving $\phi\in(0,\pi/2)$ is found from the equation
$$3\cos\phi-\frac{m}{m+2M}(\cos\phi)^3=2.$$

The detailed proof here

#### Attached Files:

• ###### completsolution.pdf
File size:
31.4 KB
Views:
146
14. May 17, 2016

### golfrmyx

Simply use your mind and imagine an answer first. There are variables to the answer. the weight of m compared to the weight of M makes a difference in the speed of the increase of speed of rotation and thus the point of separation. That is why Cub Scout derby cars must be limited to the weight and it is very important to be as close to the maximum as possible. The heavier the rolling object is the faster it accelerates also depending on it's density of mass.

15. May 17, 2016

### golfrmyx

A gold marble will roll faster down a 10 degree slope than a glass marble of the same diameter.

Also consider, if mass m is placed exactly on the top of the arc of hoop M, it will not start rolling. Then you have to factor in the initiating force of push off.

16. May 17, 2016

### Staff: Mentor

Not if it has the same coefficient of friction. Both force from gravity and inertia are proportional to mass, the mass cancels in the equations.
Take the limit for the push going to zero.

17. May 17, 2016

### golfrmyx

Friction. Surrounding atmosphere and surface rolled on. Go get your marbles and give them a try. A steel one against a glass one and perhaps a plastic one will give you a close enough answer.

18. May 17, 2016

### Jenab2

I did the problem numerically.

Assuming:
M = 1 kilogram
R = 50 centimeters
m = 10 grams
g = 9.806 m sec⁻²

Edited. I did the problem wrong. I assumed that the hoop was spinning about an axis and that its center of mass did not gather any translational momentum. I'll try again.

Last edited: May 17, 2016
19. May 18, 2016

### wrobel

Beautiful! ok. Just use your mind and bring your version of the solution to initial problem with formulas and detailed comments. Everyone can shoot the breeze

20. May 18, 2016

### Jenab2

A hoop of mass M and radius R, initially at rest, has very, very near its top a point mass, m.

The point mass will fly off the hoop when the magnitude of the centrifugal acceleration (positive) equals that of the sum of the forces pushing the point mass toward the hoop's center (negative).

The centrifugal acceleration = ω²R
The radial component of acceleration of gravity = −g cos θ
The radial component of the hoop's translational acceleration = −RΩ sin θ

ω²R − g cos θ − RΩ sin θ = 0

where
ω is the angular speed of the hoop
g is the acceleration of gravity
Ω is the angular acceleration of the hoop
θ is the angle by which the hoop has turned since the initial time

The moment of inertia for a hoop of mass M and radius R about its axis of symmetry,

I = MR²

The angular acceleration, Ω, is equal to the torque, τ, divided by the moment of inertia.

Ω = τ/I

The torque is the component of the gravitational force on the mass m that is perpendicular to the vector from the hoop's center to the mass m, multiplied by the moment arm length, R.

τ = mgR sin θ

So the angular acceleration, as a function of θ, is

Ω = (m/M) (g/R) sin θ

We are looking for the θ and ω that satisfy
ω²R/g − [cos θ + (m/M) sin²θ] = 0

Let's consolidate the constants.

k = (m/M) (g/R)
Ω = k sin θ

Assume a small, but non-zero, initial angular displacement, θ₀ > 0.
Assume an initial angular speed of zero, ω₀ = 0.
Assume a very small time step, Δt.

Ω₀ = k sin θ₀
ω₁ = ω₀ + Ω₀ Δt
θ₁ = θ₀ + ½ (ω₀+ω₁) Δt

Ω₁ = k sin θ₁
ω₂ = ω₁ + Ω₁ Δt
θ₂ = θ₁ + ½ (ω₁+ω₂) Δt

Ω₂ = k sin θ₂
ω₃ = ω₂ + Ω₂ Δt
θ₃ = θ₂ + ½ (ω₂+ω₃) Δt

And so on, until

ωᵢ²R/g − [cos θᵢ + (m/M) sin²θᵢ] = 0

Last edited: May 18, 2016