- #1

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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

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- Thread starter guv
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- #1

- 65

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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

ProfuselyQuarky

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

- #3

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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

wrobel

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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

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- #5

wrobel

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

- #6

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

- #7

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##\int f_s R dt = \Delta L = I \omega ##

##\int f_s dt = \Delta p = m v_{bg,x} - M v_{hg} ##

- #8

wrobel

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- #9

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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

wrobel

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- #11

wrobel

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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.

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- #13

wrobel

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- #15

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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

mfb

Mentor

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

Take the limit for the push going to zero.Then you have to factor in the initiating force of push off.

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- #18

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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.

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.

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- #19

wrobel

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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

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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

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

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- #21

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I also got a very similar equation, a quadratic equation, but my equation

included not only g, R, and V (ω) but also μ

Is μ

- #22

wrobel

Science Advisor

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1) The particle m can slide on hoop without any friction

2) The particle m can not slip till it falls from the hoop

I solved the problem in the first statement. The second statement is much easier than the first one, it is of high school level indeed.

So you consider the problem in the second statement. Your assertion is wrong. You did not take into account that the center of hoop moves. In the statement #2 the particle m describes a cycloid, not the circle.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

- #23

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One of the forces pushing the point mass toward the center of the hoop arises from the fact that the center of the hoop moves. I did take that force into account, and not merely the force from the component of gravity in the direction of the center of the hoop.There are two critical statements of the problem:

1) The particle m can slide on hoop without any friction

2) The particle m can not slip till it falls from the hoop

I solved the problem in the first statement. The second statement is much easier than the first one, it is of high school level indeed.

So you consider the problem in the second statement. Your assertion is wrong. You did not take into account that the center of hoop moves. In the statement #2 the particle m describes a cycloid, not the circle.

- #24

wrobel

Science Advisor

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O, I seeω²R − g cos θ − RΩ sin θ = 0

I have another question. What about the torque produced by the force of friction between the floor and the hoop?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

- #25

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There is no answer to a problem with infinity of variables. The variables must all be stated and known. This problem states some of them but not all.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

However after sleeping on this problem and giving it a lot of thought, my opinion is that the object m, regardless of all variables, will depart contact when the radian of M at m reachers gravitational level.

- #26

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I did not reply or add the "Then you have to factor in the initiating force of push off." Do not know how that got there, copied from an above answer. The friction would not be the same as the weight of gold increases friction in contact with sloping surface. Gravitational pull on heavier gold will overcome atmospheric friction on the same size diameters surface areas. Both must be calculated for each material to calculate speed of each for length traveled. Push off force could be entered if used and would need to be the same for each which would cause the heavier gold to start off slower or could use the same push off speed, thus requiring calculating difference of push off force.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.

- #27

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He did not ask for the angle of leaving. He asked for the point of leaving. Either way, I am betting your answer is wrong.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

- #28

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Very good answer but you left two components out. One being the friction of the hoop M on the surface it is rolling on that I assume you consider also zero and the other being the friction holding the "point object m" to the outer surface of M which is important.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

Consider for comparison if the point object m is round instead and thus rolls as well. This becomes a slightly different equation but works very similar.

I still believe the answer to both situations is at gravitational level radian, regardless of weight of each M and m.

- #29

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He did not ask for the angle of leaving. He asked for the point of leaving. Either way, I am betting your answer is wrong.

The angle is equivalent to that position, see the picture in post 3.

- #30

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Huh? Point mass m moving toward center of hoop? Impossible. It is on a surface strong enough to keep it on the circumference of the hoop until it falls away from the hoop outer surface. The hoop is in motion toward the direction point mass m is effectively pulling it by pushing with gravity force downward.One of the forces pushing the point mass toward the center of the hoop arises from the fact that the center of the hoop moves. I did take that force into account, and not merely the force from the component of gravity in the direction of the center of the hoop.

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