Acceleration of a Ball Rolling in a Funnel

In summary, the ball will oscillate up and down the funnel like a planet in an elliptical orbit due to the transient equalization of the orbit radius.
  • #1
QQB
10
1

Homework Statement


A ball of mass m is given an initial velocity ⃗v and entered into a funnel. Model the acceleration of the ball over a period of time.

Note: I don't have a full problem statement because this isn't really a homework question (but I figured it was homework-like enough to post here instead of the general physics subforum). Is this the wrong place to post it?

Homework Equations


Fg = mg
Fnet=Σ ⃗F=m ⃗a
Centripetal acceleration = v2/r

The Attempt at a Solution


If the initial velocity is too high, I know the inertia of the ball will cause it to expand its orbital radius... the ball will move away from the center of the funnel (up and outwards) until equilibrium is achieved.

At equilibrium, I know that:
1. The vertical component of the normal force (Fn) will balance the force of gravity (net vertical force = 0).
2. The horizontal component of Fn will be the force required to achieve stable centripetal acceleration at the current velocity and radius (m*(v^2)/r).

But how can I calculate how fast the ball accelerates away from the center of the funnel in this situation (when initial velocity is higher than equilibrium conditions)? Is the normal force constant?

Also what about how kinetic energy is increased by traveling into (down) the funnel?
 
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  • #2
QQB said:
until equilibrium is achieved.
It s unlikely that it would reach equilibrium (in the sense of reaching a constant height and speed) if work is conserved.
Treating it as a rolling ball will make it fiendishly complicated. Try starting with a frictionless particle. What equations can you write down?
 
  • #3
haruspex said:
It s unlikely that it would reach equilibrium (in the sense of reaching a constant height and speed) if work is conserved.
Treating it as a rolling ball will make it fiendishly complicated. Try starting with a frictionless particle. What equations can you write down?

Fg = mg
Σ ⃗Fy = -Fg + (Fn)*cos(θ) where θ is the angle of the funnel slope and y is the axis perpendicular to the ground
Σ ⃗Fx = -(Fn)*sin(θ) where x is the axis parallel to the ground
If Σ ⃗Fx ≠ -m*(v2/r) then I think the orbit radius will adjust until they are equal... but I don't know how to describe the motion of the ball/particle during this time. I am also not sure what Fn is.

I also know that:
KE = (½)*m*v2
PE = mgh
KE1 + PE1 = KE2 + PE2 so if the ball drops lower into the funnel, it will also gain a bit of kinetic energy... I'm not really sure how this affects the motion of the ball. Is the extra velocity just pointed towards the bottom of the funnel (meaning that the "orbiting/centripetal velocity" isn't affected)?

I don't think conservation of angular momentum (I*ω) is at play here since there is still impulse from gravity and the normal force...
 
  • #4
QQB said:
then I think the orbit radius will adjust until they are equal
They will be transiently equal, but I doubt they would stay that way. It would oscillate up and down the funnel, like a planet in an elliptical orbit.

At any given instant, we have a velocity (magnitude and direction), an acceleration component in the same direction, and one normal to that direction. That last is the centripetal acceleration. Consider a plane containing the velocity vector and the centripetal acceleration vector. This cuts the funnel in an ellipse. So, transiently, the path is along that ellipse. This allows us to find the instantaneous radius of curvature as a function of the centripetal acceleration vector.
Putting all that together, we have the three ΣF=ma equations for the three directions in space, the three acceleration component unknowns, and the velocity vector variable.
 
  • #5
haruspex said:
They will be transiently equal, but I doubt they would stay that way. It would oscillate up and down the funnel, like a planet in an elliptical orbit.

At any given instant, we have a velocity (magnitude and direction), an acceleration component in the same direction, and one normal to that direction. That last is the centripetal acceleration. Consider a plane containing the velocity vector and the centripetal acceleration vector. This cuts the funnel in an ellipse. So, transiently, the path is along that ellipse. This allows us to find the instantaneous radius of curvature as a function of the centripetal acceleration vector.
Putting all that together, we have the three ΣF=ma equations for the three directions in space, the three acceleration component unknowns, and the velocity vector variable.

Is this what you mean?

upload_2017-1-16_22-21-55.png


Where the X is the velocity into the page (or out, depending on which way the particle is going around the funnel) as well as the the acceleration component that's in the same direction as the velocity. Also, I think the velocity could be in any direction into the page, but it will always be parallel to the funnel surface.

If this is what you meant, I realize now that I'm even more confused than I thought, because I originally just assumed centripetal acceleration is parallel to the ground (horizontal). Now I can't think of why I assumed that.
 
  • #6
QQB said:
assumed centripetal acceleration is parallel to the ground
Well your diagram shows it that way, but indeed it need not be. You need to consider the tangent plane to its current motion (i.e. the one containing both its velocity and acceleration vectors). The centripetal acceleration is in that plane, perpendicular to the funnel surface.
 
  • #7
haruspex said:
Well your diagram shows it that way, but indeed it need not be. You need to consider the tangent plane to its current motion (i.e. the one containing both its velocity and acceleration vectors). The centripetal acceleration is in that plane, perpendicular to the funnel surface.

Hmm... so I decided to try writing down equations using this coordinate system:
upload_2017-1-17_20-26-11.png


(The x-axis is positive into the page.)

Σ ⃗Fz = (Fn) - (Fg)cos(θ) = m(v2/r)
Σ ⃗Fy = (Fg)sin(θ) + (y component of friction if there is any) = m*(ay)
Σ ⃗Fx = (x component of friction if there is any) = m*(ax)

Is that right? And then the radius r would be determined by the shape of the funnel surface.
And I think we wouldn't need to consider energy gains from gravitational potential since it's already accounted for by including the force of gravity.

I'm not sure how to solve these equations without using a computer simulation. I'm also not sure how to handle a case where there is friction... oh wait! I can just use the negative of the velocity unit vector right?
 
  • #8
haruspex said:
perpendicular to the funnel surface.
Sorry, that was misleading. I was right to say that the centripetal acceleration lies in the plane containing its velocity and acceleration vectors, but it will not in general be perpendicular to the surface. It will be perpendicular to the velocity vector.
QQB said:
Σ ⃗Fz = (Fn) - (Fg)cos(θ) = m(v2/r)
No, as noted above.
QQB said:
the radius r would be determined by the shape of the funnel surface.
Yes, but you have to be careful to consider the curvature in the plane of motion.
QQB said:
where there is friction... oh wait! I can just use the negative of the velocity unit vector right?
Yes, but you could only represent something like a cubic particle, i.e. one that cannot roll. Otherwise, as soon as there is friction, you have the moment of inertia of the rolling particle to worry about.
 
  • #9
haruspex said:
At any given instant, we have a velocity (magnitude and direction), an acceleration component in the same direction, and one normal to that direction. That last is the centripetal acceleration. Consider a plane containing the velocity vector and the centripetal acceleration vector. This cuts the funnel in an ellipse. So, transiently, the path is along that ellipse. This allows us to find the instantaneous radius of curvature as a function of the centripetal acceleration vector.
Putting all that together, we have the three ΣF=ma equations for the three directions in space, the three acceleration component unknowns, and the velocity vector variable.
haruspex said:
Sorry, that was misleading. I was right to say that the centripetal acceleration lies in the plane containing its velocity and acceleration vectors, but it will not in general be perpendicular to the surface. It will be perpendicular to the velocity vector.

How do I set up the equations without knowing the direction of the centripetal acceleration (except that it's some direction normal to the velocity)?

I tried writing it using parametric notation for a vector/point on a plane:
Σ ⃗Fz = (Fn) - (Fg)cos(θ) = m*(P + s*[z component of some vector N that's perp to velocity] + t*[z component of some vector B that's perp to velocity])
Σ ⃗Fy = (Fg)sin(θ) + (y component of friction if there is any) = m*(ay) + m*(P + s*[y component of vec N] + t*[y component of vec B])
Σ ⃗Fx = (x component of friction if there is any) = m*(ax) + m*(P + s*[x component of vec N] + t*[x component of vec B])
Magnitude of centripetal acceleration = v2/r = magnitude of (P + s*[vector N] + t*[vector B])

Where P is some point on the plane containing the centripetal acceleration, s is a parameter and t is a parameter.

But this seems super complicated now the unknowns are Fn, s, t, ay and ax?
Am I getting further and further away from the right track...
 
  • #10
:s
 

Related to Acceleration of a Ball Rolling in a Funnel

What is the acceleration of a ball rolling in a funnel?

The acceleration of a ball rolling in a funnel can be calculated using the equation a = (v^2 - u^2)/2h, where v is the final velocity, u is the initial velocity, and h is the height of the funnel. The acceleration will vary depending on the initial velocity and height of the funnel.

How does the shape of the funnel affect the acceleration of the ball?

The shape of the funnel can greatly affect the acceleration of the ball. A wider funnel will result in a smaller acceleration, while a narrower funnel will result in a larger acceleration. This is due to the change in height and velocity of the ball as it rolls down the funnel.

What role does gravity play in the acceleration of a ball rolling in a funnel?

Gravity is the force that causes the ball to accelerate as it rolls down the funnel. The acceleration due to gravity is typically constant and can be represented as g = 9.8 m/s^2. This means that the ball's acceleration will increase by 9.8 m/s^2 for every second it rolls down the funnel.

Is the acceleration of a ball rolling in a funnel affected by the mass of the ball?

The mass of the ball does not affect its acceleration as it rolls down the funnel. This is because the force of gravity acting on the ball is directly proportional to its mass, so the acceleration remains constant regardless of the ball's mass.

What factors can affect the acceleration of a ball rolling in a funnel?

The acceleration of a ball rolling in a funnel can be affected by various factors such as the shape and size of the funnel, the initial velocity of the ball, and the force of gravity. External factors such as air resistance and friction can also impact the acceleration of the ball.

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