Rolling down a sphere: slipping vs. separating

In summary, the conversation discusses a problem in Taylor: Classical Mechanics involving a puck sliding without friction down a sphere in a uniform gravitational field. The problem is to determine at which height the puck takes off from the sphere. This leads to a more general problem involving a fixed sphere, a gravitational field, an object with mass, moment of inertia, and circular cross section, and static friction between the sphere and the object. The question is to determine whether the object starts to slip on the sphere or takes off from it first, and at which angle this occurs. The conversation includes a discussion of the angle at which slipping sets in and a differential equation for the angular displacement of the object while rolling. The curvature of the ball and its effect on the
  • #1
jds17
7
1
There is a nice problem in Taylor: Classical Mechanics of a puck sliding without friction down a sphere in a uniform gravitational field (problem 4.8).

The question there was at which height the puck takes off from the sphere, which is not hard to solve using conservation of energy.

This problem inspired me to try to solve the following problem: Let there be given a fixed sphere of radius ##R##, a gravitational field of strength

##g##, an object of mass ##m##, moment of inertia ##I## and circular cross section of radius ##r## (e.g. a hollow or solid sphere, a cylinder, a disk or

a hoop) which is initially placed on the top of the sphere. We assume static friction between the sphere and the object, the coefficient of static

friction being ##\mu##. We nudge the object a little so that it starts to roll down the sphere.

The question is: Determine what happens first and at which angle:

the object starts to slip on the sphere (and energy starts to dissipate via heat, so there is not much left to say after that moment) or
the object takes off from the sphere

What I could find out so far ist the following:

The angle at which slipping sets in, assuming that the object reaches this angle before detaching from the sphere.
A differential equation the angle ##\theta## measuring the angular displacement from the top (see image) satisfies while the object is rolling.

I would love to get your comments on the correctness of my derivations and be very happy if you could find out more about this problem since obviously

the main question remains unanswered. (Numerical experiments are welcome if they shed a light on the problem, I have not yet tried this.) There exist

similar problems where an object rolls down an incline, but these are much simpler since the angle the surface of the incline makes with the horizon is

the same everywhere, whereas in the case of the sphere it changes from point to point.

So here is my analysis so far:

(A) The angle at which slipping sets in:

Consider a moment in time when the object has rolled around the sphere for a certain angle ##\theta## (see the image). At that time, let ##f## be the

magnitude of the force of static friction, and ##a \vec e## the acceleration of the center of mass of the rolling object, where ##\vec e## is the unit

vector of the momentary direction of motion.

We have ##m a = m g \sin(\theta) - f##. The only torque on the object comes from the static friction, so ##I \dot \omega = r f##. Since the body is

rolling, the distance traveled by a point on its circumference is the same as the displacement of its center, so ##a = r \dot \omega##.

Using these two relations, we get ##m a = m g \sin(\theta) - \frac{I \dot\omega}{r} = m g \sin(\theta) - \frac{I a}{r^2}##, ##(m + \frac{I}{r^2}) a = m

g \sin(\theta)##, ##a = \frac{g \sin(\theta)}{1+\frac{I}{mr^2}}##.

We can now calculate ##f## as follows: ##f = m g \sin(\theta) - m a = m g \sin(\theta) (1-\frac{1}{1+\frac{I}{mr^2}}) = m g \sin(\theta) (1-\frac{mr^2}

{mr^2+I}) = m g \sin(\theta) \frac{I}{mr^2+I}##.

Now, since by definition of ##\mu##, the force of static friction ##f## is constrained as follows: ##f \le \mu m g \cos(\theta)##, we conclude from the

expression we found for ##f## that ##\tan(\theta) \le \mu \frac{mr^2+I}{I}##, i.e. ##\theta \le \arctan(\mu(1+\frac{mr^2}{I})) =: \theta_{max}##.

For example, we get the following values for the angle ##\theta_{max}## at which slipping must set in:

hoop: ##I = m r^2, \theta_{max} = \arctan(2\mu)##
solid ball: ##I = \frac{2}{5} mr^2, \theta_{max} = \arctan(\frac{7}{2}\mu)##
spherical shell: ##I=\frac{2}{3} mr^2, \theta_{max} = \arctan(\frac{5}{2}\mu)##

(B) A differential equation for ##\theta##:

We consider a moment of time at which the object has been rolling on the sphere since the beginning of its motion. Choosing the gravitational potential

energy to be ##0## at ##y = 0##, the initial energy of the system is ##E_i = mgR## and the energy at the time when an angle ##\theta## has been reached

is

##E_{\theta} = \frac{1}{2}m v^2 + \frac{1}{2} I \omega^2 + mgR\cos(\theta) = \frac{1}{2}m v^2 + \frac{1}{2} I \frac{v^2}{r^2} + mgR\cos(\theta)##,

so using conservation of energy we get ##2mgR(1-\cos(\theta)) = (m+\frac{I}{r^2})v^2##.

Since ##v^2 = R^2 \dot\theta^2##, we get ##2mg(1-\cos(\theta)) = (m+\frac{I}{r^2})R\dot\theta^2##, so defining the constant ##c := \sqrt(\frac{2mgr^2}

{R(mr^2+I)})## we arrive at the differential equation

##\dot\theta = -c \sqrt(1-\cos(\theta)), \theta(0) = 0##

which holds until the object starts to slip or until it takes off from the sphere, whichever happens first. Feeding this differential equation to Maxima

yielded a very complicated solution from which I could not read anything off. Maybe you can shed a light on the behavior of the solution to this

nonlinear equation.

Snapshot.jpg
 
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  • #2
jds17 said:
the object starts to slip on the sphere (and energy starts to dissipate via heat, so there is not much left to say after that moment)
You can still calculate how it moves, but it will need a differential equation, while the situation before allows to get the motion as function of the angle directly.

Your condition for static friction should take the curvature of the ball into account - centrifugal or centripetal force depending on your coordinate system. This will reduce the force and therefore lead to earlier slipping.

The condition for slipping is a small but positive normal force, the condition for taking off is zero normal force. It will always start to slip before it takes off.
 
  • #3
Thank you for your reply!

mfb said:
Your condition for static friction should take the curvature of the ball into account - centrifugal or centripetal force depending on your coordinate system. This will reduce the force and therefore lead to earlier slipping.

Since the rolling object only touches the surface in one point, I don't see why the curvature should play a role for the magnitude of the static friction, which is modeled as being opposed to the motion and having maximum value ##\mu N##, where ##N## is the magnitude of the normal force of the surface on the rolling object. So I don't think the curvature influences when slipping happens.

mfb said:
The condition for slipping is a small but positive normal force, the condition for taking off is zero normal force. It will always start to slip before it takes off.

Yes, you are right, while taking off, for a very short period of time there will be slipping because the normal force decreases to zero. I did not think of this slipping, I must admit. But I would count this case as "separation before slipping". The question remains whether there will be a moment when the velocity is too low for the ball to detach, while high enough that slipping sets in.
 
  • #4
jds17 said:
Since the rolling object only touches the surface in one point, I don't see why the curvature should play a role for the magnitude of the static friction, which is modeled as being opposed to the motion and having maximum value μN, where N is the magnitude of the normal force of the surface on the rolling object.
This magnitude depends on the velocity. Imagine a setup where we constrain the motion to the surface, and let it move around with a very high velocity: it will "try" to move outwards everywhere, which corresponds to a negative M here.
See the condition for the slipping object to leave the surface - same idea. You have to take the velocity and curvature into account. The object is not accelerating in a straight line orthogonal to the contact point, its acceleration is more complicated.
 

FAQ: Rolling down a sphere: slipping vs. separating

1. What is the difference between "slipping" and "separating" when rolling down a sphere?

Slipping and separating refer to two different types of motion that can occur when a sphere is rolling down an incline. Slipping refers to the motion where the sphere is in contact with the incline and is also rotating. Separating, on the other hand, refers to the motion where the sphere loses contact with the incline and is only moving through its center of mass.

2. How can you determine if a sphere will slip or separate when rolling down an incline?

The motion of the sphere depends on the angle of the incline and the coefficient of friction between the sphere and the incline. If the angle of the incline is steep or the coefficient of friction is low, the sphere is more likely to slip. If the angle of the incline is shallow or the coefficient of friction is high, the sphere is more likely to separate.

3. What factors affect the amount of slipping or separating that occurs when a sphere rolls down an incline?

The amount of slipping and separating that occurs when a sphere rolls down an incline is affected by the angle of the incline, the coefficient of friction, the radius of the sphere, and the mass of the sphere. A steeper incline, lower coefficient of friction, larger radius, and greater mass will result in more slipping, while a shallower incline, higher coefficient of friction, smaller radius, and lower mass will result in more separating.

4. How does the surface of the incline affect the motion of a rolling sphere?

The surface of the incline can also affect the motion of a rolling sphere. A rougher surface will result in more friction and potentially less slipping, while a smoother surface will result in less friction and potentially more slipping. However, the angle of the incline and the coefficient of friction are still the primary factors that determine if the sphere will slip or separate.

5. Can a sphere both slip and separate when rolling down an incline?

Yes, it is possible for a sphere to both slip and separate when rolling down an incline. This can occur if the angle of the incline is between the critical angles for slipping and separating, or if the coefficient of friction is between the critical values for slipping and separating. In this case, the sphere may start by slipping and then transition to separating, or vice versa, during its descent.

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