Rolling motion on inclined plane: centre of mass acceleration, friction, torque

In summary, the conversation discusses the concept of rolling motion and how it is affected by friction on an inclined plane. The equations for rotational and translational acceleration are derived and the role of friction in reducing translational acceleration is explained. The conversation also touches on the relationship between rolling resistance and the total force exerted on an object.
  • #1
mr_sparxx
29
4
Hi,

When a circular section object (cylinder, sphere of radius r) rolls down an inclined plane (at angle theta) it experiences both linear and angular acceleration due to gravity. I have read multiple times that:

mg sin(theta) - friction = m a

where friction = Torque/r

From there on, everything is clear:

moment of inertia * angular acceleration = friction * radius of ball = Torque

a= angular acceleration * r

and finally:

a = (mg sin(theta)) / (m + moment of inertia * r^-2);

friction = moment of inertia * a * r^-2

But, going back to the first equation there is a question that has always profoundly bothered me:

mg sin(theta) - friction = m a

Why does friction appears here? Isn't it a force applied at distance r in the contact point with the surface and directed down, parallel to it?

I understand that friction is needed to avoid slipping so it should exist. Moreover, that the weight must be reduced by some force to represent the final rotational energy. I mean, if we had just

mg sin(theta) = m a

and presume rotation then we would have the same final kinetic energy as a pure traslation with no friction (slipping) but with an impossible extra rotational energy.

Summarizing, I recognize that:

At centre of mass:
mg sin(theta) - Unknown force = m a

At contact point:
Torque = moment of inertia * angular acceleration = friction * radius of ball

But how do we get Unknown force = friction?

Thanks in advance!
 
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  • #2
mr_sparxx said:
But, going back to the first equation there is a question that has always profoundly bothered me:

mg sin(theta) - friction = m a

Why does friction appears here? Isn't it a force applied at distance r in the contact point with the surface and directed down, parallel to it?
Friction is directed up the incline, opposite to gravity. Yes, it is applied at a distance r from the center. So?

I am unclear as to what's bothering you. Without friction, the object would slide without rolling. When friction acts it does two things: It exerts a torque and thus causes rolling; It exerts a force that reduces the translational acceleration.
 
  • #3
Doc Al said:
It exerts a force that reduces the translational acceleration.

What bothers me is that I don't understand why this force, applied at a distance r from the center, reduces translational acceleration as if it were applied directly at the center of mass.
(I feel I am missing something very fundamental : ) )

Thanks for your prompt response!
 
  • #4
Note also that rolling involves static fricton. It's also possible for an object to be rotating and sliding at the same time, and with sufficient incline angle versus friction, the object continues to slide and never transitions into rolling.
 
  • #5
mr_sparxx said:
What bothers me is that I don't understand why this force, applied at a distance r from the center, reduces translational acceleration as if it were applied directly at the center of mass.
(I feel I am missing something very fundamental : ) )
Yes, you are missing something fundamental. The translational acceleration of the center of mass of an object (given by Newton's 2nd law) depends only on the net force on the object. It does not depend on where the forces are applied on the object. (Of course rotational acceleration certainly depends on where the forces are applied.)
 
  • #6
Doc Al said:
Yes, you are missing something fundamental. The translational acceleration of the center of mass of an object (given by Newton's 2nd law) depends only on the net force on the object. It does not depend on where the forces are applied on the object. (Of course rotational acceleration certainly depends on where the forces are applied.)

Nice! That was what I was looking for... I think I will regard (rotational) dynamics from a completely different perspective now :shy:.

Thank you very much for your help!
 
  • #7
Is this also a way to look at it:?

Some of the pot. energy the object has let's say, at the top of the inclined plane, is turned into rotational kinetic energy because friction exists? So not all the energy goes into linear or translational kinetic energy. Therefore the acceleration at the center of mass cannot just be sin(theta)*g ??

sorry for butting in.
 
  • #8
pgardn said:
Is this also a way to look at it:?

Some of the pot. energy the object has let's say, at the top of the inclined plane, is turned into rotational kinetic energy because friction exists? So not all the energy goes into linear or translational kinetic energy. Therefore the acceleration at the center of mass cannot just be sin(theta)*g ??
Yes, perfectly correct.
 
  • #9
I posted this in another thread;

The previous two posters are very correct. The principle outlined can be expressed mathamatically without friction.

In the rolling case for a hard solid ball of uniform density;

m*g*h = rotational K.E + linear K.E

mgh = (1/2)*I*w^2 + (1/2)*m*v^2

I = moment of inertia
w = angular velocity of ball
v = linear speed
r = radius of ball

now a rolling ball of diameter r turning at frequency has a linear speed v of;
v = 2*pi*r*f = w*r

so w = v/r

Therefore:
mgh = (1/2)*I*w^2 + (1/2)*m*v^2

becomes:+ m*g*cos(theta)*b/r)*d
= (1/2)*((2/5)*m*r^2)*(v/r)^2 + (1/2)*m*v^2
g*h = (1/2)*((2/5)*r^2)*(v/r)^2 + (1/2)*v^2
= (7/10)*v^2

v= sqrt((10/7)*g*h

So if h = 1m, d = 1m
v = 3.72 m/s
 
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  • #10
bm0p700f said:
mgh = (1/2)*I*w^2 + (1/2)*m*v^2 + m*g*cos(θ)*(b/r)*d
Rolling resistance is related to the total force between surfaces, not just the normal force. If a car moving slowly enough to ignore aerodynamic drag is rolling down a hill at constant speed due to braking, or a high rolling resistance factor, then the total force at the tires is m g and not m g cos(θ).

Only in the case of a frictionless inclined plane, will the total force exerted by the plane be equal to the normal force = m g cos(θ), as the the plane exerts zero force in the direction of the plane.
 
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  • #11
I get what you are saying about the rolling resistance. What I don't get is the exact formulation of the total frictional force (not aerodynamic c drag) that the ball experiences. If I can do this for a ball it will work for a wheel or any other rolling object.

Given I approached this problem from the principle of conservation of energy work must be done against friction. Given this is not a frictionless incline then I would need to include the frictional forces up the hill. Thus would the work done against friction up the hill be = 2*pi*r*f*N
where N is the number of revolutions, r is the radius and f the static friction up the hill.
Could this be right or am barking up the wrong tree again. How would rolling resistance then be added in? Could it be a direct addition of m*g*cos(theta)*b/r? Theta is included as it on a hill and the normal force to the plane will be reduced by cos(theta). I feel I am missing something.

Thanks for pointing out my errors.
 
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  • #12
bm0p700f said:
Given I approached this problem from the principle of conservation of energy work must be done against friction. Given this is not a frictionless incline then I would need to include the frictional forces up the hill. Thus would the work done against friction up the hill be = 2*pi*r*f*N
where N is the number of revolutions, r is the radius and f the static friction up the hill.
Could this be right or am barking up the wrong tree again. How would rolling resistance then be added in? Could it be a direct addition of m*g*cos(theta)*b/r? Theta is included as it on a hill and the normal force to the plane will be reduced by cos(theta). I feel I am missing something.
If you ignore the complication of rolling friction and only worry about static friction, then there's no work done against friction and mechanical energy is conserved.
 
  • #13
bm0p700f said:
Could it be a direct addition of m*g*cos(θ)*b/r?
No, because the rolling resistance is some function of the total force exerted by the plane onto the object, and this force is a function of the objects acceleration. If the object isn't accelerating, then the force from the plane equally opposes the force from gravity, so the force exerted by the plane onto the object = m g (in a vertical direction). In this case the force in the direction of the plane is m g sin(θ) uphill, and the force perpendicular to the plane is m g cos(θ). The other limit is if the plane is frictionless, in which case the force parallel to the plane is zero, while the force perpendicular to plane remains at m g cos(θ), so the total force is m g cos(θ).

m g cos(θ) <= total_force <= m g

In your simplified case, the force from rolling resistance = crr total_force, where crr is the coefficient of rolling resistance.
 
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  • #14
So it variable. Thanks.
 
  • #15
It is variable, but you could just consider rolling resistance to be some constant force parallel to the inclined plane.

linear force = gravitational component - friction force related to anguar accleration - rolling resistance

define c = rolling resistant force / m
define fa = surface (friction) force related to angular acceleration
define fc = force related to rolling resistance = m c
define fg = gravity force in direction of plane = m g sin(θ)

m a = m g sin(θ) - m (Ia/Ih) a - m c
a (1 + Ia/Ih) = g sin(θ) - c
a = (Ih / (Ia + Ih)) (g sin(θ) - c)

fa = m (Ia/Ih) a
fa = m (Ia/Ih) (Ih / (Ia + Ih)) (g sin(θ) - c)
fa = m (Ia / (Ia + Ih)) (g sin(θ) - c)
 
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1. What is the relationship between the angle of the inclined plane and the acceleration of the object's center of mass?

The acceleration of an object's center of mass on an inclined plane is directly proportional to the sine of the angle of the inclined plane. This means that the steeper the incline, the greater the acceleration of the object's center of mass will be.

2. How does friction affect the rolling motion on an inclined plane?

Friction plays a crucial role in the rolling motion on an inclined plane. Without friction, the object would simply slide down the plane. However, friction acts in the opposite direction of motion and creates a torque that causes the object to roll down the inclined plane instead.

3. What is torque and how does it relate to the rolling motion on an inclined plane?

Torque is the rotational force applied to an object. In the context of rolling motion on an inclined plane, torque is created by the force of friction acting on the object. This torque causes the object to rotate and roll down the inclined plane.

4. Can the acceleration of an object's center of mass on an inclined plane ever be greater than the acceleration due to gravity?

No, the acceleration of an object's center of mass on an inclined plane can never be greater than the acceleration due to gravity. This is because the force of gravity is always acting in the downward direction, while the acceleration on an inclined plane is affected by both the force of gravity and the normal force of the incline.

5. How does the distribution of mass affect the rolling motion on an inclined plane?

The distribution of mass affects the rolling motion on an inclined plane by changing the object's moment of inertia. Objects with more mass concentrated towards the center of mass will have a smaller moment of inertia and will accelerate faster down the inclined plane. On the other hand, objects with more mass distributed towards the edges will have a larger moment of inertia and will accelerate slower down the inclined plane.

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