Does it really work? [Answered - No]

[ABV]

For the same force?

 

For the same time?

 

Transferred to the rolling body, or to the ground, or both?

 

The rolling bodies have same mass, initial velocity and different momet of inertia.

 

same force

 

different time

 

a momentum from rolling bodies to the ground.

[J.C.MacSwell]

The rolling bodies have same mass, initial velocity and different momet of inertia.

 

same force

 

different time

 

a momentum from rolling bodies to the ground.

 

There will be the same net transfer of linear momentum, assuming they have the same final velocity as well.

[ABV]

There will be the same net transfer of linear momentum, assuming they have the same final velocity as well.

Without slippery?

Diagram is simple. The little human stopping a rolling body with force Ffr. This little human is transferring a mometum from rolling body to the ground.

 

Why I'm asking?

Because an extra moment of inertia of rolling body will gain extra momentum through static friction.

[D H]

You are failing to look at the whole picture, ABV. Let's look at the whole picture. The object is slowing down because of the external force applied by the human. However, the object rolls without slipping. Its angular velocity is also decreasing. There must be a torque acting on the object as well. This torque comes from the contact with the ground. This force is directed opposite to the force applied by the human (but is lesser in magnitude). Denoting direction in which the object is moving as the positive x axis, then

 

[math]\aligned

\vec v &= v\hat x && \quad\text{Object center of mass velocity} \\

\vec F_r &= -F_r\hat x && \quad\text{Force applied by the human} \\

\vec F_c &= -\,\frac I r \dot \omega \hat x = -\,\frac I {r^2} \dot v \hat x

&& \quad\text{Constraint force} \\

\vec F_t &= \vec F_r + \vec F_c

= -\left(F_r + \frac I {r^2} \dot v\right) \hat x

&&\quad\text{Total force on the object}

\endaligned[/math]

 

After a bit of work, the object has a constant acceleration equal to

 

[math]

\dot v = -\,\frac {F_r} {M+2I/r^2}

[/math]

 

The stopping time and distance given this constant (negative) acceleration are

 

[math]\aligned

t_{stop} &= -\,\frac {v_0}{\dot v}

= \frac {v_0\left(M+2I/r^2\right)} {F_r} \\

d_{stop} &= -\,\frac 1 2\, \frac {v{_0}^2}{\dot v}

= \frac 1 2 \,\frac {v{_0}^2\left(M+2I/r^2\right)} {F_r}

\endaligned[/math]

[J.C.MacSwell]

Without slippery?

Diagram is simple. The little human stopping a rolling body with force Ffr. This little human is transferring a momentum from rolling body to the ground.

 

Without slipping

 

Why I'm asking?

Because an extra moment of inertia of rolling body will gain extra momentum through static friction.

 

Why gain?

 

Why not lose?

 

The body gains or loses momentum from the surface from the static friction only as the surface loses it or gains it respectively.

Edited June 9, 2009 by J.C.MacSwell

[ABV]

You are failing to look at the whole picture, ABV. Let's look at the whole picture. The object is slowing down because of the external force applied by the human. However, the object rolls without slipping. Its angular velocity is also decreasing. There must be a torque acting on the object as well. This torque comes from the contact with the ground. This force is directed opposite to the force applied by the human (but is lesser in magnitude). Denoting direction in which the object is moving as the positive x axis, then

 

[math]\aligned

\vec v &= v\hat x && \quad\text{Object center of mass velocity} \\

\vec F_r &= -F_r\hat x && \quad\text{Force applied by the human} \\

\vec F_c &= -\,\frac I r \dot \omega \hat x = -\,\frac I {r^2} \dot v \hat x

&& \quad\text{Constraint force} \\

\vec F_t &= \vec F_r + \vec F_c

= -\left(F_r + \frac I {r^2} \dot v\right) \hat x

&&\quad\text{Total force on the object}

\endaligned[/math]

 

After a bit of work, the object has a constant acceleration equal to

 

[math]

\dot v = -\,\frac {F_r} {M+2I/r^2}

[/math]

 

The stopping time and distance given this constant (negative) acceleration are

 

[math]\aligned

t_{stop} &= -\,\frac {v_0}{\dot v}

= \frac {v_0\left(M+2I/r^2\right)} {F_r} \\

d_{stop} &= -\,\frac 1 2\, \frac {v{_0}^2}{\dot v}

= \frac 1 2 \,\frac {v{_0}^2\left(M+2I/r^2\right)} {F_r}

\endaligned[/math]

 

Thank you for kinematic equation.

But why [math]

\dot v = -\,\frac {F_r} {M+2I/r^2}

[/math]

 

after

[math]\vec F_t = \vec F_r + \vec F_c

= -\left(F_r + \frac I {r^2} \dot v\right) \hat x

[/math]

 

it would be

[math]

\dot v = -\,\frac {F_r} {M+I/r^2}

[/math]

[D H]

Thank you for kinematic equation.

But why [math]

\dot v = -\,\frac {F_r} {M+2I/r^2}

[/math]

 

Argghh! I am so stupid!

There must be a torque acting on the object as well. This torque comes from the contact with the ground. This force is directed opposite to the force applied by the human ...

The force that generates this constraint force must be in the same direction as the resistive force applied by the human. To see why, define coordinates as follows:

• [math]\hat{\boldsymbol x}[/math] is parallel to the plane on which the object is rolling and points in the direction in which the object is moving.
  
• [math]\hat{\boldsymbol y}[/math] is perpendicular to the surface on which the object rolling and points from the object / surface contact point to the object's center of mass.
  
• [math]\hat{\boldsymbol z}[/math] completes a right-hand system: [math]\hat{\boldsymbol z} = \hat{\boldsymbol x} \times \hat{\boldsymbol y}[/math].

 

The object's velocity is changing and the object is rolling without slipping. Some torque must exist on the object to keep to object rolling without slipping. The force applied by the human exerts zero torque on the object. From where does this torque arise? The only other possible source is friction at the contact point.

 

A force [math]\vec F[/math] applied at a point [math]\vec r[math] away from the center of mass results in a torque [math]\vec{\tau} = \vec r \times \vec F[/math]. With the above reference system, the position of the contact point is [math]\vec r = -r \hat{\boldsymbol y}[/math]. The constraint force is of the form [math]\vec F_c = -F_c \hat{\boldsymbol x} + F_n \hat y[/math]. The normal force F_n does not contribute to the torque. Here F_c can be any number. I used a negative sign because doing so will make F_c be positive. The torque is

 

[math]\vec{\tau} = \vec r \times \vec F_c = rF_c \hat{\boldsymbol y}\times \hat{\boldsymbol x} = -rF_c \hat{\boldsymbol z}[/math]

 

To keep the object rolling without slipping the torque must be equal to [math]\vec{\tau} = I\dot{\vec \omega} = \frac I r \dot v \hat{\boldsymbol z}[/math]. (Note that since [math]\dot v < 0[/math] the angular acceleration points in the -z direction.) Equating the two

 

[math]F_c = -\,\frac I {r^2} \dot v[/math]

 

Once again, since [math]\dot v < 0[/math], F_c is positive.

 

Continuing with the corrections to my previous post,

 

[math]\aligned

\vec F_r &= -F_r\hat x && \quad\text{Force applied by the human} \\

\vec F_c &= \frac I {r^2} \dot v \hat x

&& \quad\text{Constraint force} \\

\vec F_t &= \vec F_r + \vec F_c

= -\left(F_r - \frac I {r^2} \dot v\right) \hat x

&&\quad\text{Total force on the object}

\endaligned[/math]

 

The object's kinetic energy at any point in time is

 

[math]T = \frac 1 2 Mv^2 + \frac 1 2 I \omega^2[/math]

 

Since the object is rolling without slipping, [math]\omega = v/r[/math] and thus

 

[math]T = \frac 1 2 \left(M + \frac I{r^2}\right) v^2[/math]

 

The change in kinetic energy over a short period of time [math]\Delta t[/math] is

 

[math]\Delta T \approx \left(M + \frac I{r^2}\right) v \Delta v[/math]

 

The work done on the object is during this time period is

 

[math]\Delta W \approx \vec F_t \cdot \vec v \Delta t

= -\left(F_r - \frac I {r^2} \dot v\right) v \Delta t[/math]

 

Applying the work-energy theorem, taking the limit [math]\Delta t \to 0[/math], and solving for [math]\dot v[/math] yields

 

[math]\dot v = F_r/M[/math]

 

In other words, the acceleration is independent of the moment of inertia. The stopping time and stopping distance are the same for both objects. Both objects have exactly the same linear momentum. The linear momentum transfered to the surface is the same in both cases. This is the only result that makes sense. A result that violates conservation of momentum means one of two things: (1) The system boundaries do not enclose a closed system, or (2) A mistake was made.

 

There is a lesson to be learned here: It is easy to make mistakes, even on seemingly simple problems.

[J.C.MacSwell]

Argghh! I am so stupid!

 

The force that generates this constraint force must be in the same direction as the resistive force applied by the human. To see why, define coordinates as follows:

• [math]\hat{\boldsymbol x}[/math] is parallel to the plane on which the object is rolling and points in the direction in which the object is moving.
  
• [math]\hat{\boldsymbol y}[/math] is perpendicular to the surface on which the object rolling and points from the object / surface contact point to the object's center of mass.
  
• [math]\hat{\boldsymbol z}[/math] completes a right-hand system: [math]\hat{\boldsymbol z} = \hat{\boldsymbol x} \times \hat{\boldsymbol y}[/math].

 

The object's velocity is changing and the object is rolling without slipping. Some torque must exist on the object to keep to object rolling without slipping. The force applied by the human exerts zero torque on the object. From where does this torque arise? The only other possible source is friction at the contact point.

 

A force [math]\vec F[/math] applied at a point [math]\vec r[math] away from the center of mass results in a torque [math]\vec{\tau} = \vec r \times \vec F[/math]. With the above reference system, the position of the contact point is [math]\vec r = -r \hat{\boldsymbol y}[/math]. The constraint force is of the form [math]\vec F_c = -F_c \hat{\boldsymbol x} + F_n \hat y[/math]. The normal force F_n does not contribute to the torque. Here F_c can be any number. I used a negative sign because doing so will make F_c be positive. The torque is

 

[math]\vec{\tau} = \vec r \times \vec F_c = rF_c \hat{\boldsymbol y}\times \hat{\boldsymbol x} = -rF_c \hat{\boldsymbol z}[/math]

 

To keep the object rolling without slipping the torque must be equal to [math]\vec{\tau} = I\dot{\vec \omega} = \frac I r \dot v \hat{\boldsymbol z}[/math]. (Note that since [math]\dot v < 0[/math] the angular acceleration points in the -z direction.) Equating the two

 

[math]F_c = -\,\frac I {r^2} \dot v[/math]

 

Once again, since [math]\dot v < 0[/math], F_c is positive.

 

Continuing with the corrections to my previous post,

 

[math]\aligned

\vec F_r &= -F_r\hat x && \quad\text{Force applied by the human} \\

\vec F_c &= \frac I {r^2} \dot v \hat x

&& \quad\text{Constraint force} \\

\vec F_t &= \vec F_r + \vec F_c

= -\left(F_r - \frac I {r^2} \dot v\right) \hat x

&&\quad\text{Total force on the object}

\endaligned[/math]

 

The object's kinetic energy at any point in time is

 

[math]T = \frac 1 2 Mv^2 + \frac 1 2 I \omega^2[/math]

 

Since the object is rolling without slipping, [math]\omega = v/r[/math] and thus

 

[math]T = \frac 1 2 \left(M + \frac I{r^2}\right) v^2[/math]

 

The change in kinetic energy over a short period of time [math]\Delta t[/math] is

 

[math]\Delta T \approx \left(M + \frac I{r^2}\right) v \Delta v[/math]

 

The work done on the object is during this time period is

 

[math]\Delta W \approx \vec F_t \cdot \vec v \Delta t

= -\left(F_r - \frac I {r^2} \dot v\right) v \Delta t[/math]

 

Applying the work-energy theorem, taking the limit [math]\Delta t \to 0[/math], and solving for [math]\dot v[/math] yields

 

[math]\dot v = F_r/M[/math]

 

In other words, the acceleration is independent of the moment of inertia. The stopping time and stopping distance are the same for both objects. Both objects have exactly the same linear momentum. The linear momentum transfered to the surface is the same in both cases. This is the only result that makes sense. A result that violates conservation of momentum means one of two things: (1) The system boundaries do not enclose a closed system, or (2) A mistake was made.

 

There is a lesson to be learned here: It is easy to make mistakes, even on seemingly simple problems.

 

The human force is backwards trying to decelerate but puts no torque on the object (as you stated). It acts through the center of mass.

 

However:

 

The friction force of the ground is opposite that of the human's. It pushes it forwards or better stated constrains it not to slip backwards at the contact point. It applies a torque that counters the rotation, but does no work.

 

The acceleration is dependent on the moment of inertia. The stopping time and stopping distance are not the same for both objects.

 

The linear momentum transferred to the surface is, however, the same in both cases, it just takes longer in the higher moment of inertia case.

 

It may seem counterintuitive, but that is how it works.

 

If however the human applied more force, to make up for the force of friction in each case, then the stopping time and distance would be the same for each object.

Edited June 9, 2009 by J.C.MacSwell

[ABV]

THANK YOU SO MUCH to all.

This is very good answer for me.

 

But anyway I have another IDEA.

If a rolling body got linear momentum, we have to put same linear momentum to stop it.

This is correct.

But Let's look on this one. Let's break rolling body. What will happen?

http://knol.google.com/k/alex-belov/-/1xmqm1l0s4ys/9#

 

The conditions:

A System has static friction. No rolling friction. A gravity force works as well too.

Would this system break the law?

[J.C.MacSwell]

THANK YOU SO MUCH to all.

This is very good answer for me.

 

But anyway I have another IDEA.

If a rolling body got linear momentum, we have to put same linear momentum to stop it.

This is correct.

But Let's look on this one. Let's break rolling body. What will happen?

http://knol.google.com/k/alex-belov/-/1xmqm1l0s4ys/9#

 

The conditions:

A System has static friction. No rolling friction. A gravity force works as well too.

Would this system break the law?

 

Why would you even suggest that it might?

[ABV]

Idea is very simple.

 

If divide a ring to small parts of n elements array with mass m then each of them conduct linear and circular movement.

 

Each piece has constant angular velocity and variable linear velocity. Once per circle each element stop on surface. Value of linear velocity of this element is zero at this time. And if break chain then all of these elements except one is continue to move. If calculate net linear moments of these element this should be equal to this ring initial linear momentum. But one of these elements is stop already and net momentum will be for n-1 elements.

 

Is this net of linear momentums for n-1 elements array is equal to the ring initial linear momentum?

[J.C.MacSwell]

Idea is very simple.

 

If divide a ring to small parts of n elements array with mass m then each of them conduct linear and circular movement.

 

Each piece has constant angular velocity and variable linear velocity. Once per circle each element stop on surface. Value of linear velocity of this element is zero at this time. And if break chain then all of these elements except one is continue to move. If calculate net linear moments of these element this should be equal to this ring initial linear momentum. But one of these elements is stop already and net momentum will be for n-1 elements.

 

Is this net of linear momentums for n-1 elements array is equal to the ring initial linear momentum?

 

So why would the elements take the paths shown (or described) after the break?

[ABV]

In this case one element attached to the surface and mass is M(suface) + m(element). Array of elements has a mass equal to (n-1)*m. It’s change initial condition. The surface is still keeping same momentum and increase own mass. Array of elements should hold same momentum, but lost one piece with mass m.

A few elements will return back small value momentums to surface and increase surface mass.

 

Will other elements with own momentums compensate surface momentum with new mass value?

Edited June 14, 2009 by ABV