Cylinder rolling on a moving board

[Jenny Physics]

Homework Statement

A rectangular board laying on the ground of mass $M$ is pulled with a force $F$ to the right, while a cylinder of mass $m$, radius $R$ rolls without slipping on the board. Assume there is no friction between the board and the floor.

Which way does the cylinder roll?
What is the acceleration $a$ of the board and $a_{c}$ of the cylinder with respect to the lab frame?

Homework Equations

Newton's second law, rolling without slipping constraint

The Attempt at a Solution

The equation of motion of the board is[/B]

$F=(M+m)a$

The equation of motion of the center of mass of the cylinder is

$ma_{c,c}=F-\mu mg$

($a_{c,c}$ is the acceleration of the cylinder in the board frame)

The equation of motion of rotation about the center of mass of the cylinder is

$\tau=\mu Rmg=I\ddot{\theta}$

The rolling without slipping condition is $\ddot{\theta}R=a_{c,c}$

Am I right so far?

 

[BvU]

Hi,

when you write $F=(M+m)a$ you assume the same acceleration of board and cylinder. I'd say that's not a given...

 

[Jenny Physics]

Hi,

when you write $F=(M+m)a$ you assume the same acceleration of board and cylinder. I'd say that's not a given...

So it should be

$F=Ma+ma_{c}$?

 

[BvU]

You can resolve the question mark: make a free-body diagram for the cylinder.

Is it wise to work in the board frame ?

 

[PeroK]

So it should be

$F=Ma+ma_{c}$?

Yes.

 

[Jenny Physics]

You can resolve the question mark: make a free-body diagram for the cylinder.

Is it wise to work in the board frame ?

You can resolve the question mark: make a free-body diagram for the cylinder.

Is it wise to work in the board frame ?

I believe when I use Newton's laws for the cylinder motion, it makes more sense to use the board frame. The relation between the cylinder acceleration in the board frame and in the lab frame should then be $a_{c,c}=a-a_{c}$

 

[Jenny Physics]

Yes.

And my other 2 equations for the cylinder are correct as well (sign wise)?

 

[PeroK]

I believe when I use Newton's laws for the cylinder motion, it makes more sense to use the board frame. The relation between the cylinder acceleration in the board frame and in the lab frame should then be $a_{c,c}=a-a_{c}$

In which reference frames do Newtons laws apply?

 

[Jenny Physics]

In which reference frames do Newtons laws apply?

Inertial frames (which the board frame is not). I can however use Newton's laws in non inertial frames using $F$ as a fictitious force.
That means that the equation for the translation of the cylinder should really be (in the board frame)

$ma_{c,c}=-F-\mu mg$ ($a_{c,c}$ is the acceleration of the cylinder's cm in the board frame)

 

[PeroK]

Inertial frames (which the board frame is not). I can however use Newton's laws in non inertial frames using $F$ as a fictitious force.
That means that the equation for the translation of the cylinder should really be (in the board frame)

$ma_{c,c}=-F-\mu mg$ ($a_{c,c}$ is the acceleration of the cylinder's cm in the board frame)

As @BvU suggested, I would work in the lab frame. The board is accelerating (but you don't know the acceleration). If you are working in the board frame, then you need to apply a fictitious force to the cylinder in the opposite direction. This force is not $F$. The fictitious force is proportional to the mass of the cylinder. It should be:

$f = -m_c a_b$

Where $a_b$ is the acceleration of the board.

But, it must be simpler to stay in the lab frame. You are asked for answers in the lab frame, in any case.

 

[Jenny Physics]

As @BvU suggested, I would work in the lab frame. The board is accelerating (but you don't know the acceleration). If you are working in the board frame, then you need to apply a fictitious force to the cylinder in the opposite direction. This force is not $F$. The fictitious force is proportional to the mass of the cylinder. It should be:

$f = -m_c a_b$

Where $a_b$ is the acceleration of the board.

But, it must be simpler to stay in the lab frame. You are asked for answers in the lab frame, in any case.

Then working always in the lab frame the 3 equations of motion would be

Board: $F=Ma_{b}+ma_{c}$
Cylinder: $-\mu mg=ma_{c}$ and $\mu R mg=I\ddot{\theta}=I\frac{(a_{b}-a_{c})}{R}$
with all accelerations relative to the lab frame?

 

[PeroK]

Then working always in the lab frame the 3 equations of motion would be

Board: $F=Ma_{b}+ma_{c}$
Cylinder: $-\mu mg=ma_{c}$ and $\mu R mg=I\ddot{\theta}=I\frac{(a_{b}-a_{c})}{R}$
with all accelerations relative to the lab frame?

Yes. As a hint, you have used $\mu mg$, but you might as well just use some unknown force here: $f$, say. The details of that force aren't important. That you get rolling without slipping simply means that $\mu$, hence $f$ is large enough in this case.

Can you finish it off now?

 

[haruspex]

Then working always in the lab frame the 3 equations of motion would be

Board: $F=Ma_{b}+ma_{c}$
Cylinder: $-\mu mg=ma_{c}$ and $\mu R mg=I\ddot{\theta}=I\frac{(a_{b}-a_{c})}{R}$
with all accelerations relative to the lab frame?

Not sure about the signs. What do you expect the signs of a_c and a_b-a_c to be?
Using μ here implies the cylinder is the cylinder is on the point of slipping. As @PeroK writes, better just to put f for the frictional force.

 

[Jenny Physics]

Not sure about the signs. What do you expect the signs of a_c and a_b-a_c to be?
Using μ here implies the cylinder is the cylinder is on the point of slipping. As @PeroK writes, better just to put f for the frictional force.

If I make it $f=ma_{c}$ does that mean that I should make $Rf=I\ddot{\theta}=I\frac{a_{b}-a_{c}}{R}$ or should it be $-Rf=I\ddot{\theta}=I\frac{a_{b}-a_{c}}{R}$ ?
Intuitively I expect the cylinder to roll counterclockwise if $F$ is to the right but that might be true only for certain values of $F$ or of the other parameters. So $a_{c}-a_{b}$ would be positive (counterclockwise) and $a_{c}$ would be positive. But then $a_{c}>a_{b}$ which probably doesn't make sense.

 

[haruspex]

If I make it $f=ma_{c}$ does that mean that I should make $Rf=I\ddot{\theta}=I\frac{a_{b}-a_{c}}{R}$ or should it be $-Rf=I\ddot{\theta}=I\frac{a_{b}-a_{c}}{R}$ ?

To get the signs right by rigorous argument, one needs to be quite careful in the definitions.
I shall take forces and accelerations as positive right and θ as positive anticlockwise.
From f = ma_c, I see you are taking f as the force on the cylinder rather than the force the cylinder exerts on the board.
The torque a positive f exerts on the cylinder is anticlockwise, so $Rf=I\ddot{\theta}$. (Note that this is valid even if f turns out negative; the point is that a positive force would lead to a positive angular acceleration.)
A positive angular acceleration for the rolling cylinder would contribute a negative linear acceleration: $a_c=a_b-R\ddot{\theta}$.

Or you can cheat and just consider the question I posed in post #13. You would expect the cylinder to accelerate in the same direction as the board, but not as fast: a_b>a_c>0. From this you can see that your equation in post #11 has the wrong sign.

 

[Jenny Physics]

To get the signs right by rigorous argument, one needs to be quite careful in the definitions.
I shall take forces and accelerations as positive right and θ as positive anticlockwise.
From f = ma_c, I see you are taking f as the force on the cylinder rather than the force the cylinder exerts on the board.
The torque a positive f exerts on the cylinder is anticlockwise, so $Rf=I\ddot{\theta}$. (Note that this is valid even if f turns out negative; the point is that a positive force would lead to a positive angular acceleration.)
A positive angular acceleration for the rolling cylinder would contribute a negative linear acceleration: $a_c=a_b-R\ddot{\theta}$.

Or you can cheat and just consider the question I posed in post #13. You would expect the cylinder to accelerate in the same direction as the board, but not as fast: a_b>a_c>0. From this you can see that your equation in post #11 has the wrong sign.

Solving the equations I get the following accelerations

$a_{b}=\frac{(mR^{2}+I)F}{I(M+m)+mMR^{2}}$

and

$a_{c}=\frac{IF}{I(M+m)+mMR^{2}}$

so indeed $a_{b}>a_{c}>0$ and it seems the cylinder will roll counterclockwise but move forward in the same direction of the board.

 

[PeroK]

Solving the equations I get the following accelerations

$a_{b}=\frac{(mR^{2}+I)F}{I(M+m)+mMR^{2}}$

and

$a_{c}=\frac{IF}{I(M+m)+mMR^{2}}$

so indeed $a_{b}>a_{c}>0$ and it seems the cylinder will roll counterclockwise but move forward in the same direction of the board.

Those look correct to me. Note that there is a key factor of $\frac{mR^2}{I}+1$. If you call that factor $\alpha$, say, then the equations simplify somewhat.

Also, that allows you to calculate $\alpha$ for various objects: cylinders, spheres etc. and plug that into your common equation for the accelerations.