Direction of friction acting on a rolling cylinder

[Pushoam]

Homework Statement

A cylinder of mass M and radius R rolls without slipping on a plank
that is accelerated at rate A. Find the acceleration of the cylinder.

Homework Equations

The Attempt at a Solution

Physical force acting on the system : F_phy = N + mg + f_r


w.r.t. plank frame i.e non - inertial frame,

1) the cylinder has pure rolling motion.

2)Net force acting on the cylinder : F_(cy,n-in) = F_phy + F_pseudo
3) |a_(cy,n-in) | = |α_(cy,n-in)| R

What is the direction of friction ?


The cylinder is rolling clockwise. So, α and Γ are along - $\hat z$ direction.
Now, the only force providing torque is the friction force.
f_r Ξ f_r $\hat x$

Now, the torque about the center of mass is Γ = R(- $\hat y$) × f_r $\hat x$ = |Γ|(- $\hat z$)
Rf_r $\hat z$ = |Γ|(- $\hat z$)

f_r = -|f_r |
This implies that the friction is acting towards left.
But the bottom part of the cylinder has a tendency to move towards left and friction force will oppose this relative motion. So, the friction will act toward right.

I am getting two contradictory answers. Where is it wrong?

 

[Orodruin]

I suggest that, instead of looking at the problem in the original inertial frame, you consider the problem in the rest frame of the plank (which is an accelerated frame). You can then transform the result back to the original frame to find the acceleration $a$.

 

[Pushoam]

I suggest that, instead of looking at the problem in the original inertial frame, you consider the problem in the rest frame of the plank (which is an accelerated frame). You can then transform the result back to the original frame to find the acceleration aaa.

I have done it. Please see the OP.

 

[Orodruin]

So then it should be a simple matter of relating this to how the friction force acts on a cylinder rolling without slipping on an inclined plane. However, the fact remains that you do not need to do the force analysis.

 

[Pushoam]

So then it should be a simple matter of relating this to how the friction force acts on a cylinder rolling without slipping on an inclined plane.

The plane is not inclined.

However, the fact remains that you do not need to do the force analysis.

I think what you suggest is :

But the bottom part of the cylinder has a tendency to move towards left and friction force will oppose this relative motion. So, the friction will act toward right.

But, I want to know why does this force or torque analysis give different result (as I often do this mistake)?

 

[Orodruin]

The plane is not inclined.

It is in the accelerated frame. The inertial force is not orthogonal to the plane. The problem is exactly equivalent to a cylinder rolling on a plane in the x-direction with the gravitational field equal to $-g\hat y - A \hat x$.

You can even notice this effect if you walk in a train when it is accelerating or decelerating. Walking in the forward direction when the train accelerates is just like going uphill.

But, I want to know why does this force or torque analysis give different result (as I often do this mistake)?

$(-\hat y) \times \hat x = \hat x \times \hat y = \hat z \neq - \hat z$

 

[Pushoam]

I think I got the mistake.

It is in this statement:

The cylinder is rolling clockwise. So, α and Γ are along - $\hat z$ direction.

"The cylinder is rolling clockwise " implies that the angular velocity is along - $\hat z$ direction. It doesn't give any information about the direction of α and Γ.
So, I have to find direction of friction in the following way:

But the bottom part of the cylinder has a tendency to move towards left and friction force will oppose this relative motion. So, the friction will act toward right.

So, the friction is f_r = f_r $\hat x$, f_r >0 N
Now, the torque about the center of mass is :
Γ = R(-$\hat y$) × f_r $\hat x$ = Rf_r $\hat z$ = Iα $\hat z$ = ($\frac {MR^2} 2$) (a _(cy,n-in)) /R $\hat z$
Rf_r = ($\frac {MR^2} 2$) (a _(cy,n-in)) /R
f_r = ½ M|a _(cy,n-in) |
Note that as α is anticlockwise, a _(cy,n-in) = |a _(cy,n-in) | (-$\hat x$)

From F_phy = N + mg + f_r , we have
f_r = Ma_i, w.r.t. inertial frame

So, a_i = ½ |a _(cy,n-in) |

From F_(cy,n-in) = F_phy+ F_pseudo we have
|a _(cy,n-in) | (- $\hat x$) = a_i ($\hat x$) - A ($\hat x$)
-3 a_i = -A
a_i = ⅓ A
a _(cy,n-in) = -2(⅓ A)

Is this correct?

 

[Hiero]

Rf_r = ($\frac {MR^2} 2$) (a _(cy,n-in)) /R

You seem to claim the angular acceleration of the cylinder is the linear acceleration over R?

Think about if the linear acceleration of the cylinder was zero... wouldn't the cylinder still need to rotate in order to not slip?

 

[Hiero]

Oh, sorry, I did not realize you were working partially in the inertial frame and partially in the accelerated frame.

I agree with your result, a=A/3

 

[Pushoam]

You seem to claim the angular acceleration of the cylinder is the linear acceleration over R?

w.r.t. plank frame,
The cylinder is purely rolling.
In pure rolling motion,
ds = R dθ
v = $\frac {ds} {dt}$ = R $\frac{dθ} {dt}$ = Rω
a = $\frac {dv} {dt}$ = R $\frac{dω} {dt}$ = Rα