# Direction of friction acting on a rolling cylinder

Tags:
1. Jul 21, 2017

### Pushoam

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. The attempt at a solution
Physical force acting on the system : Fphy = N + mg + fr

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) = Fphy + Fpseudo
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.
fr Ξ fr $\hat x$

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

fr = -|fr |
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.

2. Jul 21, 2017

### Orodruin

Staff Emeritus
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 $a$.

3. Jul 21, 2017

### Pushoam

I have done it. Please see the OP.

4. Jul 21, 2017

### Orodruin

Staff Emeritus
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.

5. Jul 21, 2017

### Pushoam

The plane is not inclined.
I think what you suggest is :
But, I want to know why does this force or torque analysis give different result (as I often do this mistake)?

6. Jul 21, 2017

### Orodruin

Staff Emeritus
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.

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

7. Jul 21, 2017

### Pushoam

I think I got the mistake.

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

From Fphy = N + mg + fr , we have
fr = Mai, w.r.t. inertial frame

So, ai = ½ |a (cy,n-in) |

From
F(cy,n-in) = Fphy+ Fpseudo we have
|a (cy,n-in) | (- $\hat x$) = ai ($\hat x$) - A ($\hat x$)
-3 ai = -A
ai = ⅓ A
a (cy,n-in) = -2(⅓ A)

Is this correct?

8. Jul 21, 2017

### Hiero

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?

9. Jul 21, 2017

### 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

10. Jul 21, 2017

### Pushoam

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