# Rolling motion of plank and cylinders

1. Oct 10, 2005

### Kenny Lee

This will be a long thread... hope you'll take the time to read, cause I really need help.

A plank with a mass M = 6.00 kg rides on top of two identical solid cylindrical rollers that have R = 5.00 cm and m = 2.00 kg. The plank is pulled by a constant horizontal force F of magnitude 6.00 N applied to the end of the plank and perpendicular to the axes of the cylinders (which are parallel). The cylinders roll without slipping on a flat surface. There is also no slipping between the cylinders and the plank. Find the acceleration of both the plank and rollers.

Answer is 0.8 ms-2 for the plank and 0.4 ms-2 for the cylinders.

I can solve the problem cause I asked some other guys... but I don't really understand the method. It goes something like this:

Draw a freebody diagram of the roller, such that there are two frictional forces. So that means there is a resultant torque, which we can equate to N's second law in angular form. --- 1 eq, and two variables.
Note: supposing the system is accelerating to the right; the direction of the friction at the rollers both point towards the RIGHT. That's one point where I'm stuck. My rationale for this is as follows: relative to the plank, motion of the roller is towards the left, and since friction acts in direction opposite to motion, friction at the top of the wheel therefore points towards the right. Now, this frictional force produces an ACW torque, so at the bottom of the wheel, there is another frictional force acting towards the right. That makes sense? I dunno. I tend to make these things up just so I'm happy with everything. Need clarification.

And then, the second equation shows the relationship between the roller's and plank's acceleration. I understand this part, but I'm just gonna put it on, so people can work through it. The acceleration of plank is double that of the rollers- ---- 2nd equation - one new variable.

Then we have the planks moving relative to the cylinders, with 2F acting against it... F is friction. SO that's another equation with 2 variables... no new variables.

Then finally (this is where I get stuck)... we apply N's second law (linear form) to the rollers, with the friction at the top and bottom, equating to ma_rollers. I'm wondering why we can use this equation; would that not imply that the rollers was sorta slipping along the ground, and not rolling. I dunno if I'm making sense. But it doesn't matter really. I just need someone to help give some sort of rationale to the use of this equation. Any help would be appreciated.

ANyway, so we have four equations, with four variables. Solve simultaneously to get the result.

2. Oct 10, 2005

### arildno

This will be a long answer, I hope you'll take the time to read it, cause I really think it will help..

We first set up ALL variables and conditions for the rollers:
1. Symmetry
There is no reason for the two rollers to be treated any differently from each other.
2. Newton's 2.law for the center of mass of the cylinder:
$$F_{ground}\vec{i}+F_{top}\vec{i}=ma_{roller}\vec{i} (A)$$
Here, $F_{ground},F_{top}$ are (scalarized) forces acting upon the roller at the ground and top point, respectively, $a_{roller}$ is the acceleration of the roller's center of mass, whereas $\vec{i}$ is the unit horizontal vector.
Note:
I have NOT assumed whether $F_{ground},F_{top}$ are positive or negative here!!

3. The moment of momentum equation with respect to the center of mass:
With the center of mass as origin, the top point has position vector $r\vec{k}$, the ground point $-r\vec{k}$.
In addition, we let the angular velocity vector be written as $\omega\vec{j}$
Again, no assumption on the sign of $\omega$ has been made!
Thus, we get, in the $\vec{j}$ direction:
$$rF_{top}-rF_{ground}=I\frac{d\omega}{dt} (B)$$
where I is the moment of inertia.

3. Kinematic condition at the top point:
Here, due to the no-slip condition, if we call the velocity of the plank $v_{plank}\vec{i}[/tex], we must have that the top point of the roller has the same velocity: $$v_{roller}+r\omega=v_{plank} (C)$$ Here, I have removed the common [itex]\vec{i}$ "factor"
4. Kinematic condition on the ground:
Due to the no-slip condition, we must have:
$$v_{roller}-r\omega=0 (D)$$

5. Derivations:
a)By combining (C) and (D), we gain the relations:
$$v_{roller}=\frac{v_{plank}}{2}, \omega=\frac{v_{plank}}{2r}(E)$$
Similar relations hold, of cours between the accelerations.

b) Dividing through with r in (B), and adding it to (A), and utilizing the relations from (E), we get:
$$2F_{top}=\frac{m}{2}(1+\frac{I}{mr^{2}})a_{plank}$$
Or,
$$F_{top}=\frac{3}{8}ma_{plank} (F)$$
where I have utilized that $I=\frac{1}{2}mr^{2}$

This concludes the analysis on the roller level.

But now, we set up Newton's 2.law for the plank:
$$F-2F_{top}=Ma_{plank}(G)$$
where F is the applied force.

That is, using equation (F), we get:
$$a_{plank}=\frac{F}{M+\frac{3m}{4}}=\frac{6}{6+\frac{3}{2}}=\frac{12}{15}=0.8$$

Last edited: Oct 10, 2005
3. Oct 10, 2005

### mukundpa

The plank is pulled to the right then friction on the plank will be in (opposit) left direction. according to the N's third law the reactionary friction on the top of the roller is towards right.
Now with this force F (on each cylinder) find torque and angular acceleration about instatenious axis of rotation, from the contect point at bottum. This gives acceleration of center of mass of the cylinder equal to 4F/3m which is greater then F/m hence the friction at bottum must be F/3 in forward direction.

The friction on the top and the bottom of a cylinder are not equal. This pair of force constitute a couple (torque) and a force. With the net force you are finding the acceleration of center of mass of the cylinder(translational) and with the net torque you are finding the angular acceleration of the cylinder about it axis(passing through center of mass)(rotational). The combined motion is given by only rotation about the bottom( instantaneous axis of rotation)

4. Oct 10, 2005

### Kenny Lee

=) whoa. Thanks a heap guys. a whole week I've been stuck on the problem; now, its settled. phew. I really, really appreaciate it. Thanks.

Last edited: Oct 10, 2005
5. Oct 10, 2005

### mukundpa

As the motion of both cylinders is identical. there is a short cut to solve this problem.

Take torque about the surface (Contect point) = F_ext * 2R
moment of inertia(at that instant) about contect point
for cylinders = 2*(3/2)mR^2
for plank = M(2R)^2
Total I = (3m + 4M)R^2
angular acceleration about point of contect alfa = torque/I
acceleration of cy. = alfa*R = 2F_ext/(3m + 4M) (R eliminated)
of plank = alfa*2R = 4F_ext/(3m + 4M) :yuck:

6. Oct 19, 2009

### seewhy

Hi,

Sorry to dig this thread up, but I've encountered the exact some problem and I'm trying to understand arildno's solution.

In equation (B), he mentions that
$$rF_{top}-rF_{ground}=I\frac{d\omega}{dt} (B)$$

However, by drawing the free body diagram for the roller, shouldn't the equation instead be

$$rF_{top}+rF_{ground}=I\frac{d\omega}{dt} (B)$$?

7. Nov 23, 2010

### flyingpig

bump I need help on this too lol

8. Nov 27, 2010

### issacnewton

To understand this problem, you need a complete understanding of the pure rolling motion.
I found excellent material on the net. Read these articles at cnx.org

Code (Text):
http://cnx.org/content/m14384/latest
http://cnx.org/content/m14311/latest/
http://cnx.org/content/m14371/latest/
http://cnx.org/content/m14385/latest/
[PLAIN]http://cnx.org/content/m14391/latest/
[/PLAIN] [Broken]

All of these articles are written by Sunil Kumar Singh. Try to read articles on other areas of physics by him. They are all excellent.

Once you read above articles, you will get A VERY GOOD understanding of the pure rolling motion.

Now coming to the problem at hand,lets say that the plank is moving right. Now lets focus our attention on the right roller , at the point of contact between the plank and the roller.
If the roller doesn't move, then the plank will start sliding to the right. Now if the roller doesn't move then the roller is sliding to the left RELATIVE TO THE plank. Kinetic friction exerted by the plank on the roller will oppose this and the direction of the kinetic friction will be the the right. Now this process happens for a fraction of a second. As a result, the roller starts rolling. So once roller starts rolling, kinetic friction drops to zero,since there is no longer a relative motion between the roller and the plank. (Remember that in pure rolling motion, the point of contact between the rolling body and the ground is always at rest relative to the ground). But there is a force F applied to the right on the plank. So to prevent the sliding between the roller and the plank, there will exist a friction force on the roller to the right. Now this is no longer a kinetic friction but the static friction since the point of contact between the roller and the plank is at rest relative to the plank.
Ok we got the direction of the static friction on the roller at the top, which is rightwards. Now lets think what happens between the roller and the ground. Now the force which at the top of the roller is at right. This force is not passing through the center of mass, so it will cause a torque around the axis passing through the COM. So this torque will cause the roller to rotate clockwise. If the roller rotates in clockwise direction, then there will be sliding at the point of contact between the roller and the ground. So ground will exert a force in the opposite direction (forward direction) to oppose the relative motion between the roller and the ground. This force exerted by the ground is the static frictional force ,since , during the rolling, the point of contact between the roller and the ground is at rest relative to the ground.

So we established two static frictional forces which act on the roller. One at the top and one at the bottom. And they are both directed rightward. These static frictional forces maintain the rolling motion by preventing the sliding. The whole role of the static frictional forces here is to prevent the sliding (or in other language, relative motion between the surfaces).

Now lets say that the force at the top is f1 and the one at the bottom is f2. In vector form,

$$\vec{f_1}=f_1 \hat{i}\quad \mbox{and}\quad \vec{f_2}=f_2 \hat{i}$$

now set up the equation of motion, for the roller. Here I am concentrating on the right roller only since the equations for the left roller will be identical.

$$\vec{f_1}+\vec{f_2}=m\vec{a_{cm}}$$

where m is the mass of the roller and acm is the acceleration of the center of mass. now equation for the rotational motion.

$$\vec{\tau_{net}}=(f_2 R)(\hat{k})+(f_1 R)(-\hat{k}) =I\vec{\alpha}$$

now for the rolling motion, the the velocity of the point of contact between the roller and the ground relative to the roller , has the same magnitude but opposite direction as the velocity of the center of mass of the roller relative to the ground. Similar relationship holds for the accelerations. So we can write the relationship.

$$\vec{a_{cm}}=\vec{r}\times \vec{\alpha}$$

where, $\vec{r}$ is the position vector of the point of contact between the roller and the ground, with respect to the center of mass, which is the origin of our coordinate system.

$$\vec{r}=R(-\hat{j})$$

so,

$$\vec{a_{cm}}=R(-\hat{j})\times \vec{\alpha}$$

but

$$\vec{\alpha}=\frac{R(f_2-f_1)}{I}\, \hat{k}$$

so we have

$$\vec{a_{cm}}=R(-\hat{j})\times \frac{R(f_2-f_1)}{I}\, \hat{k}$$

$$\vec{a_{cm}}=\frac{R^2 (f_1-f_2)}{I}\, \hat{i}$$

but for the roller, which is in a cylindrical form, moment of inertia about the axis passing through the COM is

$$I=\frac{mR^2}{2}$$

so plugging this we get

$$\vec{a_{cm}}=\frac{2(f_1-f_2)}{m}\, \hat{i}$$

now we use this in our translational equation of motion given above.

$$\vec{f_1}+\vec{f_2}=m\vec{a_{cm}}$$

$$\vec{f_1}+\vec{f_2}=2(f_1-f_2)(\hat{i})$$

which can be simplified to

$$\vec{f_1}=3\vec{f_2}$$

now let a be the acceleration of the plank. Since the point of contact between the roller and the plank is moving twice as fast as the center of mass, relative to the ground, we have that

$$\vec{a}=2\vec{a_{cm}}=\frac{4(f_1-f_2)}{m}(\hat{i})$$

using the relation between the f1 and f2, which we got above, we can write this as

$$\vec{a}=\frac{8}{3m}\vec{f_1}$$

we had all our attention on the roller on the right. Similar equations will hold for the roller
on the left. If f3 and f4 are the friction forces at the top and the bottom ,for the left roller, then we have that( because both rollers are the same in mass and
I)

$$\vec{f_1}=\vec{f_3}\quad \mbox{and}\quad \vec{f_2}=\vec{f_4}$$

Now lets set up the equation of motion for the plank. f1 and f3
are the friction forces exerted by the plank on the rollers. By Newton's third law, the planks
will exert the forces f1 and f3 on the plank in the opposite direction.
so the equation of motion for the plank is

$$M\vec{a}=\vec{F}-\vec{f_1}-\vec{f_3}$$

now we use relations between a and f1 , f3

$$M\vec{a}=\vec{F}-\frac{3m}{8}\vec{a}-\frac{3m}{8}\vec{a}$$

so we get

$$\vec{a}=\frac{\vec{F}}{(M+\frac{6m}{8})}$$

and

$$\vec{f_1}=\vec{f_3}=\frac{\vec{F}}{(\frac{8M}{3m}+2)}$$

and

$$\vec{f_2}=\vec{f_4}=\frac{\vec{F}}{(\frac{8M}{m}+6)}$$

that solves the whole problem. Its important in such problems in rotational dynamics to
keep track of the signs. So I used vector notation all the way. I don't like the use of the scalar
equations in many introductory books on physics. On the contrary, in books like
"Engineering Mechanics-Statics by R.C.Hibbeler" and "Engineering Mechanics-Dynamics by R.C.Hibbeler" thorough use of vector notation is maintained. Its understandable since
unlike physicists , the mechanics is the bread and butter of the engineers so they can't
afford to go slightly wrong . For beginners in mechanics, I STRONGLY
recommend a thorough use of vector notation.

I have one question about the Latex typesetting though. When I write $$\vec{a_{cm}}$$ the arrow for the vector doesn't appear exactly on the top of 'a'. Why is that ?
Anyway I can improve this ?

Thanks

Last edited by a moderator: May 5, 2017