# Cylinder on a decelerating truck

1. May 30, 2015

### Karol

1. The problem statement, all variables and given/known data
A truck with a cylinder of mass M and moment of inertia $I=kMR^2$ on top has initial velocity v0 and decelerates with deceleration B. the cylinder is attached with a rope to a weight m. the coefficient of friction is μ.
The outer radius of the cylinder is R and the smaller is r. The cylinder moves without sliding and there isn't friction between the weight m and the truck.
What is the relation between the angular acceleration of the cylinder and the linear acceleration of the weight.
What are the forces and the moments that act on the weight and the cylinder.
What is the linear acceleration of the cylinder in the reference frame of the truck.
What is the condition on B so that the cylinder will move to the right relative to the truck.

2. Relevant equations
Torque and moment of inertia: $M=I\ddot{\theta}$

3. The attempt at a solution
The relation between the accelerations:
$$\frac{\ddot{x}}{r}=\frac{\ddot{y}}{r}=\ddot{\theta}$$
The tension in the rope is T. the forces:
$$\left\{ \begin{array}{l} MBR+Tr=kMR^2\cdot \frac{\ddot{\theta}}{r} \\ mg-T=m\ddot{y} \end{array} \right.$$
$$\rightarrow MBR+m\left( g+\ddot{y} \right)r=kMR^2\cdot \frac{\ddot{\theta}}{r}$$
$$\ddot{y}=\frac{r\left( MBR+mgr \right)}{kMR^2+mr^2}$$
The linear acceleration of the cylinder in reference to the truck:
$$\frac{\ddot{x}}{R}=\ddot{\theta}=\frac{\ddot{y}}{r}=\frac{MBR+mgr}{kMR^2+mr^2}$$
The condition on B:
$$\left\{ \begin{array}{l} mg-T=\ddot{y}m=\ddot{\theta}mR\rightarrow T=m\left( g-R\ddot{\theta} \right) \\ Tr-MBR=kMR^2\cdot \ddot{\theta} \end{array}\right.$$
$$\ddot{\theta}=0 \rightarrow B=\frac{mr}{MR}g$$

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2. May 30, 2015

### haruspex

For the first part, it is not clear whether the acceleration of the mass is wanted relative to the truck or relative to the vround. Asuuming it's relative to the truck, I believe R should feature in the answer. If the cylinder rotates an angle theta, how far does it move relative to the truck? How far does a horizontal piece of the string move?

3. May 31, 2015

### Karol

The relation between the accelerations:
$$\frac{\ddot{y}}{R}=\ddot{\theta}$$
The tension in the rope is T. the forces:
$$\left\{ \begin{array}{l} MBR+Tr=kMR^2\cdot \ddot{\theta} \\ mg-T=m\ddot{y} \end{array} \right.$$
$$\rightarrow MBR+m\left( g+\ddot{y} \right)r=kMR^2\cdot \ddot{\theta}$$
$$\ddot{y}=-\frac{mrg+MBR}{mr-kMR}$$
The linear acceleration of the cylinder in reference to the truck: $\ddot{x}=\ddot{y}$
The condition on B:
$$\left\{ \begin{array}{l} mg-T=\ddot{y}m=\ddot{\theta}mR\rightarrow T=m\left( g-R\ddot{\theta} \right) \\ Tr-MBR=kMR^2\cdot \ddot{\theta} \end{array}\right.$$
$$\ddot{\theta}=0 \rightarrow B=\frac{mr}{MR}g$$
It comes out also from $\ddot{y}=0$

4. May 31, 2015

### haruspex

No.
Forget the motion of the truck for the moment and consider the cylinder rolling an angle theta, to the right say. The centre of the cylinder has moved $R\theta$ to the right, so the point on the cylinder where the rope first contacts it has also moved $R\theta$ to the right. But there is now more rope wound on the cylinder. How much more? So how far has the rope moved?
Check the answer by considering the extreme cases, r=R, and r=0.
That's not right either. Again, let B=0 for the moment. Which way will the cylinder roll? Is the tension in the string the only force exerting a torque on the cylinder?

5. Jun 1, 2015

### Karol

The relation between the accelerations:
$$\ddot{x}=R\ddot{\theta}-r\ddot{\theta}=\ddot{\theta}(R-r)$$
The moment of inertia round the point of contact with roof:
$$I_a=kMR^2+MR^2=MR^2(k+1)$$
The tension in the rope is T. in the frame of the truck d'alambert:
$$\left\{ \begin{array}{l} Tr-MBR=kMR^2(k+1)\cdot \ddot{\theta} \\ mg-T=m\ddot{y} \end{array} \right.$$
But i don't know to do it in laboratory frame since the forces are:
And they rotate the cylinder in the same direction, to the left, if i take moments round the center.
And how do i take the sign of $\ddot{\theta}$? i never knew to do that. do i decide that counterclockwise is positive? but there are two forces, so which one is dominant? if θ rotates in it's direction

Last edited: Jun 1, 2015
6. Jun 2, 2015

### haruspex

I assume the first k on the right is a typo.
How do you get Tr on the left? Isn't this supposed to be the torque the rope generates about the point of contact with the roof?

7. Jun 3, 2015

### Karol

$$T(R-r)-MBR=MR^2(k+1)\cdot \ddot{\theta}$$
But still how do i solve the forces in the laboratory frame? see the other sketch

8. Jun 3, 2015

### haruspex

It's almost the same. You just have to include the truck's acceleration in the total angular acceleration resulting from the torque. I.e. it is a real acceleration instead of a fictitious force. If you have trouble convincing yourself of the validity of this, you can put in a friction force and take moments about the cylinder's centre instead.

9. Jun 5, 2015

### Karol

D'alambert:
$$\left\{ \begin{array}{l} T(R-r)-MBR=MR^2(k+1)\cdot \ddot{\theta} \\ mg-T=m(R-r)\ddot{y} \end{array} \right.$$
$$\rightarrow \ddot{\theta}=\frac{mg(R-r)}{MR^2(k+1)+m(R-r)^2}$$
Laboratory:
$$R\ddot{\theta}=B\rightarrow \ddot{\theta}=\frac{B}{R}$$
$$\left\{ \begin{array}{l} T(R-r)=\left[MR^2(k+1)-\frac{B}{R} \right]\ddot{\theta} \\ mg-T=m(R-r)\ddot{y} \end{array} \right.$$
But i don't get the same result.
It's strange that i made moments round the center for $R\ddot{\theta}=B$ but for T i made moments around the point of contact. is it allowed to combine them into one equation?

10. Jun 5, 2015

### haruspex

How do you get that? Isn't that saying the centre of the cylinder does not accelerate?

11. Jun 5, 2015

### Karol

To simplify suppose there is no rope, only cylinder. if it were a box and the deceleration increases gradually the friction force f would grow until it would reach $mg\mu$. as long as $f<mg\mu$ the deceleration of the box is B and it stays in place.
If the cylinder was mass-less $R\ddot{\theta}=B$ would be true but with mass the cylinder will rotate slower, but how slow? what's the friction force? if B is big the cylinder will also slide.

12. Jun 6, 2015

### haruspex

Look for a relationship between B, R, r, $\ddot\theta$, $\ddot y$.

13. Jun 6, 2015

### Karol

I see only kinematics, is it true? no gravitation, Mg?

14. Jun 6, 2015

### haruspex

No, sorry, I got that wrong. (Making a lot of mistakes today.)
If you want to take moments about the cylinder's centre then you will need to introduce the force of friction between the cylinder and truck. This contributes both to the rotational acceleration and the linear acceleration in the lab frame. It's the latter that relates to B, R, r, $\ddot\theta$, $\ddot y$

15. Jun 6, 2015

### Karol

The acceleration of the center: $B-R\ddot{\theta}$
Moments round the center:
$$M\left( B-R\ddot{\theta} \right)R+Tr=kMR^2\ddot\theta$$
But the friction f and T act in the same direction, relative to the center and it's not true, B rotates counter clockwise and T clockwise.
If i could take moments round the contact point for T and round the center for f it would be better.

16. Jun 6, 2015

### haruspex

I have trouble keeping track of which way positive is being defined for these variables. For the equation in post #7, with which I agree, I believe B is defined as positive to the right and theta as positive clockwise.
On that basis, if a is the linear acceleration of the cylinder's mass centre, to the right, in an inertial frame, $a = B+R\ddot{\theta}$. If the frictional force on the cylinder is F to the right,
$F+T=Ma=MB+M\ddot\theta R$, $kMR^2\ddot\theta=-FR-Tr$.
I believe that putting those together leads to the same equation as in post #7.

17. Jun 8, 2015

### Karol

Inertial frame:
$$\left\{ \begin{array}{l} F+T=MB+M\ddot\theta R \\ kMR^2\ddot\theta=-FR-Tr \\ mg-T=m\ddot\theta (R-r) \end{array} \right.$$
$$\rightarrow \ddot\theta=\frac{mg(R-r)-MRB}{MR^2(k+1)+m(R-r)^2}$$
Truck's frame:
$$\left\{ \begin{array}{l} T(R-r)-MBR=MR^2(k+1)\cdot \ddot{\theta} \\ mg-T=m\ddot\theta (R-r) \end{array} \right.$$
It gives the same

18. Jun 8, 2015

### Karol

The equation in post #7 took moments round the contact point so if a force acts to the right θ is positive clockwise. but if i took moments round the center θ would have been positive counter clockwise.
I just want to ask how do i know which way θ is positive, and i assume that for every equation it depends on the pivot point, the axis, right?

19. Jun 9, 2015

### haruspex

The positive direction for theta, as a variable, is however you choose to define it. You just need to be consistent. The choice is independent of your choices for the linear movements etc.
The value computed for theta may be positive or negative. If you defined it positive clockwise but the value of $\ddot{\theta}$ comes out negative then you know that in fact it rotates counterclockwise.

20. Jun 9, 2015

### Karol

Doesn't the positive direction of the force define θ?
The equation in post #7:
$$T(R-r)-MBR=MR^2(k+1)\cdot \ddot{\theta}$$
I could say positive θ is counter clockwise and still the equation would be the same!