# A rolling cylinder problem

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1. Dec 8, 2016

### doktorwho

1. The problem statement, all variables and given/known data
With the equations of motion in polar coordinates given by $r(t)=pcos{kt^2}, φ(t)=kt^2$ determine the velocity intensity of a point $M$ on the circumference of a cylinder which is rolling without friction on a horizontal plane at time $t$ is the velocity of the center of cylinder at time $t$ is $v_c$
2. Relevant equations
3. The attempt at a solution

I first drew a picture of a cylinder and noted that the angle $φ$ is the angle from the horizontal to the point $M$ and i need to use another angle which is more suited for this rolling motion and that would be the angle $\theta$ which is measured from the radius to the vertical.
From here $\theta=2φ=2kt^2$. Since i know how the angle changes, i can calculate the angular velocity and acceleration. $ω=\dot \theta = 4kt, α=\ddot \theta = 4k$. Since the function of $r$ is dependent only on $cos(kt^2)$ i gues whats in front must be the amplitude therefore the radius should be $R=p/2$? Is this correct thinking?
The velocity and acceleration are $v=wR=2pkt, a=Rα=2pk$, i can find the the $a_n=8pk^2t^2$ but how do i find the velocity at that point. Kinda stuck on that?

2. Dec 8, 2016

### haruspex

Not following that.
Where is the origin for these coordinates? Seems like it must be moving at the same velocity as the centre of the circle, and r varies between 0 and p, so I guess the origin is, at each instant, the point of contact and p is the diameter. That makes the reference frame non-inertial.
It also makes φ the angle M and the circle's centre subtend at the point of contact.

3. Dec 9, 2016

### doktorwho

The problem does not have a diagram but i tried drawing it

4. Dec 9, 2016

### haruspex

Ok, that's what I deduced about your view from what you wrote. But the equations say r=p cos(φ), so where is p in your diagram?
Seems to me φ should be the angle to the vertical and p is the diameter of the circle.

What is the angular velocity at time t? So what is vc as a function of t?

5. Dec 10, 2016

### doktorwho

Well my function of $r$ is at its peak when $cos{kt^2}=1$ so the p is a diameter and $R=p/2$, oh yeah so because of this the angle must be measure from the vertical so that the $p$ part is met? If the equation was $r(t)=psin{kt^2}$ then it would be like this right?
My angluar velocity is then, since the angle $\theta$ is now the twice of $\phi$, $w=4kt$. $v_c$ is constant at all times so it is not dependent on $t$?

6. Dec 10, 2016

### haruspex

I meant to ask what this means:
i wondered if that was a typo, and you meant without slipping (in which case vc is not constant).
It seems odd to describe it as rolling if it is slipping.

7. Dec 10, 2016

### doktorwho

Yeah xD, it just rolls without slipping, wring words sorry. It rolls so the $v_c$ is constant. But still how does this relate to the velocity of some point on the circumference? Visualising it i would think that the points on the circumference must move faster than $V_c$?

8. Dec 10, 2016

### haruspex

No, you have an increasing rotation rate, so if it is not slipping vc must be increasing.

9. Dec 10, 2016

### doktorwho

Lets imagine this. At some point in time the velocity of the center of the cyliner is $v_c$. I then conclude that the velocity of the top of the cylinder (the hightest point) must also have the velocity $+v_c$ and the lowest point ( the one where the contact is ) must have the velocity $-v_c$. Therefore the change $-2v_c$ so there must be a point of zero velocity in between. Regarding that in a angle form of our chosen system i would conclude that the velocity of the point on the circumference is $v_p=v_c*cos{\theta}$. Does that seem correct and if not, what seems wrong?

Last edited: Dec 10, 2016
10. Dec 10, 2016

### haruspex

I do not understand your reasoning. Why would it have the same velocity? What you do know is that the point of contact with the ground is instantaneously stationary.

11. Dec 11, 2016

### doktorwho

Well then is the point of contact is stationary at time $t$ and the centers speed is $v_c$ at time $t$ the only reasonable image of this would be that the highest point has a velocity of $2v_c$. Then all the other points on the circumference would have a velocity of $v_p=2v_ccosθ$ where $cosθR=h-p/2$, $h$ is the height of the point measured from the ground. This case includes the uppmost velocity of the cylinder at point $t$ which is $v_p=2v_ccos{\pi /2}$ but not sure about the velocity when $h=p/2$

Last edited: Dec 11, 2016
12. Dec 11, 2016

### haruspex

You cannot mean that. h=p cos2θ?

13. Dec 11, 2016

### doktorwho

Sorry what i meant was $R*cosθ=h-p/2$ How did you get $h=pcos^2{θ}$?

Last edited: Dec 11, 2016
14. Dec 11, 2016

### haruspex

You did not define θ so I had to guess from your vp formula.
Instantaneously, the cylinder is rotating about what point? What is its rotation rate in terms of vc and p?

15. Dec 11, 2016

### doktorwho

Instantaneously, its rotating around the point of contact. Rotation rate? $\frac{2v_c}{p}$? Im not sure what im calculating now..

16. Dec 11, 2016

### haruspex

Right, and right.
So if some point of the cylinder is at distance x from the point of contact with the ground, what is its speed?

17. Dec 11, 2016

### doktorwho

$\frac{2v_c}{p}*\frac{h}{cosθ}$?

Last edited: Dec 11, 2016
18. Dec 11, 2016

### haruspex

Right (again, guessing at your definition of θ), but you can express x in terms of θ and p, rather than θ and h.