# Force of a bar to stop a rotating cylinder

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1. Aug 5, 2015

### whatdoido

Hi, I'm having conceptual problems for solving this one.

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

I drew a picture of this problem, which should show up below. Lengths x,a and b were given in the original picture.

A rotating closed cylinder is braked with a bar. Its brake shoe is pressed in respect of point B without friction. The bar AB presses the cylinder. Force $F_1$ is used in point A.

Calculate the needed force when cylinder stops from a rotational speed of 1200 RPM clockwise in 12 seconds. Cylinder's mass is 14 kg, radius 6,5 cm and frictional coefficient 0,33.

$r=6,5 cm= 0,065m$
$\mu=0,33$
$x=7,0 cm=0,07m$
$a=19 cm=0,19m$
$b=1,2 cm=0,012m$
$m=14kg$
$\Delta t=12s$
$n=1200 RPM=20r/s$

2. Relevant equations
$M=J\alpha$
$M=Fr$
$J=\frac{1}{2}mr^2$
$\sum M=0$

3. The attempt at a solution
Well I know how to calculate the force needed to stop the rotating cylinder

$M=J\alpha\hspace{30mm}M=\mu Nr$
$J\alpha=\mu Nr$
.
.
.
$N=\frac{mr2\pi n}{t \mu}$
$=28,877... N$

$F_1$ can be calculated with torques I'm pretty sure, but this is the spot where I am stuck at. Something like $F_2 x=F_1 (x+a)$ won't do. I should somehow take in the consideration the clockwise movement of cylinder (in part b of this problem it is supposed to calculate $F_1$ in counter-clockwise movement). Cylinder has a torque $F_3$, right? So I thought that is what I am supposed to use, but I'm not sure how to proceed.. I tried using some angles and make it work, but I cannot think something that would make me think "oh of course! That's how I can solve it" and give me the correct answer, eventually.

2. Aug 5, 2015

### Dr. Courtney

How can the sum of the torques be zero if rotatational motion is being slowed to a stop?

3. Aug 5, 2015

### haruspex

Quite so. What point are you taking moments about? What other force acting on the bar has a moment about that point?

(There is something not quite right in the question, though. A brake shoe has a curved surface. The 'normal' acts in different directions around the curve. While the net force is F2, the frictional torque depends on the integral of the frictional force around the curve, so will be greater than $\mu F_2$. To answer the question fully you would need to know the angle the shoe subtends at the cylinder's centre.)

4. Aug 6, 2015

### whatdoido

Okay, then the best point would be probably between A and B. Let's call it C.

There is a moment $F_1 a$ at least. Then I think the other moment is cylinder's moment, which is $M=J \alpha=0,309...Nm$, right? The moment $F_2 x$ should make the cylinder stop rotating. $F_2 x$ creates a moment $F_3 b$ and $M$ resists that moment. So the moments acting in point C are $F_1 a$, $F_3 b$ and $M$. Or is this how it is?

Yeah, the curve made me think. But since it has no special dimensions, I ignored it.

5. Aug 6, 2015

### haruspex

That would leave you with an unknown force at the fulcrum, B. What point does that suggest?

6. Aug 6, 2015

### whatdoido

Yes, you are right.. So B then, right? But I need to change the equation $F_2 x=F_1 (x+a)$ somehow. Dr. Courtney mentioned that the sum of torques should not be zero.

7. Aug 6, 2015

### haruspex

The sum of torques acting on the cylinder cannot be zero, but the bar does not rotate, so the sum of torques on that will be zero.

8. Aug 8, 2015

### whatdoido

Okay, I got a bit confused earlier. But is the sum of the torques on cylinder $\mu xF_2$?

Then I could calculate the needed moment by $\mu xF_2=J\alpha+F_3b$

I tried that, but still the answer is wrong a little bit

9. Aug 8, 2015

### haruspex

The force tangential to the cylinder is clearly $F_3=\mu F_2$. But I don't see how you bring x into it to get the torque. How far is the frictional force from the cylinder's centre?

10. Aug 8, 2015

### whatdoido

$6,5 cm (=r)$ I would think

11. Aug 8, 2015

### haruspex

Right, so what torque does it apply to the cylinder?

12. Aug 9, 2015

### whatdoido

$F_3 r$, right?

By the way, can I write a torque $F_3b$ in relation to $B$? Like $F_2x+F_3b-F_1(x+a)=0$ and since $F_3=\mu F_2$, it would make $F_2x+\mu F_2b=F_1(x+a)$

Last edited: Aug 9, 2015
13. Aug 9, 2015

### haruspex

Yes to both.

14. Aug 9, 2015

### whatdoido

And $J\alpha=\mu F_2r$
$F_2=\frac{mr\pi n}{t \mu}$
So I get
$F_1=\frac{F_2(x+\mu b)}{x+a}=\frac{\frac{mr\pi n}{t \mu}(x+\mu b)}{x+a}=4,107..N$
$≈4,1N$
That is the correct answer according to my book.

In part b the movement of cylinder is counter-clockwise, thus I figured that I just subtract the frictional moment of cylinder because less torque is needed

$F_2x+\mu F_2b-\mu M-F_1(x+a)=0$
where $M=J \alpha$

This gives me $F_1≈3,7N$ which it should be

Did I get it correctly?

15. Aug 9, 2015

### haruspex

You don't seem to have used the same method that was successful in part a.
By what logic do you subtract $\mu M$?
Think about which way F3 acts. Is it the same as before? Will its contribution to the moments about B be the same?

16. Aug 9, 2015

### whatdoido

I see! $F_3$ decreases the moment in relation to B.

That would give:
$F_2x-\mu F_2b=F_1(x+a)$

Now that I think about it, subtracting $\mu M$ does not make so much sense.. My logic was that cylinder's moment decreases B's moments. But $F_3r$ is cylinder's moment, so I should have thought about it more. I just saw a convenient answer..

17. Aug 9, 2015

### haruspex

OK!

18. Aug 9, 2015

### whatdoido

Alright, thanks for help

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