# Homework Help: Finding acceleration of cylinder

1. May 7, 2013

### Saitama

1. The problem statement, all variables and given/known data
(see attachment)

2. Relevant equations

3. The attempt at a solution
Can I use the instantaneous axis of rotation for this question?

I placed it at the point of contact of cylinder and the ground. Moment of inertia about it is: $4mR^2/2+4mR^2+mx^2=6mR^2+mx^2$, where m is the mass of dog and x is the distance of it from the axis. x can be calculated using the cosine rule i.e
$$\cos(150^o)=\frac{R^2+R^2-x^2}{2R^2} \Rightarrow \frac{x^2}{2R^2}=1+\frac{\sqrt{3}}{2} \Rightarrow x=\sqrt{2+\sqrt{3}}R$$

Toque about the axis is $mgx\sin(15^o)=I\alpha=Ia/R$. Substituting the values of x and I, I don't get the right answer.

Any help is appreciated. Thanks!

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2. May 7, 2013

### TSny

Only the cylinder rotates about the point of contact, the dog is moving along a horizontal line. So, when you set up $\tau = dL/dt$ about the point of contact, you might want to separate out the contributions of the cylinder and dog to $L$. So,

$\tau = dL_{cyl}/dt + dL_{dog}/dt$

3. May 7, 2013

### Saitama

Does that mean it is fine to use the instantaneous axis of rotation?

$$L_{cyl}=6mR^2\omega=6mRv$$
$$L_{dog}=mxv$$
Are these correct?

4. May 7, 2013

### TSny

The expression for $L_{dog}$ is not quite correct. Recall, for a particle $L = rpsin\phi = r_{\bot} p$ where $r_{\bot}$ is "moment arm" for p.

5. May 7, 2013

### Saitama

Is it $mxv\sin(15^o)$?

6. May 7, 2013

### TSny

No, did you use the correct trig function here? $r_{\bot}$ is the perpendicular distance from the point of contact to the line of motion of the dog.

7. May 7, 2013

### Saitama

Oh yes, sorry about that. As the dog moves horizontally, r_{\bot} is $x\cos(15^o)$, so $L_{dog}=mxv\cos(15^o)$.

Differentiating the expression for $L_{cyl}$ and $L_{dog}$ w.r.t time
$$\frac{dL_{cyl}}{dt}=6mRa$$
$$\frac{dL_{dog}}{dt}=mxa\cos(15^o)$$
Hence,
$$\tau=I\alpha=6mRa+mxa\cos(15^o)$$
Should I plug in the values of I and x from my first post?

8. May 7, 2013

### TSny

Looks like you left out the cos(15) in mxa. [EDIT: I see you already fixed that.] Yes, you can plug in your value of x. But, you might just want to write $r_{\bot}$ directly in terms of R and the angle of 30 degrees. I think it will be a little nicer. But it should all come out the same in the end.

9. May 7, 2013

### Saitama

I edited my previous post. :)

But the $a$ cancels out in my torque equation.

10. May 7, 2013

### TSny

I thought you were using mgxsin(15) for the torque? xsin(15) can be written in a nice way in terms of R and the 30 degrees.

11. May 7, 2013

### Saitama

This may be a dumb question but why can't I use $I\alpha$?

And I can use a calculator to evaluate sin(15) so I don't need to worry about expressing it in terms of 30 degrees. :)

12. May 7, 2013

### TSny

The general law is $\tau = dL_{sys}/dt$. For the cylinder, you can write $dL/dt = I\alpha$ while for the dog you found $dL/dt = mxacos(15)$.

So, you have the right hand side of $\tau = dL/dt$ figured out. You just need to think about what to put in for the torque for the left hand side. And I think you had that figured out in your first post.

Aww.

Last edited: May 7, 2013
13. May 7, 2013

### Saitama

Thanks TSny, $a$ comes out be 0.6177 m/s^2. Is this right?

Is it possible to do this question by taking torque about the CM of the cylinder? I tried it before switching to the instantaneous axis but I was stuck on the direction of friction. How can I find it?

14. May 7, 2013

### TSny

Pretty close to what I got. Might be some round off error (which can be avoided by writing things in terms of R and 30 degrees and not plugging in numbers until the end. R will cancel out.) See if you can show that the answer can be expressed as

a = gsin30/(7+cos30)

The friction force at the ground is the only external force acting on the system horizontally. So, that should tell you the direction of the the friction force from the ground. You need to be careful if you want to take torques about the CM of the cylinder, since the CM of the cylinder is not the same as the CM of the system. However, the CM of the cylinder and the CM of the system have the same acceleration in this problem. So, I think you would be ok in taking torques about the CM of the cylinder.

Of course, you can always treat the cylinder and dog separately. You would then draw free body diagrams for each. For the cylinder, you could take torques about the CM of the cylinder. The algebra is more involved now because you will have to take into account the normal and friction force between the dog and cylinder.

15. May 7, 2013

### Saitama

I reached that answer. xin(15) is equivalent to Rsin(30) and xcos(15) is equivalent R(1+cos(30)). This was obvious from the diagram. I should have done that before.

Thanks you.

Okay, so the direction of friction on the cylinder from the ground is in the right direction ($f_1$) and from the dog, its tangential in the direction opposite to the motion of dog ($f_2$).

This will give me 2 equations:
$$f_1=4ma$$
$$f_1R-f_2R=I\alpha$$

Am I doing this right? :)

Last edited: May 7, 2013
16. May 7, 2013

### TSny

If you're going to treat the cylinder alone, then there will be horizontal components of the normal and friction forces from the the dog that will need to be included in ƩFx = (4m)a

For the torque equation for cylinder, make sure you have the correct direction for the frictional torque from the dog (due to $f_2$). I'm not sure whether you're taking clockwise or counterclockwise as positive rotational direction. [EDIT: Never mind, I thought you had a + sign rather than a - sign in the torque equation. But, still, it would be nice to know your sign convention for the rotation.]

17. May 7, 2013

### Saitama

I am treating them separately. I will rewrite everything again. Please see my attachment. The direction of forces are shown. I have not shown the normal reaction from the ground on the cylinder. :)

$f_2=mg\sin(30^o)$, $N=mg\cos(30^o)$

$$\sum F_x=f_1+f_2\cos(30^o)-N\sin(30^o)=4ma$$

Is this correct?
I feel that direction of f_2 is incorrect.

#### Attached Files:

• ###### fbd.png
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18. May 7, 2013

### TSny

Diagram looks good.

These equations are not correct. To get the correct equations for f2 and N, you will need to apply 2nd law to the dog.

This looks correct.

I think you have the direction for f2 on the cylinder correct.

19. May 7, 2013

### Saitama

The forces acting on the dog are friction (f_2), mg and normal reaction from the cylinder
For the tangential motion, $mg\sin(30^o)-f_2=ma$ and for the horizontal motion, $N\cos(60^o)-f_2\cos(30^o)=ma$. Is this correct now?

20. May 7, 2013

### TSny

For the tangential direction, you'll need the component of acceleration in the tangential direction.

Horizontal equation looks good.

21. May 7, 2013

### Saitama

Why? Isn't it a, I mean, the acceleration is $\alpha R$ which is $a$? Or is it $(a+a\cos(30^o))$?

22. May 7, 2013

### TSny

What is the direction of the acceleration of the dog?

23. May 7, 2013

### Saitama

Won't it be resultant of the two accelerations, one in horizontal direction and the other in tangential direction?

24. May 7, 2013

### TSny

No. You're standing on the ground as this contraption comes by. It's pitch dark except for a small light on Fido's nose and a small light carried by the center of the cylinder. What direction of acceleration do you see for the light on Fido's nose? How does it compare to the acceleration of the center of the cylinder?

Last edited: May 7, 2013
25. May 7, 2013

### Saitama

I still don't get it. The light on Fido's nose performs circular motion about centre of cylinder with a tangential acceleration of a and moves in horizontal direction again with an acceleration of $a$.

If someone sits on the centre of cylinder, he/she would see that Fido's nose moves about the centre in a clockwise manner.