Understanding Rolling without Slipping: Solving for Acceleration and Rotation

[uriwolln]

Homework Statement

http://img406.imageshack.us/img406/9971/setupq.jpg In the diagram, there is a rope wrapped around a cylinder which is connected to a mass m.
The cylinder is on cart, so the cart accelerates to the right at a value of a.

The Attempt at a Solution

so far I got this:
lets say the whole system is moving right, the mass m is falling down, which means the cylinder will make a CCW rotation. (I think I am right up to this part)
so the equations are as follows:
mg- T = ma1
With relation to the point of contact P.
Thus I=(1/2)MR^2 + MR^2. R x T= I (alpha)
*The rope as well not slipping*
And because the cylinder is rolling I have a=R x (alpha)

This is where I am stuck. I have no idea how to connect the moving plane with the roll of the cylinder. And another thing, I get so confused with when does the equation a=R x alpha
get to be positive, or negative.HELP!

 

[tiny-tim]

hi uriwolln!

(have an alpha: α and an omega: ω )

I have no idea how to connect the moving plane with the roll of the cylinder. And another thing, I get so confused with when does the equation a=R x alpha
get to be positive, or negative.

a = rα only applies to the relative motion between the cylinder and the cart …

call the displacement of the cart x, and the displacement of the cylinder c …

then the relative position is c - x, the relative speed is c' - x' etc, and c'' - x'' = rα

 

[uriwolln]

Hey tiny-tim,
Thanks for the reply. I thought there was going to be no help for me :)
So tell me please, will I do the same thing, let's say, on the point above the CM of the cylinder where the rope connects to the cylinder.
Will it be (c for cylinder and r for rope): c - r ,c' - r', c'' - r''= (alpha)r

and if so, how do I know which variable do I subtract from the other?
In my mind, I am thinking that the cylinder is moving faster than the rope, so that is why I should do c -r. And the same thing for the cart? cart moves faster so thus c - x ?

 

[tiny-tim]

So tell me please, will I do the same thing, let's say, on the point above the CM of the cylinder where the rope connects to the cylinder.
Will it be (c for cylinder and r for rope): c - r ,c' - r', c'' - r''= (alpha)r

sorry, but you're really confusing me

write out the actual equations that solve the problem ​

 

[uriwolln]

:)
OK, here goes.
So, I assume here positive direction is to the right. CW is negative.
I have got:
mg- T = ma1

With relation to the point of contact P.
Thus I=(1/2)MR^2 + MR^2.
2R x T= I (alpha)

C will denote cart's acceleration. X the cylinder, and Ro the rope.
So if I get the notion of the displacement right, I will have

C - (- X)= -R(alpha) and Ro - X = -R(alpha)

have I got this right?

 

[tiny-tim]

hi uriwolln!

(what happened to that α i gave you? and try using the X² icon just above the Reply box )

With relation to the point of contact P.
Thus I=(1/2)MR^2 + MR^2.
2R x T= I (alpha)

L = Iω generally only works for the centre of mass or the centre of rotation …

P is not the centre of rotation …

so you need to use L = I_c.o.mω + r x mv_c.o.m.

 

[uriwolln]

LOL,
I need to learn how to write those characters in forum style... :)
Anyways,
I know that P is not the center of rotation, but I am using that point as a reference so I do not count the friction force that acts as a torque.
It's not wrong to use it like this right?

And did I get the displacement part right?

 

[tiny-tim]

I know that P is not the center of rotation, but I am using that point as a reference so I do not count the friction force that acts as a torque.
It's not wrong to use it like this right?

You can use P (to avoid using the friction), but then you must use the formula I gave you …

L = I_c.o.mω + r x mv_c.o.m.

 

[uriwolln]

WOW!
now I got confused a bit. What is then the difference between the equation you gave
L = Ic.o.mω + R x mVc.o.m.

and the one I wrote?
Does not the equation I wrote imply the momentum of inertia around point P, still using the center of the cylinder as the axle.

p.s I've tried using the new syntax :)

Edit: After thinking for a moment, (Does not happen too much :) ), the two equations are the same. taking v=wr
then L=w(Ic.o.m + R²M). Which are the same.

but PLZ again, did I get the displacement equations right?

 

[tiny-tim]

(you can just click the B icon for bold, instead of using a new font; and the X₂ icon will give you subscripts )

Does not the equation I wrote imply the momentum of inertia around point P, still using the center of the cylinder as the axle.

p.s I've tried using the new syntax :)

Edit: After thinking for a moment, (Does not happen too much :) ), the two equations are the same. taking v=wr
then L=w(Ic.o.m + R²M). Which are the same.

no, only if rv_c.o.m = r²ω …

in fact v_c.o.m = rω + v_P

 

[uriwolln]

but since the cylinder is rolling and not slipping, then v_P=0?

 

[tiny-tim]

but since the cylinder is rolling and not slipping, then v_P=0?

it's not slipping on the cart …

but the cart is moving ​

 

[uriwolln]

so what will V_p be equal to?

or am I too confused? :)

 

[tiny-tim]

why does it matter?

just use the formula I gave you …

L = I_c.o.mω + r x mv_c.o.m.

 

[uriwolln]

Sorry tiny-tim for being slow on this. PLZ do not lose hope on me :)

OK, so I got the formula you gave me, but how do I connect it to the system?
I mean, where can I use the parameter L, in the equations I wrote earlier?

If I just use it to find I, then I still have another parameter to worry about, which I do not know how...

 

[tiny-tim]

I mean, where can I use the parameter L, in the equations I wrote earlier?

If I just use it to find I, then I still have another parameter to worry about, which I do not know how...

i don't understand … you know what I_c.o.m is

now use τ_P = dL_P/dt = I_c.o.mα + r x ma_c.o.m.

 

[uriwolln]

OK, now I understand. The thing is, I was trying to solve the problem, and compare with the solution. But still they were different. So just to be on the safe side, I work out the equations from the center of the cylinder, and so I will include the friction.

Thanks, I've learned a lot.

One final question on the matter though, so I know I finally got it right. In this diagram:
http://img23.imageshack.us/img23/1324/setupfk.jpg

whereby the straight line is a rope.
So in order to figure out the displacement equations, it should be in regard to point P, right?
So now I have V_cylinder-V_rope=Rw
In this case Rw is positive because it goes CCW, am I right up to now?

 

[tiny-tim]

OK, now I understand. The thing is, I was trying to solve the problem, and compare with the solution. But still they were different. So just to be on the safe side, I work out the equations from the center of the cylinder, and so I will include the friction.

yes, if in doubt, it's always safe to use the centre of mass!

So in order to figure out the displacement equations, it should be in regard to point P, right?
So now I have V_cylinder-V_rope=Rw
In this case Rw is positive because it goes CCW, am I right up to now?

no …

if the cylinder moves to the right, the rope will unroll, so the mass will go down for two reasons: i] the cylinder is nearer the pulley, ii] there's more rope …

so the mass descending corresponds to CW rotation

 

[uriwolln]

AWESOME!

Ok, I get this part.
But, because I chose left side to be positive, means
V_rope and V_cylinder are actually negative, so it becomes:
-V_rope + V_cylinder=-R(omega)

And now, in order for me to do the relative velocity of the cart to the cylinder I go:
V_cart-V_cylinder=-R(omega)
Whereby V_cart and V_cylinder are to the positive direction

 

[tiny-tim]

But, because I chose left side to be positive …

why did you do that? was it for some religious or political reasons?

you can see the mass will pull the cylinder to the right (and the cart is accelerating to the right), so why make it difficult for yourself??

 

[uriwolln]

why did you do that? was it for some religious or political reasons?

Had to laugh out of this one :)

So right is positive. :)
So because at the point where the cart and the cylinder meet, the equation of
V_cart-V_cylinder=-R(omega)
will actually be,

V_cart+V_cylinder=-R(omega)
because V_cylinder at that point moves to the left, so it is negative.

Is my assumption correct?

 

[tiny-tim]

So because at the point where the cart and the cylinder meet, the equation of
V_cart-V_cylinder=-R(omega)
will actually be,

V_cart+V_cylinder=-R(omega)
because V_cylinder at that point moves to the left, so it is negative.

why are you even trying to guess which way the cylinder will go?

The only sensible thing is to measure both V_cart and V_cylinder in the same direction (eg to the right) …

anything else is confusing and likely to lead to mistakes

(and if V_cylinder turns out to be negative, ok it's negative, big deal)

 

[uriwolln]

So I can use either?
(1) V_cylinder-V_cart=R(omega)
or
(2) V_cart-V_cylinder=R(omega)

and for the rope part as well:
(3) V_rope-V_cylinder=R(omega)
or
(4) V_rope-V_cylinder=R(omega)

How do I know what will R(omega) in each of the equation 1,2,3,4. I mean, positive or negative?

 

[tiny-tim]

How do I know what will R(omega) in each of the equation 1,2,3,4. I mean, positive or negative?

(what happened to that ω i gave you ? )

you don't need to know, it will come out when you solve the equations …

the important thing is to use the same notation for everything … eg right and clockwise (or anticlockwise) are always positive … so you don't make mistakes

i really think you should stop planning now, and start actually writing the equations out, and solving them ​

 

[uriwolln]

ALRIGHT, one last shot! :)
This is the new setup, so it will be easier for me:
http://img830.imageshack.us/img830/8643/setupd.jpg A cylinder is on a rug which is pulled at a constant acceleration a as shown. There is also a drag force of F= -bv. I need to find f_min (friction) of the carpet that will allow rolling without slipping.

The lower part of the diagram, have the general motion I think the cylinder will do. As well as, I have placed the force of friction to work to the right.

Equations as follows:
(1) f-bv=ma_cylinder
(2) Rf=I_c.o.m(alpha) ----> (I lost the alpha) :)
V_rel=ωR {here CCW is positive} so: V_carpet-V_cylinder=ωR
Thus: (3) a_carpet-a_cylinder=R(alpha)
using (2) and (3) I get : a_cylinder= a-2f/m

plugging it into (1) I get f-bv=ma -2f
So f=(ma+bv)/3
Am I right?

 

[tiny-tim]

i'm confused … is it a + 2f/m or a - 2f/m ?

 

[uriwolln]

Sorry I must have edited while you posted.
f=(ma+bv)/3
a_cylinder= a-2f/m
and so I took a_cylinder and plugged it into (1)

 

[tiny-tim]

ah!

yes, i think that's ok

 

[uriwolln]

AWESOME!
Thank you for your patience!
I think I get it now :)