Irodov: Tape going down a ramp

• A13235378
First, a note of caution in using that equation.##F=\frac{dp}{dt}##, where F is an external force applied to a system, is only valid if p is the momentum of a closed system. So if you expand it to ##F=m\frac{dv}{dt}+v\frac{dm}{dt}## then ##\frac{dm}{dt}=0## and we recover the familiar F=ma.I found it helpful first to consider a simpler problem: the spool rolling along a horizontal surface. As the radius shrinks, the angular velocity must increase relative to the linear velocity.

A13235378

Homework Statement
A flexible strip of length L is wound and its free end is fixed by a nail to the surface of an inclined plane that forms an angle θ with the horizontal. Then, the tape is allowed to unwind as it descends the slope.

How long will the tape be completely unrolled?
Relevant Equations
I = mR^2 /2
Hello everbody,

I found the following resolution:

We can consider the winding of the wire as a thin disc.

Like this,

For a given instant t, we will have M (t) mass of the disc and R (t) radius of the disc.

Analyzing linear dynamics:

$$M(t)gsen\theta - T = M(t).a$$

Analyzing the rotational dynamics:

Resulting torque = moment of inertia. angular acceleration (modularly)

Considering the center of the disk as the point of rotation, the only force that will make torque is traction.

$$T.R(t) = I \alpha = I . \frac{a}{R(t)}$$

The moment of inertia for a thin disc is given by:

$$I=\frac{M(t).R(t)^2}{2}$$

Replacing

$$T=\frac{M (t)R (t)^2.a}{2}$$

Substituting T, we find that the linear acceleration will be:

$$a=\frac{2}{3}.g.sen\theta$$

Realize that this linear acceleration is constant over time.

So, we can apply a cinematic:

$$S=\frac{at^2}{2}$$

Replacing S with L and isolating t:

$$t=\sqrt{\frac{3L}{gsin\theta} }$$

My question is:

Why in the analysis of the translation dynamics did you not consider the contribution of the system's mass variation? Since this is a variable mass system, shouldn't we use that in the tangential direction $$F = \frac{Mdv}{ dr} + \frac{udm}{dt}$$ where $$u$$ is the relative velocity of mass ejection in relation to the analyzed body?

Delta2
A13235378 said:
We can consider the winding of the wire as a thin disc.
It occurs to me that this is not a thin disc. It is just a tiny bit heavier on the down-hill side than on the up-hill side.

Which means that the following is not completely correct.
A13235378 said:
Considering the center of the disk as the point of rotation, the only force that will make torque is traction.
It further occurs to me that there is an insight that would allow this effect to be addressed cleanly: Consider the trajectory of the center of mass of the remaining spooled tape.

A13235378 said:
shouldn't we use that in the tangential direction $$F = \frac{Mdv}{ dr} + \frac{udm}{dt}$$ where $$u$$ is the relative velocity of mass ejection in relation to the analyzed body?
First, a note of caution in using that equation.
##F=\frac{dp}{dt}##, where F is an external force applied to a system, is only valid if p is the momentum of a closed system. So if you expand it to ##F=m\frac{dv}{dt}+v\frac{dm}{dt}## then ##\frac{dm}{dt}=0## and we recover the familiar F=ma.

In the present case, the subtlety is that although ##v=r\omega##, it is not true that ##a=r\alpha##.
I found it helpful first to consider a simpler problem: the spool rolling along a horizontal surface. As the radius shrinks, the angular velocity must increase relative to the linear velocity.

NTesla
@A13235378 , are you still working on this? Were you able to correct your equation for the relationship between a and α?

Here's how I dealt with the horizontal version:
Tape length = L, thickness h, initial radius r0, initial velocity v0, initial angular velocity ω0.
Lh=πr02
v0=r0ω0

##\frac{dr}{d\theta}=-h##
##\dot r=\frac{dr}{d\theta}\frac{d\theta}{dt}=-h\omega##
##v=r\omega##

Differentiating
##a=\frac d{dt}v= \dot r\omega+r\alpha=-h\omega^2+r\alpha##

If T is the tension in the tape on the ground (or, equivalently, static friction)
##T=-ma=mh\omega^2_mr\alpha##
and for torque:
##Tr=\frac 12mr^2\alpha##
Whence
##h\omega^2=\frac 32r\alpha##
Substituting for α and ω from prior equations
##\frac{\ddot r}{\dot r}=-\frac 23\frac{\dot r}r##

Integrating
##\dot r=Ar^{-\frac 23}##
From initial conditions, ##A=-h\omega_0r_0^{\frac 23}##

Integrating again
##\frac 35r^{\frac 53}=c+At##
##\frac r{r_0}=1-\frac{5h\omega_0t}{3r_0}=1-\frac{5\pi v_0 t}{3L}##

When the tape runs out, r=0, ##t=\frac{3L}{5\pi v_0}##.

Looks persuasive, but that answer is much less than L/v0; I feel sure it should be more than L/v0. So maybe I have an error somewhere.

What does Irodov give as the answer in the slope case?

Lnewqban
haruspex said:
If T is the tension in the tape on the ground (or, equivalently, static friction)
##T=-ma=mh\omega^2_mr\alpha##

I set up the problem almost exactly how you did, but I did something a little different here. I thought about the variable mass equation ##\vec{F}_{\text{ext}} + \vec{v}_{\text{rel}} \frac{dm}{dt} = m\vec{a}##, let the system be the cylinder, and considered that the "exhaust velocity" ##\vec{v}_{\text{rel}}## of the tape just about to leave the system and to come into contact with the table, with respect to the centre of mass of the cylinder, being ##-r\omega \hat{x}##. So in addition to the tension from the tape, you also have this extra term due to mass transfer out of the system [in the derivation of the variable mass equation, the external force acts on both the system and the mass ##-dm## that leaves in a time interval ##dt##]. What do you think?

In any case, it might not make too much of a difference so long as ##\dot{m}/m## is small...

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Lnewqban
haruspex said:
##\frac{dr}{d\theta}=-h##
Should this be ##\large \frac{dr}{d\theta} = -\frac{h}{2\pi}##?

Lnewqban and haruspex
TSny said:
Should this be ##\large \frac{dr}{d\theta} = -\frac{h}{2\pi}##?
Ah, thank you. I changed my notation while transcribing, not a wise move.
I believe that changes my final expression to ##\frac{6L}{5v_0}##, which is very reasonable.

Lnewqban and TSny
etotheipi said:
I set up the problem almost exactly how you did, but I did something a little different here. I thought about the variable mass equation ##\vec{F}_{\text{ext}} + \vec{v}_{\text{rel}} \frac{dm}{dt} = m\vec{a}##, let the system be the cylinder, and considered that the "exhaust velocity" ##\vec{v}_{\text{rel}}## of the tape just about to leave the system and to come into contact with the table, with respect to the centre of mass of the cylinder, being ##-r\omega \hat{x}##. So in addition to the tension from the tape, you also have this extra term due to mass transfer out of the system [in the derivation of the variable mass equation, the external force acts on both the system and the mass ##-dm## that leaves in a time interval ##dt##]. What do you think?

In any case, it might not make too much of a difference so long as ##\dot{m}/m## is small...
You could be right. It can be hard figuring out when and how to use that variable mass equation.
In the present case, what exactly is Fext, and how does it relate to the torque?

To try to resolve this, consider a small step as occurring in two parts: first, the spool advances and rotates as though there were no friction/tension. So now it is moving a tad too fast for its rotation rate.
Next, a small horizontal impulse dI from the ground brings the two back into synch:

##m.dv=-dI##
(That m has reduced slightly is a second order small quantity there, and again for m and r below.)
##\frac 12mr^2.d\omega=r.dI##
Hence ##r.d\omega=-2.dv##
Before: ##v=r\omega##
After: ##(v+dv)=(r+dr)(\omega+d\omega)=r\omega+\omega.dr+r.d\omega##
Pulling that all together: ##3dv=\omega.dr##. That's the same as I had before.

etotheipi and Lnewqban