A question from kleppner's book

[MathematicalPhysicist]

on page 108, question 2.34:
A mass m whirls around on a string which passes through a ring, ngelect gravity.
initially the mass is distance r0 from the centre and is revolving at angular velocity w0.
the string is pulled with constant velocity V starting at t=0 so that the radial distance to the mass decreases.
obtain a differential equation for w(t).

here what i did:
v'=w0r0
r(t)=r0+Vt
v'=w(t)*r(t)
w(t)r(t)=w0r0

but from the text this isn't correct, can someone help me here?

 

[WigneRacah]

1) Why the tangential velocity of the mass should be the same at any time?
2) You should to impose the angular momentum conservation law.

 

[MathematicalPhysicist]

do you know a method which do not employ the angular momentum cl?

 

[WigneRacah]

If you cannot use angular momentum cl, thus you have to use cinematics.

Write the expressions for radial and tangential velocity (I mean the component which is orthogonal to the radial one) in polar coordinates and do the same for acceleration.
Take into account that it is a central motion (so acceleration only has the radial component).

Note that by saying that tangential acceleration is zero you are implicitly imposing the angular momentum conservation law.

 

[MathematicalPhysicist]

here what i did so far:
v=dr/dtr+rdr/dt
a=d^2r/dt^2r+2w^2$$\theta$$+r$$d\theta/dt$$$$\theta$$-rw^2$$\theta$$

now how do i proceed?

 

[WigneRacah]

here what i did so far:
v=dr/dt<b>r<\b>+rd<b>r<\b>/dt
a=d^2r/dt^2<b>r<\bold>+2w^2$$<b>\theta<\b>$$+r$$d\theta/dt$$$$<b>\theta<\b>$$-rw^2$$<b>\theta<\b>$$

now how do i proceed?

Well, begin by imposing that the radial component of the velocity is V and the tangential component of the acceleration is zero ...

 

[MathematicalPhysicist]

you mean that:
dr/dt=V
and dr^2/dt^2=0?

 

[WigneRacah]

you mean that:
dr/dt=V
and dr^2/dt^2=0?

I mean:

$$\frac{d \rho}{d t} = - V$$

$$2 \frac{d \rho}{d t} \omega + \rho \frac{d \omega}{ dt} = 0$$.


You should be able to go on, now ...

EDIT: here $$\rho$$ is r.

 

[WigneRacah]

here what i did so far:
v=dr/dtr+rdr/dt
a=d^2r/dt^2r+2w^2$$\theta$$+r$$d\theta/dt$$$$\theta$$-rw^2$$\theta$$

now how do i proceed?

I think you didn't perform the correct derivation for the velocity and acceleration in polar coordinates ...

 

[MathematicalPhysicist]

isnt v=d(rr)/dt
and a=dv/dt
?
according to these two equations i wrote the above.

 

[WigneRacah]

Velocity and acceleration in polar coordinates are:

$$v_r= \frac{d r}{d t}$$

$$v_\theta = r \frac{d \theta}{d t}$$

$$a_r = \frac{d^2 r}{d t^2} - r (\frac{d \theta}{d t})^2$$

$$a_\theta = 2 \frac{d r}{d t} \frac{d \theta}{d t} + r \frac{d^2 \theta}{d t^2}$$

with $$\omega = \frac{d \theta}{d t}$$.

 

[MathematicalPhysicist]

i know this is what i wrtoe in v.

 

[WigneRacah]

Ok. What's the problem then?
You have all you need to solve the question.

 

[MathematicalPhysicist]

im having just one little problem, in the answer key it's written that:
if Vt=r0/2 then w=4w0.

while when i do the calculation:
$$\int_{0}^{t}2Vdt=\int_{w0}^{w(t)}\frac{dw}{w}$$
which is:
2Vt=ln(w(t)/w0)
w(t)=(e^2Vt)*w0
which is clearly not the same as what is written in the answer key, can you point me where did i go wrong here?

thanks in advance.

 

[WigneRacah]

The integrand function on the left hand side of your equation is wrong ...

From the second equation in post n. 8 you get:

$$2 \int_{r_0}^r \frac{d \rho}{\rho} = - \int_{\omega_0}^{\omega} \frac{d \omega}{\omega}$$.

 

[MathematicalPhysicist]

but from this equation we get:
2(ln(r-r0))=-ln(w-w0)
ln(r-r0)^2=ln(1/(w-w0))
(r-r0)2=1/(w-w0)
if i put r=Vt=r0/2 then i get
r0^2/4=1/(w-w0) and you havnet got rid from r0.

 

[WigneRacah]

but from this equation we get:
2(ln(r-r0))=-ln(w-w0)
ln(r-r0)^2=ln(1/(w-w0))
(r-r0)2=1/(w-w0)
if i put r=Vt=r0/2 then i get
r0^2/4=1/(w-w0) and you havnet got rid from r0.

We don't get that!

I think you have some problems with the fundamental theorems of calculus or with logarithmic properties.

$$\ln r - \ln r_0 = \ln \frac{r}{r_0}$$
$$\ln \omega - \ln \omega_0 = \ln \frac{\omega}{\omega_0}$$

 

[MathematicalPhysicist]

oh, my mistake. frogot to switch them.

i know these rules of logarithms, but i did a horribel mistake here.

nevermind, it's good to this now, before the exams. (-;

 

[MathematicalPhysicist]

so idont need to use V here, or i shouldve used it here?

 

[WigneRacah]

good luck!

 

[WigneRacah]

so idont need to use V here, or i shouldve used it here?

No, you don't. For the angular velocity, you only need initial position, initial w and final position.