# Spinning block with decreasing radius

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1. May 20, 2017

### Buffu

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

A mass $m$ whirls around on a string which passes through a ring. Given that gravity is not present and initially the mass is at a distance $r_0$ from the center and is revolving at angular velocity $\omega_0$. The string is pulled with constant velocity $V$ starting $t = 0$ so that radial distance to the mass decreases. Find $\omega (t)$

2. Relevant equations

3. The attempt at a solution

There is no tangential forces on the block and the only radial is tension which is not required.

Anyhow, $r = Vt$ since the velocity is constant, $\dot r = V$ and $\ddot r= 0$.

Pluggin those in,

$\vec a = \hat r(\ddot r - r \dot \theta^2 ) + \hat \theta (r \ddot \theta + 2\dot r \dot \theta)$

$\vec a = -\hat r (Vt \dot\theta^2) + \hat \theta (Vt \ddot \theta + 2V\dot \theta)$

Since there are no tangential forces,

$Vt \ddot \theta + 2 V\dot \theta = 0 \iff t \ddot \theta = - 2\dot \theta$

Since $\dot \theta = \omega$

$\displaystyle t \dot \omega = -2 \omega \iff {\dot \omega \over \omega} = {-2 \over t} \iff \int^{\omega}_{\omega_0} {d\omega \over \omega} = \int^t_0 {-2 \over t}dt$

Now the second integral has the closed form $\ln |t|$ which is not defined at $t = 0$,

Where did I go wrong ?

2. May 21, 2017

### TSny

Does this give the correct value of $r$ at $t = 0$?

3. May 21, 2017

### Buffu

No. But why should it since after $t = 0$ the rope is pulled, no ?

4. May 21, 2017

### ehild

The radius decreases, so it is not Vt.

5. May 21, 2017

### zwierz

Angular momentum about the hole is conserved

Last edited: May 21, 2017
6. May 21, 2017

### haruspex

t=0 is the instant at which the pulling starts. At that time, r should be its initial value still.

7. May 21, 2017

### Buffu

Ok I did it.

At any instance $r = r_0 - Vt$

Which gives $\vec a = -(r_0 - Vt)\hat r + \hat \theta((r_0 - Vt ) \dot \theta + -2V\ddot\theta)$

So, we get $(r_0 - Vt ) \dot \theta = 2V\ddot\theta$

Substituting for omega,

$\dfrac{\dot \omega}{\omega} = \dfrac{2V}{r_0 - Vt}$

Integrating from $\omega_0 \to \omega$ and LHS and from $0 \to t$ on right,

I get $\omega = \dfrac{\omega_0 r_0^2}{(r_0 - Vt)^2}$

Is this correct ?

Second part of the question was to find the radial force on the block,

Which I think is $F_r =m(r_0 - Vt)\omega^2$ and I plug from above in this formula,right ?

8. May 21, 2017

### Buffu

Ok I did it.

At any instance $r = r_0 - Vt$

Which gives $\vec a = -(r_0 - Vt)\hat r + \hat \theta((r_0 - Vt ) \dot \theta + -2V\ddot\theta)$

So, we get $(r_0 - Vt ) \dot \theta = 2V\ddot\theta$

Substituting for omega,

$\dfrac{\dot \omega}{\omega} = \dfrac{2V}{r_0 - Vt}$

Integrating from $\omega_0 \to \omega$ and LHS and from $0 \to t$ on right,

I get $\omega = \dfrac{\omega_0 r_0^2}{(r_0 - Vt)^2}$

Is this correct ?

Second part of the question was to find the radial force on the block,

Which I think is $F_r =m(r_0 - Vt)\omega^2$ and I plug from above in this formula,right ?

9. May 21, 2017

### haruspex

Yes, that all looks right.

10. May 21, 2017

### Buffu

Will this solve the question without solving a DE ?

11. May 21, 2017

### zwierz

I guess so

12. May 21, 2017

### Buffu

Ok I will try.