Kleppner Kolenkow 2.34 (mass whirling around on a string)

[geoffrey159]

Homework Statement

A mass m whirls around on a string which passes through a ring. Neglect gravity. Initially, the mass is distance $r_0$ from the center and is revolving at angular velocity $\omega_0$. The string is pulled with constant velocity $V$ starting at $t=0$ so that the radial distance to the mass decreases. Find $\omega(t)$ and string tension applied to the mass.

Hint: $Vt = \frac{r_0}{2}$ implies $\omega = 4\omega_0$

Homework Equations

Newton equations of movment in tangential and radial directions.

The Attempt at a Solution

Firstly, because of the junction between the mass and the string, I find that the radial speed of the mass is $\dot{r}(t) = -V$. Therefore, the radial distance should be r(t) = r_0 - Vt for $0\le t \le \frac{r_0}{V}$, and 0 for any later time.
Then, using Newton's second law in tangential direction, I get that a_{\theta}=0 at any time, and therefore
\frac{\dot{\omega}}{\omega} = \frac{2V}{r_0-Vt}.
This can be rewritten
\frac{d}{dt}(ln(\omega) ) = -2 \frac{d}{dt}(ln(r_0-Vt)) ,
which gives
\omega(t) = \omega_0 {(\frac{r_0}{r_0-Vt})}^2.
Finally, computing this result into the equation of movement for radial direction gives me that the tension in the string is :
T(t) = \frac{m\omega_0^2r_0^4}{{(r_0-Vt)}^3}

4. Question
Even though it works for the hint, there is something very wrong with my proof because angular velocity and string tension are unbounded as $r(t) \rightarrow 0$. It doesn't make sense to me. Thanks for taking the time.

 

[Bystander]

What's your objection to your result beyond "not making sense?"

 

[geoffrey159]

Hi, thanks for replying.
It seems unlikely to me that angular speed and string tension tend to infinity as $r(t) \rightarrow 0$. That's why I think I'm wrong, very wrong :-)

 

[Bystander]

Angular momentum is conserved --- if your "ice skater" is not just anorexic but achieved the physique of a stick figure, and pulls stick arms into the point that they're parallel and co-located with the stick body, r in I = mr², the angular moment of inertia, goes to zero, I goes to zero, and angular momentum = I x omega pretty much demands an infinite rotational speed.

 

[geoffrey159]

I don't know what is momentum yet, I'm a beginner in physics

 

[Bystander]

Have you had to watch the Olympic figure skating to keep other people in the house happy enough to let you watch the hockey?

 

[geoffrey159]

Huh, yes maybe. I've not been watching TV lately. What do you mean?

 

[Bystander]

When ice skaters go into a spin for a big finish, they pull their arms in along their rotational axis, and the spin rate increases quite dramatically. Conservation of angular momentum. If you've got a young enough physics prof/instructor, he/she'll stand on a rotating turntable and do the same thing as a demonstration.

 

[geoffrey159]

By dramatically, you mean infinitly ?

 

[Bystander]

They never get to the zero radius for their angular moments of inertia, so, "no."

 

[geoffrey159]

So what you say is that it'll be clear when I'll study angular momentum ? The resulat seems correct to you then ?

 

[Bystander]

It looks great --- you dun good.

 

[geoffrey159]

Thank you for your explanations and for giving some of your time.
Geoffrey

 

[johnnyhgrace]

Hey, hopefully someone sees this. I had the same problem for a homework problem and I got the answer wrong. In the solutions, my instructor says the following.

Since there is no tangential acceleration therefore we have:
$$\dot{r}\omega +2r\dot{\omega}=0$$

And then proceeds to OP's correct answer. I don't understand where the two comes from? I did:
$$v_t =\omega r$$
$$\frac{d v_t}{dt} = \frac{d}{dt}(\omega r)=\dot{r}\omega +r\dot{\omega}=0$$

What did I do wrong?

 

[kuruman]

Hey, hopefully someone sees this. I had the same problem for a homework problem and I got the answer wrong. In the solutions, my instructor says the following.

Since there is no tangential acceleration therefore we have:
$$\dot{r}\omega +2r\dot{\omega}=0$$

And then proceeds to OP's correct answer. I don't understand where the two comes from? I did:
$$v_t =\omega r$$
$$\frac{d v_t}{dt} = \frac{d}{dt}(\omega r)=\dot{r}\omega +r\dot{\omega}=0$$

What did I do wrong?

Derive the general expression for the acceleration starting with $$\mathbf{a}=\frac{d^2\mathbf{r}}{dt^2}=\frac{d^2(r\mathbf{\hat r})}{dt^2}$$and apply it to the specific problem that you have. If you still need help, please start a new thread. Recycling old threads is frowned upon.

 

[haruspex]

It's because the acceleration in the tangential direction is not the same as the rate of change of tangential speed. Just like the radial component is not the same as the rate of change of radial speed.
See e.g. https://proofwiki.org/wiki/Acceleration_Vector_in_Polar_Coordinates