- #1
geoffrey159
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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 [itex] r(t) = r_0 - Vt [/itex] 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 [itex] a_{\theta}=0 [/itex] at any time, and therefore
[itex] \frac{\dot{\omega}}{\omega} = \frac{2V}{r_0-Vt} [/itex].
This can be rewritten
[itex] \frac{d}{dt}(ln(\omega) ) = -2 \frac{d}{dt}(ln(r_0-Vt)) [/itex] ,
which gives
[itex] \omega(t) = \omega_0 {(\frac{r_0}{r_0-Vt})}^2 [/itex].
Finally, computing this result into the equation of movement for radial direction gives me that the tension in the string is :
[itex] T(t) = \frac{m\omega_0^2r_0^4}{{(r_0-Vt)}^3} [/itex]
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.