- #1
PFuser1232
- 479
- 20
Homework Statement
A disk rotates with constant angular velocity ##ω##, as shown (attached image). Two masses, ##m_A## and ##m_B##, slide without friction in a groove passing through the center of the disk. They are connected by a light string of length ##l##, and are initially held in position by a catch, with mass ##m_A## at distance ##r_A## from the center. Neglect gravity. At ##t = 0## the catch is removed and the masses are free to slide.
Find ##\ddot{r}_A## immediately after the catch is removed, in terms of ##m_A##, ##m_B##, ##l##, ##r_A##, and ##ω##.
Homework Equations
$$\sum_{}^{} \vec{F}_r = m(\ddot{r} - r \dot{\theta}^2) \hat{r}$$
3. The Attempt at a Solution
The particles are constrained to move according to ##r_A + r_B = l##. Differentiating twice with respect to time, we get ##\ddot{r}_A = -\ddot{r}_B##. Each particle experiences a force of magnitude ##T## where ##T## is the tension in the string. Writing down the equations of motion:
$$-T = -m_A(\ddot{r}_A - r_A \omega^2)$$
$$-T = -m_B(\ddot{r}_B - r_B \omega^2)$$
Eliminating ##T## and substituting ##r_B = l - r_A## and ##\ddot{r}_B = -\ddot{r}_A## we get:
$$\ddot{r}_A = \omega^2 \frac{m_A r_A + m_B r_A - m_B l}{m_A + m_B}$$
Is this answer correct?
Also, before ##t = 0##, were ##m_A## and ##m_B## acted upon by an additional force ##F## such that ##|F - T| = mr\omega^2##? Or was the string fixed?
It seems to me that my physical intuition was wrong (again). I don't understand why the ##\ddot{r}## terms appear in the equations of motion. Why can't the masses remain at a fixed distance from the circle?
My last question is really about the polar coordinate system and the constraint I used above, namely ##l = r_A + r_B##.
If the disk was big enough, and the string was entirely on one side of the circle (the origin no longer lies on the string), then we would be forced to interpret the radial coordinate of the particle closer to the pole as negative, otherwise the sum of the radial coordinates would not equal the length of the string, right?
Sometimes, I find the non-uniqueness of polar coordinates really frustrating