Two masses connected by spring rotate around one axis

  • #1

Homework Statement



Take the x-axis to be pointing perpendicularly upwards.

Mass ##m_1## slides freely along the x-axis. Mass ##m_2## slides freely along the y-axis. The masses are connected by a spring, with spring constant ##k## and relaxed length ##l_0##. The whole system rotates with constant angular velocity ##\omega## around the x-axis. Determine the Lagrangian in terms of generalized coordinates ##x## and ##y##.

Homework Equations



$$L = T - U$$
$$F=mg$$
$$F=-kx$$
$$P.E. = mgh$$
$$P.E. = -\frac{1}{2}kx_e^2$$


The Attempt at a Solution



So ##m_1## is affected by gravity so we have ## -m_1gx ##. The potential in the string is ##\frac{1}{2}k(d-l_0)^2## where ##d^2 = x^2 + y^2##.

So $$U = -m_1gx + \frac{1}{2}k(d-l_0)^2$$

$$T = \frac{1}{2}m_1\dot{x}^2 + \frac{1}{2}m_2\dot{y}^2$$

Is this correct? It feels wrong, but I don't know why. I think my ##T## is wrong though. Shouldn't it be zero? But if it is, I cannot get any eom, later on. I am confused by the fact that I am working within a non-inertial frame, the rotating one.
 
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Answers and Replies

  • #2
Orodruin
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Why do you think the kinetic energy should be zero? Also, how do you take into account the fact that the system is rotating? What will this add to your energies?

While you are working with what is coordinates in a non-inertial frame, nothing stops you from thinking about how they relate to coordinates in an inertial frame.
 
  • #3
Just to close this thread.

The solution is that you consider the inertial frame of reference and write your Lagrangian in cylinderical coordinates. Then it turns out that you can reformat it into ##x## and ##y## since the angular speed is a constant.
 

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