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wdednam
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Another Lagrangian problem: Bead sliding along a horizontal rotating ring
A horizontal ring of mass M and radius a rotates freely about a vertical axis passing through a point on its circumference. If a bead of mass m slides along the ring without friction, what is the Lagragian function of the bead-ring system? Also write down the Kinetic energy of the bead if the ring rotates at a constant angular velocity \omega about the vertical axis.
T = \displaystyle{\frac{1}{2}}I\dot{\theta}^2 + \displaystyle{\frac{1}{2}}m\bold{v}^2
L = T - V
I = I_{CM} + MR^2 = 2Ma^2 in this case
\bold{v}^2 = \bold{v_{radial}}^2 + \bold{v_{tangential}}^2
Okay, since the motion is horizontal, the gravitational potential energy of the system is constant and the Lagragian function of the system will simply be the kinetic energy of the system: L = T. Now, I can write down an expression for the kinetic energy of the ring (it is simply Ma^2\dot{\theta}^2) and the radial kinetic energy of the bead (\displaystyle{\frac{1}{2}}m\bold{v}^2 = \displaystyle{\frac{1}{2}}m\dot{r}^2 right?), but I don't know what the tangential component of the bead's velocity will look like. Is it simply the algebraic sum of the bead's and ring's tangential velocities? I'd appreciate any help, thanks.
Homework Statement
A horizontal ring of mass M and radius a rotates freely about a vertical axis passing through a point on its circumference. If a bead of mass m slides along the ring without friction, what is the Lagragian function of the bead-ring system? Also write down the Kinetic energy of the bead if the ring rotates at a constant angular velocity \omega about the vertical axis.
Homework Equations
T = \displaystyle{\frac{1}{2}}I\dot{\theta}^2 + \displaystyle{\frac{1}{2}}m\bold{v}^2
L = T - V
I = I_{CM} + MR^2 = 2Ma^2 in this case
\bold{v}^2 = \bold{v_{radial}}^2 + \bold{v_{tangential}}^2
The Attempt at a Solution
Okay, since the motion is horizontal, the gravitational potential energy of the system is constant and the Lagragian function of the system will simply be the kinetic energy of the system: L = T. Now, I can write down an expression for the kinetic energy of the ring (it is simply Ma^2\dot{\theta}^2) and the radial kinetic energy of the bead (\displaystyle{\frac{1}{2}}m\bold{v}^2 = \displaystyle{\frac{1}{2}}m\dot{r}^2 right?), but I don't know what the tangential component of the bead's velocity will look like. Is it simply the algebraic sum of the bead's and ring's tangential velocities? I'd appreciate any help, thanks.
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