Lagrangian Equation

Question

A hoop with radius r is rolling, without slipping, down an inclined plane of length l and with angle of inclination 'a'. Assign appropriate generalized coordinates to the system. Determine the constraints, if any. Write down the Lagrangian equations for the system. Hence or otherwise determine the velocity of the hoop at the bottom of the inclined plane.

Answer 1

Define the generalized coordinates $\theta$ and $x$ as the hoop angular displacement (rotation) and the linear displacement along the plane, respectively, with $x$ pointing downward.

See that, since the hoop can't slip, the tangential velocity of the hoop in contact to the plane must be equal to the plane velocity, i.e., 0. Therefore, the "forward" velocity of the hoop $\dot{x}$ must be equal to the "backward" velocity $r \dot{\theta}$. Then, you have the constraint $\dot{x} = r \dot{\theta}$, that ensure that the hoop is rolling, not slipping.

The kinetic energy of the hoop is given by the sum of the kinetic energy of each "kind" of motion: $$ T = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} J \dot{\theta}^2 $$ The first term corresponds to the energy of the linear movement ($m$ being the mass) and the second term to the angular movement ($J$ is the moment of inertia).

The potential energy is simply $$ V = -mgx \sin a. $$ Therefore, the Lagrangian is $L = T-V$, or $$ L = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} J \dot{\theta}^2 + mgx \sin a. $$ Introducing the constraint $\dot{x} = r \dot{\theta}$, $$ L = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} \frac{J}{r^2} \dot{x}^2 + mgx \sin a. $$

From the Lagrange equation, that, for our case, is $$ \frac{d}{dt} \frac{\partial L}{\partial \dot{x}} - \frac{\partial L}{\partial x} = 0, $$ we have $$ \left( m + \frac{J}{r^2} \right) \ddot{x} + mg\sin a = 0. $$

We can obtain the final velocity of the hoop through the solution of this equation, but we can use the conservation of the total energy $E = T+V$ instead. See that $$ T = \left( m + \frac{J}{r^2} \right) \frac{\dot{x}^2}{2}, $$ then the quantity $$ E = \left( m + \frac{J}{r^2} \right) \frac{\dot{x}^2}{2} - mgx\sin a $$ is constant for all the trajectory. In the initial condition, $x = 0$ and $\dot{x} = v_0$, then $$ E = \left( m + \frac{J}{r^2} \right) \frac{v_0^2}{2}. $$ In the end of the plane, $x=l$ and $\dot{x} = v_{end}$. Then, $$ E = \left( m + \frac{J}{r^2} \right) \frac{v_0^2}{2} = \left( m + \frac{J}{r^2} \right) \frac{v_{end}^2}{2} - mgl\sin a $$ $$ v_{end}^2 = v_0^2 + 2\left( 1 + \frac{J}{r^2m} \right)^{-1} gl\sin a $$ The moment of inertia of a hoop is $r^2m$, therefore,

$$ v_{end} = \sqrt{v_0^2 + gl\sin a} $$