Question

Write down the Lagrangian for a cylinder (mass $m$, radius $R$, and moment of inertia $I$ ) that rolls without slipping straight down an inclined plane which is at an angle $\alpha$ from the horizontal. Use as your generalized coordinate the cylinder's distance $x$ measured down the plane from its starting point. Write down the Lagrange equation and solve it for the cylinder's acceleration $\ddot{x}$. Remember that $T=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2,$ where $v$ is the velocity of the center of mass and $\omega$ is the angular velocity.

Short answer

The cylinder's acceleration is $\ddot{x}=-\frac{mg\sin (\alpha )}{m+\frac{I}{R^2}}$.

Step-by-step solution

Define the Kinetic Energy

The kinetic energy of the rolling cylinder is the sum of translational and rotational components. Using the formulas given, the translational kinetic energy is $T_{trans}=\frac{1}{2}m{v}^2$, and the rotational kinetic energy is $T_{rot}=\frac{1}{2}I{\omega }^2$. For rolling without slipping, the velocity $v=\dot{x}$ and the angular velocity $\omega =\frac{\dot{x}}{R}$. Substituting for $\omega$ into the rotational energy, we have $T_{rot}=\frac{1}{2}I{\left(\frac{\dot{x}}{R}\right)}^2$. Therefore, the total kinetic energy $T$ is $T=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}I{\left(\frac{\dot{x}}{R}\right)}^2$.

Define the Potential Energy

The potential energy $V$ of the cylinder when it has fallen a distance $x$ down the incline is $V=mgx\sin (\alpha )$, as the component of gravitational force acting along the incline is $mg\sin (\alpha )$.

Calculate the Lagrangian

The Lagrangian $L$ is obtained by taking the difference between the kinetic energy and potential energy: $L=T-V=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}I{\left(\frac{\dot{x}}{R}\right)}^2-mgx\sin (\alpha )$.

Write the Lagrange's Equation

Lagrange's equation is given by $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)-\frac{\partial L}{\partial x}=0$. First, calculate $\frac{\partial L}{\partial \dot{x}}=m\dot{x}+\frac{I}{R^2}\dot{x}$. Differentiating this with respect to time, we get $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)=(m+\frac{I}{R^2})\ddot{x}$. Next, compute $\frac{\partial L}{\partial x}=-mg\sin (\alpha )$. Use these results in Lagrange's equation: $(m+\frac{I}{R^2})\ddot{x}+mg\sin (\alpha )=0$.

Solve for the Acceleration

Rearrange the equation from Step 4 to solve for $\ddot{x}$: $\ddot{x}=-\frac{mg\sin (\alpha )}{m+\frac{I}{R^2}}$. This is the linear acceleration of the cylinder down the incline.

Key concepts

Kinetic Energy

Kinetic energy is a fundamental concept in physics that deals with the energy an object possesses due to its motion. For a rolling cylinder, kinetic energy is divided into two types: translational and rotational.

• Translational kinetic energy is the energy due to the motion of the cylinder's center of mass. It is given by the formula: $T_{trans}=\frac{1}{2}m{v}^2$, where $m$ is the mass and $v$ is the velocity.
• Rotational kinetic energy is the energy due to the cylinder spinning around its own axis. It is calculated using: $T_{rot}=\frac{1}{2}I{\omega }^2$, where $I$ is the moment of inertia and $\omega$ is the angular velocity.

In the case of a rolling cylinder without slipping, the linear velocity $v$ relates to the angular velocity $\omega$ through the radius $R$, as $\omega =\frac{v}{R}$. Hence, the total kinetic energy $T$ of the cylinder becomes: $$T=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}I{\left(\frac{\dot{x}}{R}\right)}^2$$ This combined energy tells us how much energy the cylinder has because of its overall motion.

Potential Energy

Potential energy is associated with the position of an object in a force field, typically a gravitational field. In the scenario of a cylinder rolling down an incline, the potential energy is influenced by its position along the slope.

• The potential energy $V$ of the cylinder is given by $V=mgx\sin (\alpha )$.
• Here, $x$ is the distance the cylinder has traveled down the incline.
• $\alpha$ is the angle of the incline with respect to the horizontal, while $m$ and $g$ represent the mass of the cylinder and gravitational acceleration, respectively.

Potential energy essentially quantifies the energy stored within the cylinder because of its elevated position on the slope. As it rolls downward, this energy is converted into kinetic energy, specifically the translational and rotational energies discussed before.

Lagrange's Equation

Lagrange's equation is a powerful tool in mechanics that allows us to derive equations of motion for systems. It is particularly useful when dealing with systems possessing complex constraints like the rolling cylinder. This equation links the kinetic and potential energies through the concept of the Lagrangian, $L$, defined as the difference between kinetic energy $T$ and potential energy $V$:
$$L=T-V$$ For the rolling cylinder, the Lagrangian is computed as: $$L=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}I{\left(\frac{\dot{x}}{R}\right)}^2-mgx\sin (\alpha )$$Using Lagrange's equation $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)-\frac{\partial L}{\partial x}=0$, we can derive the motion equation for the cylinder. This involves calculating the derivative of the Lagrangian with respect to velocity $\dot{x}$ and position $x$, and setting up the equation:

• The term $\frac{\partial L}{\partial \dot{x}}=m\dot{x}+\frac{I}{R^2}\dot{x}$ is differentiated over time $t$.
• Simultaneously, $\frac{\partial L}{\partial x}=-mg\sin (\alpha )$ is calculated.
• Inserting these into Lagrange's equation gives the acceleration $\ddot{x}$: $$\ddot{x}=-\frac{mg\sin (\alpha )}{m+\frac{I}{R^2}}$$

This result provides the acceleration with which the cylinder rolls down the incline. Understanding Lagrange's formulation helps simplify complex systems and is a cornerstone in classical mechanics.