Question

[Computer] Consider a massless wheel of radius $R$ mounted on a frictionless horizontal axis. A point mass $M$ is glued to the edge, and a massless string is wrapped several times around the perimeter and hangs vertically down with a mass $m$ suspended from its bottom end. (See Figure 4.28.) Initially I am holding the wheel with $M$ vertically below the axle. At $t=0,$ I release the wheel, and $m$ starts to fall vertically down. (a) Write down the Lagrangian $\mathcal{L}=T-U$ as a function of the angle $\varphi$ through which the wheel has turned. Find the equation of motion and show that, provided \(m

Short answer

Set up the Lagrangian, derive the motion equation, find equilibrium, then solve numerically.

Step-by-step solution

Analyze the System

The system is composed of a massless wheel with radius $R$ and two point masses: $M$ at the edge of the wheel, and $m$ hanging from a string wrapped around the wheel, both subject to gravity $g$. Initially, $M$ is directly below the axis.

Define the Lagrangian

The kinetic energy $T$ is due to the mass $m$ falling and the rotation of the mass $M$ around the wheel. The potential energy $U$ arises from the heights of both masses. The Lagrangian is given by $\mathcal{L}=T-U$.- Kinetic energy: $T=\frac{1}{2}M{R}^2{\dot{\varphi }}^2+\frac{1}{2}m(R\dot{\varphi }{)}^2=\frac{1}{2}(M+m)R^2{\dot{\varphi }}^2$.- Potential energy: $U=MgR(1-\cos \varphi )+mgR\varphi$.So, $\mathcal{L}=\frac{1}{2}(M+m)R^2{\dot{\varphi }}^2-(MgR(1-\cos \varphi )+mgR\varphi )$.

Find the Equation of Motion

Using the Euler-Lagrange equation $\frac{d}{dt}(\frac{\partial \mathcal{L}}{\partial \dot{\varphi }})-\frac{\partial \mathcal{L}}{\partial \varphi }=0$, we derive the equation of motion.Compute $\frac{\partial \mathcal{L}}{\partial \dot{\varphi }}=(M+m)R^2\dot{\varphi }$.Then, $\frac{d}{dt}(\frac{\partial \mathcal{L}}{\partial \dot{\varphi }})=(M+m)R^2\ddot{\varphi }$.Compute $\frac{\partial \mathcal{L}}{\partial \varphi }=MgR\sin \varphi -mgR$.Substitute into Euler-Lagrange: $$(M+m)R^2\ddot{\varphi }=MgR\sin \varphi -mgR.$$Simplifying gives:$$\ddot{\varphi }=\frac{g}{R}\left(\frac{M\sin \varphi -m}{M+m}\right).$$

Determine Equilibrium Positions

For equilibrium, $\ddot{\varphi }=0$, implying $M\sin \varphi =m$. Therefore, equilibrium positions occur at $\varphi ={\sin }^{-1}(\frac{m}{M})$. For stability, require $\varphi$ to be a minimum or maximum. Checking second derivative, stable equilibrium occurs if $m<M$.

Sketch Potential Energy for Part (b) and Interpret

Sketch $U(\varphi )=MgR(1-\cos \varphi )+mgR\varphi$ for $-\pi \le \varphi \le 4\pi$. The potential energy has periodic minima and maxima which represent equilibrium points. When $m<M$, there is one potential energy minimum corresponding to $\varphi ={\sin }^{-1}(\frac{m}{M})$.

Predict and Solve Motion Numerically for Part (c)

Set $M=g=R=1$ and $m=0.7$. Plot $U(\varphi )$ against $\varphi$. From rest, initially at $\varphi =0$, expect periodic oscillation due to the potential minima assuming small oscillations around this position. Verify this prediction by solving the derived equation of motion numerically using a suitable method (e.g., Runge-Kutta) for $0\le t\le 20$.

Repeat Calculation for Part (d) with Modified Parameters

Repeat numerical simulations with $m=0.8$. The shift in mass will alter the equilibrium positions, slightly increasing potential oscillation around the new stable position.

Key concepts

Stable Equilibrium

In Lagrangian mechanics, equilibrium positions represent states where the system experiences no net force, meaning it is at rest. An equilibrium is stable if, when slightly perturbed, the system tends to return to the equilibrium position. This happens because the potential energy is at a minimum.

For a massless wheel setup, as described in the exercise, the equilibrium points are found by setting the angular acceleration to zero, i.e., where $M\sin \varphi =m$. Hence, the position $\varphi ={\sin }^{-1}(\frac{m}{M})$ is an equilibrium point.

When $m<M$, the effect of gravity pulls the mass $M$ more strongly downward, creating a restoring force.

• This results in potential energy being minimized at these equilibrium points, indicating stability.

Any deviation from this point results in the wheel naturally trying to return to this equilibrium state, making it a stable equilibrium.

Potential Energy

Potential energy in this context arises from the gravitational forces acting on the masses attached to the wheel. It defines the energy storage due in part to their positions relative to a datum point, often taken as the lowest point in the gravitational field.

For the wheel system, the potential energy $U(\varphi )$ changes with the angle $\varphi$. It is calculated as:

• $U=MgR(1-\cos \varphi )+mgR\varphi$

The term $MgR(1-\cos \varphi )$ comes from the mass $M$ moving with the wheel, while $mgR\varphi$ accounts for the vertical motion of mass $m$.

Potential energy graphs can provide insights into motion. Minima on such graphs indicate stable equilibrium positions. The graph undulates with maxima and minima, representing various positions of high and low potential energy. When plotted for $-\pi \le \varphi \le 4\pi$, the periodic nature of the graph clearly shows the recurring points of stable equilibrium.”

Equation of Motion

The equation of motion arises from the Euler-Lagrange equation, which employs the Lagrangian to derive the dynamic behavior of the system.

The relevant equation for the wheel system:

• $$\ddot{\varphi }=\frac{g}{R}\left(\frac{M\sin \varphi -m}{M+m}\right)$$

This second-order differential equation links the angular acceleration $\ddot{\varphi }$ to the system's parameters.

When $M>m$, oscillatory behavior emerges due to this equation's nature. As it cannot be straightforwardly solved using elementary functions:

• Numerical methods, like Runge-Kutta, are employed to simulate and analyze the actual motion.

By tuning the mass values $M$ and $m$, one predicts how any disturbance from the equilibrium will evolve over time. Thus, numerical solutions verify predictions about the wheel's oscillations.