A mass \(m\) lies on a table and is connected to a string of length \(\ell\) as shown in Figure 10.45. The string passes through a hole in the table and is connected to another mass \(M\) that is hanging in the air. We assume that the string remains taught and that mass \(M\) can only move vertically. a. The Lagrangian for this setup is $$ \mathcal{L}=\frac{1}{2} M \dot{r}^{2}+\frac{1}{2} m\left(\dot{r}^{2}+r^{2} \dot{\theta}^{2}\right)+M g(\ell-r) $$ where \(r\) and \(\theta\) are polar coordinates describing where mass \(m\) is on the table with respect to the hole. Explain why the terms in the Lagrangian are appropriate. b. Derive the equations of motion for \(r(t)\) and \(\theta(t)\). c. What angular velocity is needed for mass \(m\) to maintain uniform circular motion?

Short Answer

Expert verified
The terms in the Lagrangian represent the kinetic energy of the two masses and the potential energy of the hanging mass 'M'. The Euler-Lagrange equation is used to derive the equations of motion for \( r(t) \) and \( \theta(t) \). Uniform circular motion is achieved when the net radial force is zero, by substituting this condition into the \( r \) equation of motion, the required angular velocity for maintaining uniform circular motion is obtained.

Step by step solution

01

Understanding the Lagrangian

The Lagrangian function, represented as \( \mathcal{L} \), is defined as the kinetic energy subtracted by the potential energy of a system. In our specific case, the kinetic energy of the system is the kinetic energy of the two masses. The potential energy is the gravitational potential energy of the hanging mass 'M'. The potential energy changes as 'M' moves up or down, which is represented by \( \ell - r \). Thus each term in the Lagrangian represents an aspect of the energy of the system.
02

Deriving the equations of motion

The equation of motion can be obtained using the Euler-Lagrange equation, which states that derivative of Lagrangian with respect to the general coordinates minus its derivative with respect to the time derivative of the general coordinates must be equal to zero. Two equations can be obtained by applying this to the two general coordinates \( r \) and \( \theta \).
03

Deriving expression for angular velocity

For uniform circular motion, the net radial force is zero. Applying this condition to the \( r \) equation of motion will give the expression for angular velocity.

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