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A bar of soap (mass m) is at rest on a frictionless
Chapter 7, Problem 7.33(choose chapter or problem)
A bar of soap (mass m) is at rest on a frictionless rectangular plate that rests on a horizontal table. At time t = 0, I start raising one edge of the plate so that the plate pivots about the opposite edge with constant angular velocity \(\omega\), and the soap starts to slide toward the downhill edge. Show that the equation of motion for the soap has the form \(\ddot{x}-\omega^{2} x=-g \sin \omega t\), where x is the soap's distance from the downhill edge. Solve this for x (t), given that \(x(0)=x_{0}\). [You'll need to use the method used to solve Equation (5.48). You can easily solve the homogeneous equation; for a particular solution try \(x=A \sin \omega t\) and solve for A.]
Questions & Answers
QUESTION:
A bar of soap (mass m) is at rest on a frictionless rectangular plate that rests on a horizontal table. At time t = 0, I start raising one edge of the plate so that the plate pivots about the opposite edge with constant angular velocity \(\omega\), and the soap starts to slide toward the downhill edge. Show that the equation of motion for the soap has the form \(\ddot{x}-\omega^{2} x=-g \sin \omega t\), where x is the soap's distance from the downhill edge. Solve this for x (t), given that \(x(0)=x_{0}\). [You'll need to use the method used to solve Equation (5.48). You can easily solve the homogeneous equation; for a particular solution try \(x=A \sin \omega t\) and solve for A.]
ANSWER:Step 1 of 4
The Lagrangian equation of the soap consists of both translational and rotational motion, which can be calculated as:
\(\begin{array}{l}
L=E_{1}+E_{2}-U \\
L=\frac{1}{2} m x^{2}+\frac{1}{2} l \omega^{2}-m g h
\end{array}\)
Here, m is the mass of the soap, \(\dot x\) is the velocity, \(\dot \omega\) is the angular velocity, g is the gravitational acceleration, I is the moment of inertia and h is the height of the soap.
For \(I=m x^{2}\) and \(h=x \sin \theta=x \sin \omega t\).
\(\begin{array}{l}
L=\frac{1}{2} \dot m x^{2}+\frac{1}{2}\left(m x^{2}\right) \omega^{2}-m g(x \sin \omega t) \\
L=\frac{1}{2} \dot m x^{2}+\frac{1}{2} m x^{2} \omega^{2}-(m g x \sin \omega t)
\end{array}\)