Question

Consider a frictionless puck on a horizontal turntable that is rotating counterclockwise with angular velocity \(\Omega\). (a) Write down Newton's second law for the coordinates \(x\) and \(y\) of the puck as seen by me standing on the turntable. (Be sure to include the centrifugal and Coriolis forces, but ignore the earth's rotation.) (b) Solve the two equations by the trick of writing \(\eta=x+\) iy and guessing a solution of the form \(\eta=e^{-i \alpha t} .\) [In this case \(-\) as in the case of critically damped SHM discussed in Section \(5.4-\) you get only one solution this way. The other has the same form (5.43) we found for the second solution in damped SHM.] Write down the general solution. (c) At time \(t=0,\) I push the puck from position \(\mathbf{r}_{\mathrm{o}}=\left(x_{\mathrm{o}}, 0\right)\) with velocity \(\mathbf{v}_{\mathrm{o}}=\left(v_{\mathrm{xo}}, v_{\mathrm{yo}}\right)(\) all as measured by me on the turntable). Show that $$\left.\begin{array}{l} x(t)=\left(x_{0}+v_{x} t\right) \cos \Omega t+\left(v_{y o}+\Omega x_{o}\right) t \sin \Omega t \\ y(t)=-\left(x_{0}+v_{x 0} t\right) \sin \Omega t+\left(v_{y o}+\Omega x_{o}\right) t \cos \Omega t \end{array}\right\\}.$$ (d) Describe and sketch the behavior of the puck for large values of \(t .\) [Hint: When \(t\) is large the terms proportional to \(t\) dominate (except in the case that both their coefficients are zero). With \(t\) large, write (9.72) in the form \(x(t)=t\left(B_{1} \cos \Omega t+B_{2} \sin \Omega t\right),\) with a similar expression for \(y(t),\) and use the trick of (5.11) to combine the sine and cosine into a single cosine \(-\) or sine, in the case of \(y(t) .\) By now you can recognize that the path is the same kind of spiral, whatever the initial conditions (with the one exception mentioned).]

Short answer

The puck follows a spiral path dominated by terms proportional to time.

Step-by-step solution

Newton's Second Law in Rotating Frame

In a rotating frame, we must account for fictional forces. For a puck on a rotating turntable with angular velocity \( \Omega \), the centrifugal force and Coriolis force act on it. According to Newton's second law in the rotating frame:For the \( x \)-direction:\[ m \ddot{x} = -2m\Omega \dot{y} + m\Omega^2 x \]For the \( y \)-direction:\[ m \ddot{y} = 2m\Omega \dot{x} + m\Omega^2 y \]

Solving Using Complex Representation

Introduce \( \eta = x + iy \) and assume a solution of the form \( \eta = e^{-i \alpha t} \). Substituting into the differential equations we derived, convert them into a single complex equation:\[ m \frac{d^2 \eta}{dt^2} = -i2m\Omega \frac{d\eta}{dt} + m\Omega^2 \eta \]Substitute \( \eta = e^{-i \alpha t} \) into this equation to get:\[ -m \alpha^2 e^{-i \alpha t} = -2m\Omega \cdot (-i\alpha e^{-i \alpha t}) + m\Omega^2 e^{-i \alpha t} \]

Finding the Characteristic Equation

Simplify the equation to find the characteristic equation:\[ \alpha^2 + 2\Omega \alpha i = \Omega^2 \]Solve this by finding \( \alpha \) using the quadratic formula. The quadratic is:\[ \alpha = -i\Omega \]Therefore, the solution of the form is consistent, showing evidence of critical damping-like behavior.

General Solution

The general solution for \( \eta \) given the results, using \( e^{-i\alpha t} \), is of form:\[ \eta(t) = (A + Bt)e^{i\Omega t} \]Hence, converting back to \( x \) and \( y \), we have:\[ x(t) = x_0 + v_{x0}t \cos(\Omega t) + (v_{y0} + \Omega x_0)t \sin(\Omega t) \]\[ y(t) = - (x_0 + v_{x0}t)\sin(\Omega t) + (v_{y0} + \Omega x_0)t\cos(\Omega t) \]

Large Time Behavior

For large values of \( t \), terms proportional to \( t \) dominate. Consider:\[ x(t) = t \left( B_1 \cos(\Omega t) + B_2 \sin(\Omega t) \right) \] Use trigonometric identity to express this as a single cosine function:\[ B \cos(\Omega t + \phi) \] Similarly, apply the same to \( y(t) \). This indicates the puck will move in a spiral path over time, and the dominant axis of motion rotates.

Key concepts

Newton's Second Law

Newton's Second Law is the foundation for understanding motion in physics. In its simplest form, this law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This can be written as: \[ F = ma \]When dealing with rotating frames, such as a puck on a turntable, additional forces need to be considered. These forces arise due to the rotation and include both the Coriolis and centrifugal forces. In a rotating frame with angular velocity \( \Omega \), Newton's second law must include these fictitious forces. For the puck's coordinates, the equations become more complex:- In the \( x \)-direction: - The equation includes a Coriolis force, \(-2m\Omega \dot{y}\), which results from the rotation. - A centrifugal force, \(m\Omega^2 x\), acts outward from the axis of rotation. - In the \( y \)-direction: - The Coriolis effect appears with \(2m\Omega \dot{x}\). - A similar centrifugal term, \(m\Omega^2 y\), applies.Thus, the motion equations for the puck with respect to the rotating frame combine real and fictitious forces.

Coriolis Force

The Coriolis force is a fascinating impact observed in rotating systems. Named after the French engineer Gaspard-Gustave Coriolis, this force affects moving objects within a rotating frame.It's a "fictitious" force because it arises not from any physical interaction but from the rotation of the reference frame. - **Effect**: - It causes moving objects to deflect to the right in the northern hemisphere and to the left in the southern hemisphere. - On the rotating turntable, this force causes an object moving straight to curve, altering its path sideways.- **Mathematical Representation**: - In our exercise, for an object moving in the \( x-y \) plane, the force acts perpendicular to its velocity. - It appears in the equations as \(-2m\Omega\dot{y}\) for \( x \) direction and \(2m\Omega\dot{x}\) for \( y \) direction.Understanding the Coriolis force is crucial in predicting the motion of objects in rotating systems, whether in meteorology, astronomy, or simple physics experiments.

Centrifugal Force

When we talk about centrifugal force, we're referring to another crucial concept in rotating frames. Unlike the Coriolis force, the centrifugal force feels more intuitive to us.It's the sensation of being "thrown outwards" when moving in a circle—like when you feel pushed to the side in a car taking a sharp turn.- **Nature of the Force**: - It's a "fictitious" force because it doesn't result from any physical interaction but appears due to the inertia of the object's motion. - It acts radially outward from the center of rotation.- **Mathematical Expression**: - In the context of our problem, it contributes \(m\Omega^2 x\) in the \( x \) direction and \(m\Omega^2 y\) in the \( y \) direction.- **Consequences**: - This force balances the necessities of circular motion. It affects the object's perceived path as seen by observers within the rotating system.Recognizing the role of centrifugal force helps us comprehend how rotation can modify dynamics, giving a fuller picture of motion in both natural phenomena and engineered systems.