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# Tutor profile: Jim L.

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Jim L.
Aerospace Engineer for >30 years Tutors in Math, Mechanics and Programming
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

A tube of length, L, rotates about one end in the horizontal plane with a constant angular velocity, $$\Omega$$. The rotation is about the vertical axis. A particle of mass, m, is released in the tube from a radial distance of L/2. Find the radial and transverse components of velocity as the particle exits the end of the tube.

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Jim L.

We start by setting up our coordinate systems. In the inertial frame, N, the x and y axes are found in the horizontal plane, and the z axis is vertical. We attach a reference frame to the rotating tube as well. $$\hat{e}_r$$ is a unit vector that is parallel to the tube, $$\hat{e}_\t$$ is in the horizontal plane, perpendicular to the tube, pointing in the direction of rotation, and $$\hat{e}_n$$ is normal to the horizontal plane, pointing in the same direction as the inertial z axis. The radial (r), tangential (t), and normal (n) coordinate axes will be used to develop the acceleration , velocity, and position of the particle. The angular velocity of the Tube (T) in the inertial frame (N) is said to be constant, and is expressed as: $$^N\vec{\omega}^T = \Omega \hat{e}_n$$ The position of the ball can be expressed quite simply in the r,t,n coordinate frame as: $$\vec{r}_{P/O} = r \hat{e}_r$$ Here, we define P = point coincident with particle, and O = center of rotation of the tube. We take the derivative of the position vector with respect to time to determine the velocity vector: $$^N\vec{v}_{P} = \frac{^N d r \hat{e}_r}{dt} = \dot{r} \hat{e}_r + r \dot{\hat{e}}_r = \dot{r} \hat{e}_r + r \cdot {^N\vec{\omega}^T} \times \hat{e}_r =\dot{r} \hat{e}_r + \Omega r \hat{e}_t$$ Here, we made use of the kinematic relationship for unit vectors in rotating frames: $$\dot{\hat{e}} = \omega \times \hat{e}$$ We apply the vector differentiation formula to compute the acceleration of the particle in the inertial frame: $$^N\vec{a}_{P} = \frac{^Nd(^N\vec{v}_{P})}{dt} = \frac{^Td(^N\vec{v}_{P})}{dt} + ^N\vec{\omega}^T \times ^N\vec{v}_{P}$$ Substituting ... $$^N\vec{a}_{P} = \frac{^Td(\dot{r}\hat{e}_r + \Omega r \hat{e}_t)}{dt} + (\Omega \hat{e}_n) \times (\dot{r}\hat{e}_r + \Omega r \hat{e}_t)$$ We make use of the cyclical cross product relationships: $$\hat{e}_r \times \hat{e}_t = \hat{e}_n$$, $$\hat{e}_t \times \hat{e}_n = \hat{e}_r$$, and $$\hat{e}_n \times \hat{e}_r = \hat{e}_t$$ to expand the cross product above. We also take advantage of the fact that $$\Omega$$ is constant. Assembling terms along the component directions we get the following expression for the acceleration of the particle: $$^N\vec{a}_{P} = ( \ddot{r} - \Omega^2r) \hat{e}_r + 2\Omega \dot{r} \hat{e}_t$$ The particle is subject to transverse reaction forces from the tube, but there are no forces along the radial direction. So when we write our particle dynamics equation, $$\vec{F} = m \vec{a}$$, we have: $$\left(\begin{array}{c} 0 \\ R_{transverse} \end{array} \right) = m \cdot \left(\begin{array}{c} \ddot{r} - \omega^2r \\ 2\omega \dot{r} \end{array} \right)$$ Only, the radial component matters to the solution of the tube exit conditions (at $$r = L$$). The radial component of the vector equation reduces to the following linear second order homogeneous differential equation: $$\ddot{r} - \Omega^2 r = 0$$ The general solution to this equation is of the form: $$r(t) = A e^{\Omega t} + B e^{-\Omega t}$$ Also, $$\dot{r}(t) = A \Omega e^{\Omega t} - B \Omega e^{-\Omega t}$$ We use the initial conditions to solve for the constants $$A$$ and $$B$$: $$r(0) = \frac{L}{2} = A + B$$ Also, $$\dot{r}(0) = 0 = A - B$$ This gives us: $$A = B = L/4$$. So... $$r(t) = {L \over 4} ( e^{\Omega t} + e^{-\Omega t})$$ and $$\dot{r}(t) = { \Omega L \over 4} (e^{\Omega t} - e^{-\Omega t})$$ We make note that the exponential sums appearing in our solution for r and $$\dot{r}$$ are related to the hyperbolic trigonometric functions $$cosh(x) = (e^x+e^{-x})/2$$ and $$sinh(x) = (e^x-e^{-x})/2$$ Yielding: $$\boxed{r(t) = {L \over 2} \cosh( \Omega t) }$$ and $$\boxed{\dot{r}(t) = { \Omega L \over 2} \sinh(\Omega t) }$$ To compute the exit velocity, we first compute the exit angle ($$\Omega T$$) by applying the end condition to the position equation: $$r(T) = {L \over 2} \cosh( \Omega T) = L$$ Our exit radial position condition yields: $$\cosh( \Omega T) = 2$$ The so-called Pythagorean identity for hyperbolic trigonometric functions is $$cosh^2(x) + sinh^2(x) = 1$$. Since $$\cosh( \Omega T) = 2$$, we have: $$sinh^2 (\Omega T) = 3$$ Substituting into our $$\dot{r}$$ equation gives our exit radial velocity: $$\dot{r}(T) = { \Omega L \over 2} \sinh(\Omega T) = { \Omega L \over 2} \sqrt{3}$$ Recalling our velocity vector equation from above, we can compute the exit velocity we seek: $$^N\vec{v}_{P}(T) =\dot{r}(T) \hat{e}_r + \Omega r(T) \hat{e}_t = { \Omega L \over 2} \sqrt{3}\hat{e}_r + \Omega L \hat{e}_t$$

### Subject:Calculus

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Question:

Evaluate the integral, $$\int e^{-x^2} dx$$.

Inactive
Jim L.

This is an eminently useful, but deceptively simple integral. It finds use in the evaluation of probabilities resulting from a normal distribution. The student will come to quickly realize that there is no immediately available u-substitution, nor is there an apparent antiderivative of $$e^{-x^2}$$. Integration by parts gets us nowhere fast. So, how does one compute the indefinite integral? One, must resort to a series solution. We first note that the McLaurin series of $$e^u$$ is given by: $$e^u = 1 + \frac{u}{1} + \frac{u^2}{2!} + ... + \frac{u^n}{n!} + ...$$ This series has an infinite radius of convergence. Thus, $$e^{-x^2} = 1 + \frac{-x^2}{1} + \frac{({-x^2})^2}{2!} + ... + \frac{({-x^2})^n}{n!} + ...$$ or, $$e^{-x^2} = \sum_0^\infty (-1)^n \frac{x^{2n}}{n!}$$ While we cannot integrate the function directly, we can integrate the series... $$\int e^{-x^2} dx = \int \sum_0^\infty (-1)^n \frac{x^{2n}}{n!} dx$$ We can reverse the order of the appearance of the integral and the sum (as the integral of a sum is equal to the sum of the integrals): $$\int e^{-x^2} dx = \sum_0^\infty \int (-1)^n \frac{x^{2n}}{n!} dx$$ Factoring out the constants from the integral gives us: $$\int e^{-x^2} dx = \sum_0^\infty \frac{(-1)^n}{n!} \int x^{2n} dx$$ Using the polynomial rule of integration gives us the following series solution for the desired indefinite integral: $$\int e^{-x^2} dx = \sum_0^\infty \frac{(-1)^n}{n!} \frac{x^{2n+1}}{2n+1} + C$$ The series can be shown to converge for all x by the ratio test as follows: $$| \frac{a_{n+1}}{a_n} | = \frac{x^{2(n+1)+1}}{(n+1)!(2(n+1)+1)}\cdot \frac{n!(2n+1)}{x^{2n+1}} = \frac{x^2 (2n+1)}{(n+1)(2n+3)} = \frac{x^2 (2+1/n)}{(n+1)(2+3/n)} \rightarrow \frac{x^2}{(n+1)} \rightarrow 0 < 1$$ Thus, the series appearing in the indefinite integral has an infinite radius of convergence.

### Subject:C++ Programming

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Question:

The operation of splitting a string into tokens separated by delimiters (like spaces, tabs or commas), is so common, it is supported by the string object of virtually every language. C++ stands as one of the exceptions to this. How would you write a function that accepts a string, and a single character delimiter, and returns a vector containing a list of tokens from the string?

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Jim L.

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