# Tutor profile: Kartik S.

## Questions

### Subject: MATLAB

This problem will expose you to some MATLAB tools that can be used to determine the poles and zeroes of a linear, time-invariant dynamc system. If you don't know what a dynamic system is, don't worry. This exercise will show you how to find and plot the roots of a polynomial in the complex plane. Let: $( A = \begin{bmatrix} -2 & -2 & -1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix} ~~,~~ B = \begin{bmatrix} 2 \\ 0 \\ 0 \\ \end{bmatrix} ~~,~~ C = \begin{bmatrix} 0.5 & 0 & -2 \\ \end{bmatrix} ~~,~~ D = \begin{bmatrix} 0 \end{bmatrix} ~~,~~ I = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $) $$A,B,C,D$$ are state space coefficient matrices for a particular continuous-time LTI system. If you don't know what a dynamic system is, dont worry, just take the matrices at face value. $$(a)$$ Create a symbolic variable $$s$$ using $$\texttt{syms}$$. Create $$I$$ using $$\texttt{eye}$$. $$(b)$$ Construct the following block matrix using concatenation operations: $( M = \begin{bmatrix} sI - A & B \\ -C & D \\ \end{bmatrix} $) $$(c)$$ Compute $$N(s) = \det(M)$$ using $$\texttt{det}$$. Collect \textit{all} polynomial coefficients (including any that may be zero) into an array called $$\texttt{cN}$$ using $$\texttt{coeffs}$$. You might need to refer to online documentation. $$(d)$$ Compute $$D(s) = \det(sI-A)$$ using $$\texttt{det}$$. Collect $$\textit{all}$$ polynomial coefficients (including any that may be zero) into an array called $$\texttt{cD}$$ using $$\texttt{coeffs}$$. You might need to refer to online documentation. $$(e)$$ Use $$\texttt{roots}$$ to store the roots of $$N(s)$$ in an array called $$\texttt{rN}$$. Use $$\texttt{scatter}$$ to plot these roots in the complex plane using red, circular markers. $$(f)$$ Use $$\texttt{roots}$$ to store the roots of $$D(s)$$ in an array called $$\texttt{rD}$$. Use $$\texttt{scatter}$$ to plot these roots in the complex plane using blue, cross-shaped markers. Use $$\texttt{hold}$$ to overlay these new points on the same set of axes used in the previous part.

Completed program: $$a$$ $( \texttt{% Clean Up}\\ \texttt{close all; clear; clc;} \\ \\ \texttt{% State space matrices}\\ \texttt{A = [-2 -2 -1; 1 0 0; 0 1 0];}\\ \texttt{B = [2; 0; 0];}\\ \texttt{C = [0.5, 0, -2];}\\ \texttt{D = 0;}\\ \\ \texttt{% Part (a)}\\ \texttt{I = eye(3);}\\ \texttt{syms s}\\ \\ \texttt{% Part (b)}\\ \texttt{M = [s*I-A, B; -C, D]}\\ \\ \texttt{% Part (c)}\\ \texttt{N = det(M)}\\ \texttt{cN = coeffs(N, s, 'All')}\\ \\ \texttt{% Part (d)}\\ \texttt{D = det(s*I - A)}\\ \texttt{cD = coeffs(D, s, 'All')}\\ \\ \texttt{% Part (e)}\\ \texttt{rN = roots(double(cN))}\\ \texttt{scatter(real(rN), imag(rN), 'ro')}\\ \\ \texttt{% Part (f)}\\ \texttt{rD = roots(double(cD))}\\ \texttt{hold on;}\\ \texttt{scatter(real(rD), imag(rD), 'bx')}\\ \\ \texttt{% Always label your plots!}\\ \texttt{xlabel('Real Part')}\\ \texttt{ylabel('Imaginary Part')}\\ \texttt{grid on}\\ $)

### Subject: Applied Mathematics

Set up and solve a system of linear equations in $$K_1$$, $$K_2$$, and $$K_3$$ such that $$\textit{all}$$ roots of $$P(s) = s^3+\left(2K_{1}+2\right)s^2+\left(2K_{2}+2\right)s+2K_{3}+1$$ are located at some known value, $$s = -\lambda~,\lambda > 0$$.

Equating coefficients from: $( s^3+\left(2K_{1}+2\right)s^2+\left(2K_{2}+2\right)s+2K_{3}+1 = (s + \lambda)^3 = s^3 + 3 \lambda s^2 + 3 \lambda^2 s + \lambda^3 $) Results in the system of equations: $( \left\{ \begin{array}{l} 1 = 1\\ 2K_{1}+2 = 3\lambda \\ 2K_{2}+2 = 3\lambda^2 \\ 2K_{3}+1 = \lambda^3 \\ \end{array} \right. $) Which yields: $( K_1 = \frac{3\lambda - 2}{2} ~~,~~ K_2 = \frac{3\lambda^2 - 2}{2} ~~,~~ K_3 = \frac{\lambda^3 - 1}{2} $) Alternatively, $$P(s) = (s+\lambda)^3$$ implies that $$P(s)$$, $$\frac{dP(s)}{ds}$$, and $$\frac{d^2 P(s)}{ds^2}$$ are all zero at $$s = -\lambda$$. Thus: $( \left\{ \begin{array}{l} (-\lambda)^3+\left(2K_{1}+2\right)(-\lambda)^2+\left(2K_{2}+2\right)(-\lambda)+2K_{3}+1 = 0\\ 3(-\lambda)^2+2\left(2K_{1}+2\right)(-\lambda)+2K_{2}+2 = 0\\ 6(-\lambda)+2\left(2K_{1}+2\right) = 0\\ \end{array} \right. $) These equations are linear in $$K_1$$, $$K_2$$, and $$K_3$$ and can be solved by elimination. It can be verified that the values of $$K_1$$, $$K_2$$, and $$K_3$$ reported above also satisfy these equations.

### Subject: Physics

$$1.$$ A body's moment of inertia (about some axis) measures its tendency to resist angular acceleration. The purpose of this problem is to develop some intuition for the inertia presented by an object. Consider Object 1 - a solid, uniform cylinder of mass $$M$$, with radius $$R$$, and height $$H$$. $$(a)$$ Calculate the moment of inertia about a vertical axis passing through the center of the cylinder. Recall that the moment of inertia is computed by integrating over the entire body: $( J = \int r^2 dm $) $$(b)$$ How does the moment of inertia of a solid cylinder depend on its mass? $$(c)$$ How does the moment of inertia of a solid cylinder depend on its radius? $$(d)$$ How does the moment of inertia of a solid cylinder depend on its height? $$2.$$ Suppose we obtain Object 2 - a cylindrical object with mass $$M$$, outer radius $$R$$ and height $$H$$. However, Object 2 has a hole of radius $$R_0 < R$$ drilled through it. $$(a)$$ Give an expression for the density (mass per unit volume) of Object 2. $$(b)$$ Integrate over this new body to determine the moment of inertia about the same vertical axis as in the previous problem. This axis points out of the page in the top-view. $$(c)$$ Compare the moment of inertia of Object 2 to that of Object 1. $$(d)$$ If Object 1 and Object 2 were brought to the top of a hill and allowed to roll freely to the bottom, which object would reach the bottom first? (No slip, no air resistance, etc.). Why?

$$1.$$ $$(a)$$ Define the density (mass per unit volume) of the uniform cylinder as: $( \rho = \frac{M}{\pi R^2 H} \implies dm = \rho dV $) A cylindrical coordinate system is most convenient. Choose the origin to be the center of the base and integrate over the body: $( J = \int r^2 dm = \int r^2 \rho dV = \rho \int_{0}^{H} \int_{0}^{2\pi} \int_{0}^{R} r^2 ~ r~ dr ~ d\theta ~ dh \\ = \rho \int_{0}^{H} \int_{0}^{2\pi} \int_{0}^{R} r^3~ dr ~ d\theta ~ dh = \rho \int_{0}^{H} \int_{0}^{2\pi} \frac{R^4}{4} ~ d\theta ~ dh \\ = \rho \frac{2\pi H R^4}{4} = \frac{M}{\pi R^2 H} \frac{2\pi H R^4}{4} \\ = \frac{1}{2} MR^2 $) $$(b)$$ Linearly. See answer to part $$(a)$$. $$(c)$$ Quadratically. See answer to part $$(a)$$. $$(d)$$ No dependence. See answer to part $$(a)$$. $$2.$$ $$(a)$$ $( \rho_2 = \frac{M}{\pi (R^2 - {R_0}^2) H} $) $$(b)$$ A cylindrical coordinate system is most convenient. Choose the origin to be the center of the base and integrate over the body: $( J = \int r^2 dm = \int r^2 \rho_2 dV = \rho_2 \int_{0}^{H} \int_{0}^{2\pi} \int_{R_0}^{R} r^2 ~ r~ dr ~ d\theta ~ dh \\ = \rho_2 \int_{0}^{H} \int_{0}^{2\pi} \int_{R_0}^{R} r^3~ dr ~ d\theta ~ dh = \rho_2 \int_{0}^{H} \int_{0}^{2\pi} \frac{R^4 - {R_0}^4}{4} ~ d\theta ~ dh \\ = \rho_2 \frac{2\pi H (R^4 - {R_0}^4)}{4} = \frac{M}{\pi (R^2 - {R_0}^2) H} \frac{2\pi H (R^4 - {R_0}^4)}{4} \\ = \frac{M}{ (R^2 - {R_0}^2)} \frac{ (R^4 - {R_0}^4)}{2} = \frac{M}{ (R^2 - {R_0}^2)} \frac{ (R^2 + {R_0}^2)(R^2 - {R_0}^2)}{2} \\ = \frac{1}{2}MR^2 + \frac{1}{2}M{R_0}^2 $) $$(c)$$ $$M$$, $$R$$, and $$R_0$$ are strictly positive physical quantities, so the moment of inertia of Object 2 is greater than that of Object 1. $$(d)$$ Object 1 would reach first. Since both objects have the same mass ($$M$$), the friction forces (between the hill and the objects) acting on either object are equal. Since both objects have the same outer radius ($$R$$), the torque applied on each object (by friction) will be the same. Since Object 1 has a lower moment of inertia than Object 2 (as determined in part $$(c)$$), it experiences a greater angular acceleration due to the applied torque, and reaches the bottom of the hill first.

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