Consider a L$-$shaft rotating at constant angular speed of $=10ra\frac{d}{s}.$ A $1-kg$ mass is attached to the end of the shaft by a massless rod with length of $L=1\mathrm{.}2$meters. The horizontal dimension of the shaft is $0\mathrm{.}7$ m $.$

$m{L}^2\frac{d^2(t)}{d{t}^2}-m{}^{2}L(d+$Lsin$(t))cos(t)+$mgLsin$(t)=0$

Find the equilibrium position between $0$ and $90$ degrees, and linearize the equation of motion about this equilibrium position. The linearized equation of motion should be of the form, $\frac{d^2(t)}{d{t}^2}+a(t)=b.$

What is $a?$ Use $g=9\mathrm{.}81\frac{m}{s^2}.$ Keep $3$ significant figures and write the answer in scientific notation. For example, $130\mathrm{.}2$ is written as $1\mathrm{.}30$ e $2.$