$!!!$Dont need the answer!!!! I just want to know why i only convert x$2$ to r$*$theta and not x$1!!!10$ pts $)$ A mechanical system schematic is shown below, consisting of a lever of length $2r$ and rotational inertia $J$ two masses $m_1$ and $m_2,$ and two springs $k_1$ and $k_2.$ The lever pivot is frictionless, and there is no applied torque. The springs are relaxed when $=0$ and $x_1=0.$ Include gravity in your analysis. Also use the small$-$angle approximations $(sin=,cos$~~$1).$

a$)$ How many degrees of freedom does the system have? $(1$ pt$)2$ Degrees of Freedom

b$)$ Write the kinetic energy in the system in terms of the coordinate derivatives $x_1^{}$ and ${}^{}.(2$pts$)$

$T=\frac{1}{2}J{}^{}{?}^2+\frac{1}{2}{m}_1{x}^{}{?}^2+\frac{1}{2}{m}_2{x}_2^{}{?}^2,x_2=rsin$~~$r$

$=>,T=\frac{1}{2}(J+m_2{r}^2){}^{}{?}^2+\frac{1}{2}{m}_1{x}_1^{}{?}^2,x_2^{}=r{}^{},x_2^{}=r{}^{}$

c$)$ Write the potential energy in the system in terms of the coordinates $x_1$ and $,(2$ pts$)$

$V=\frac{1}{2}{k}_1(rsin{)}^2+\frac{1}{2}{k}_2(x_1-rsin{)}^2+m_{2}g{x}_2-m_{1}g{x}_1$

$V$~~$\frac{1}{2}{k}_1{r}^2{}^2+\frac{1}{2}{k}_2(x_1-r{)}^2+m_{2}gr-m_{1}g{x}_1$