= Solution
{c}
All <forces> on the first particle lie in the <plane> spanned by $\mathbf e_r$ and $\mathbf e_z$: this holds for <gravity>, <tension>, and the <normal reaction>. Hence the azimuthal component of <Newton's second law> vanishes:
$$
m_1(r_1\ddot\theta+2\dot r_1\dot\theta)=0.
$$
Using the <product rule> gives
$$
\boxed{\dot h=\frac{d}{dt}(r_1^2\dot\theta)=r_1(r_1\ddot\theta+2\dot r_1\dot\theta)=0.}
$$
Strictly, $h$ is the axial <specific angular momentum>; the actual axial <angular momentum> is $m_1h$, which is conserved as well.
Differentiate the stated total <energy> using the <chain rule> and then substitute the two <Newton's second law> equations:
$$
\begin{aligned}
\dot E&=m_1\dot{\mathbf r}_1\cdot\ddot{\mathbf r}_1+m_1g\dot z_1+m_2\dot{\mathbf r}_2\cdot\ddot{\mathbf r}_2+m_2g\dot z_2\\
&=\mathbf N\cdot\dot{\mathbf r}_1+\mathbf T_1\cdot\dot{\mathbf r}_1+\mathbf T_2\cdot\dot{\mathbf r}_2.
\end{aligned}
$$
The <gravitational potential energy> terms have canceled the <power> of <gravity>. Also $\mathbf N$ is proportional to $\mathbf e_r+\mathbf e_z$, whose <dot product> with $\dot{\mathbf r}_1$ is $\dot r_1+\dot z_1=0$. Thus the stationary <smooth mechanical constraint> does no <work>. The remaining <tension> <powers> cancel by part (b), proving \b[$\dot E=0$ and conservation of total <energy>] while the <massless inextensible string> remains taut and contact is maintained.
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