Conservation of energy; 2 kg collar on curved rod

a) Principle of conservation of energy

The total mechanical energy of a system, kinetic plus potential, remains constant when only conservative forces (gravity, springs) do work:

$$T_1+V_1=T_2+V_2$$

Example: a pendulum bob released from rest at height $h$ above its lowest point. At the top it has only potential energy $mgh$; at the bottom only kinetic energy $\tfrac12mv^{2}$. Hence $v=\sqrt{2gh}$. During the swing the energy changes between the two forms but the total is constant (air friction neglected).

b) Collar problem

Assumed geometry (figure dimensions 150, 200, 250 mm): semicircular rod of radius $0.2$ m, A and C at the same level, B $0.2$ m below them. The spring is fixed at D on the vertical axis through B, $0.15$ m below line AC, so $L_A=L_C=\sqrt{0.2^{2}+0.15^{2}}=0.25$ m and $L_B=0.05$ m. Undeformed length $0.10$ m, $k=800$ N/m, $m=2$ kg.

Deformations: $x_A=x_C=0.15$ m, $x_B=-0.05$ m.

At B (datum at B, $h_A=0.2$ m):

$$2(9.81)(0.2)+\tfrac12(800)(0.15)^{2}=\tfrac12(2)v_B^{2}+\tfrac12(800)(0.05)^{2}$$ $$3.924+9.00=v_B^{2}+1.00\ \Rightarrow\ v_B^{2}=11.924,\ \ v_B=3.453\ \text{m/s}$$

At C: same height and same spring deformation as at A, so $\tfrac12mv_C^{2}=0$ and $v_C=0$.

Answer: $v_B\approx3.45$ m/s; $v_C=0$ for the assumed geometry (use the same energy equation with figure values if C lies at a different level).