Conservation of energy; 5 kg collar on curved rod

Conservation of energy in kinetics of particles

When a particle moves under conservative forces only (weight, elastic spring force), the sum of kinetic energy $T=\tfrac12mv^{2}$ and potential energy $V=V_g+V_e$ stays constant:

$$T_1+V_1=T_2+V_2,\qquad V_g=mgy,\quad V_e=\tfrac12kx^{2}$$

where $x$ is the spring deformation from its undeformed length. Forces that do no work (normal reaction of a smooth rod) can be ignored.

Collar problem

Assumed geometry (figure dimensions 150, 200, 250 mm): semicircular rod of radius $0.2$ m with A and C at the same level, B $0.2$ m below. The spring is fixed at D on the vertical axis through B, $0.15$ m below the line AC. Then the spring length is $L_A=L_C=\sqrt{0.2^{2}+0.15^{2}}=0.25$ m and $L_B=0.2-0.15=0.05$ m. Undeformed length $0.10$ m; $k=800$ N/m; $m=5$ kg.

Spring deformations: $x_A=x_C=0.25-0.10=0.15$ m (stretched) and $x_B=0.05-0.10=-0.05$ m (compressed).

a) At B

Released from rest at A (take B as datum, $h_A=0.2$ m):

$$0+mgh_A+\tfrac12kx_A^{2}=\tfrac12mv_B^{2}+0+\tfrac12kx_B^{2}$$ $$5(9.81)(0.2)+\tfrac12(800)(0.15)^{2}=\tfrac12(5)v_B^{2}+\tfrac12(800)(0.05)^{2}$$ $$9.81+9.00=2.5v_B^{2}+1.00\ \Rightarrow\ v_B^{2}=7.124,\ \ v_B=2.669\ \text{m/s}$$

b) At C

C is at the same height as A and the spring has the same length, so both potential energies are the same as at A:

$$\tfrac12mv_C^{2}=0\ \Rightarrow\ v_C=0$$

Answer: $v_B\approx2.67$ m/s and $v_C=0$ for the geometry assumed above. If the figure shows C at a different level or spring length, use the same energy equation with those values.