Rolling cylinder attached to a spring

Given: $m = 15$ kg, $r = 0.25$ m, $k = 400$ N/m, initial spring extension $x = 0.3$ m (released from rest). Rolling without slipping, so $\omega = v/r$.

  |/\/\/\/\/\---(o)-->       cylinder rolls; spring is
  |  k = 400                 unstretched when x = 0

Moment of inertia

$$I_G = \tfrac12 mr^2 = \tfrac12(15)(0.25)^2 = 0.4688\ \text{kg·m}^2$$

Work done

Friction at the rolling contact, the normal reaction and the weight do no work (the contact point is instantaneously at rest and the motion is horizontal). Only the spring does work, as its extension goes from 0.3 m to zero:

$$U_{1\to2} = \tfrac12 k x^2 = \tfrac12(400)(0.3)^2 = 18.0\ \text{J}$$

Kinetic energy

Position 1: at rest, $T_1 = 0$.

Position 2 (spring unstretched):

$$T_2 = \tfrac12 m v^2 + \tfrac12 I_G\omega^2 = \tfrac12 mv^2 + \tfrac12\left(\tfrac12 mr^2\right)\frac{v^2}{r^2} = \tfrac34 mv^2$$

Work-energy equation

$$0 + 18.0 = \tfrac34(15)v^2 = 11.25\,v^2$$ $$v = \sqrt{\frac{18.0}{11.25}} = 1.265\ \text{m/s} \qquad \omega = \frac{v}{r} = \frac{1.265}{0.25} = 5.06\ \text{rad/s}$$

Check of the energy split: translational $\tfrac12mv^2 = 12.0$ J and rotational $\tfrac14 mv^2 = 6.0$ J, total $18.0$ J.

Answer: v = 1.26 m/s; ω = 5.06 rad/s.