Answer to Question #89012 in Mechanics | Relativity for Adnan

Question

A 40g marble with a 5mm radius is rolling at 10 revolutions per second, towards a spring attached to a wall. The spring is initially at its equilibrium length, and has a spring constant of 12 N/m. The marble hits the spring, and then eventually comes momentarily to rest. How far has the spring recoiled when the marble is at rest? 

Answer 1

Rotational energy of marble:

$$E_r=\frac{1}{2}I\omega^2=\frac{1}{2}(\frac{2}{5}mr^2)\omega^2$$

Translational energy of marble:

$$E_t=\frac{1}{2}(mr^2)\omega^2$$

Total energy of marble:

$$E=E_r+E_t=\frac{7}{10}mr^2\omega^2$$

From the conservation of energy:

$$E=\frac{1}{2}kx^2=\frac{7}{10}mr^2\omega^2$$

$$x=r\omega \sqrt{\frac{7m}{5k}}$$

$$x=0.005(10(2 \pi)) \sqrt{\frac{7(0.04)}{5(12)}}=0.021\ m=21 \ mm$$