Question

A spring, of spring constant k, is attached to a cylinder of mass M and radius R. The cylinder rests on a smooth horizontal surface. When the spring is stretched from its equilibrium position and ‘let–go’ the cylinder starts rolling on the horizontal surface without slipping. The time period, of the resulting motion of the cylinder, is

(1) \(2\pi\sqrt{\frac{M}{2K}}\) 

(2) \(\pi\sqrt{\frac{6M}{k}}\)

(3) \(2\pi \sqrt{\frac{3M}{k}}\)

(4) \(\pi\sqrt{\frac{M}{2k}}\)

Answer 1

 (2) \(\pi\sqrt{\frac{6M}{k}}\)

Let I denote the moment of inertia of the cylinder about its own axis. We than have I = 1/2 MR².

At any instant when the spring has been stretched through a distance x from the its equilibrium position, let E be the total energy of the ‘spring–cylinder’ system. We have 

E = Potential energy of the stretched spring + Rotational K.E of the cylinder + translational K.E of the cylinder

Since dx/dt is not equal to 0, we have

The motion of the cylinder is, therefore, sample harmonic is nature. Its time period, T, is given by