A uniform cylinder of mass 5 kg and radius 0.2 m rolls smoothly over a horizontal surface in a straight line with a velocity of 2 m/s. 

Find 

(i) the speed of the particle situated at the top of the cylinder and 

(ii) kinetic energy of the rolling cylinder. 

The speed of the particle at the top of the cylinder is:

v = 2v_c = 2 × 2 = 4 m/s

The kinetic energy of the rolling solid cylinder is:

K = (1/2) I_c ω² + (1/2) Mv_c²

To evaluate this, we find the angular velocity and moment of inertia of the cylinder.

Angular velocity of the cylinder:

ω = v_c / R = 2 / 0.2 = 10 rad/s

Moment of inertia of the cylinder about its axis:

I_c = (MR²) / 2 = (5 × 0.2²) / 2 = 0.1 kg·m²

Putting values:

K = (1/2 × 0.1 × 10²) + (1/2 × 5 × 2²) = 15 J