Question

A solid cylinder of diameter D=0.92m undergoes a pure rolling motion on an inclined plane from rest at A.

The height of the incline is h=9m.

What is the speed of the center of mass of the cylinder (vG) at B?

Answer 1

method - wise, whilst there is no drawing, this seems to suit a conservation of energy approach .

the lost gravitational potential energy of the cylinder as it rolls *down* the incline becomes translational and rotation kinetic energy of the cylinder

\(mg\Delta h = \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2\) where \(v = \omega \frac{D}{2}\) and I is I for a solid cylinder...

v is the speed of the centre of mass

Answer 2

Thank you very much!!! Tried your way out and it worked out perfectly, was only considering pure rotation (don't know why).