A solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane of length $L$ and height $h$. What is the speed of its centre of mass when the cylinder reaches its bottom: (2003)
Solution:
Using conservation of energy, potential energy equals total kinetic energy: $Mgh = \frac{3}{4}Mv^2$. Solving for velocity gives $v = \sqrt{\frac{4}{3}gh}$.
Leave a Reply