# A solid homogeneous cylinder of mass 2.30 kg is released from rest at the top of an incline of...

## Question:

A solid homogeneous cylinder of mass 2.30 kg is released from rest at the top of an incline of height 2.51 m and rolls without slipping to the bottom. The ramp is at an angle of 24.5 degrees to the horizontal. Calculate the speed of the cylinder's center of mass at the bottom of the incline.

## Rolling Motion:

The motion where the object is in the pure rotation simultaneously with translation motion is known as Rolling motion. Therefore, the center of mass of the object will have both rotational as well as translational kinetic energy.

## Answer and Explanation:

Given Data

Mass of the cylinder {eq}(m) = 2.3 \ kg {/eq}

Height of the plane {eq}(H) = 2.51 \ m {/eq}

Now, the initial energy of the cylinder would be

{eq}(E_{1}) = mgH {/eq}

At the bottom, the energy of the cylinder would be

{eq}E_{2} = \dfrac{1}{2}mv^{2} + \dfrac{1}{2}Iw^{2} \\ E_{2} = \dfrac{1}{2}mv^{2} + \dfrac{1}{2}(0.5mr^{2})\left(\dfrac{v}{r}\right)^{2} \\ E_{2} = \dfrac{1}{2}mv^{2} + \dfrac{1}{4}(mv^{2}) \\ E_{2} = \dfrac{3}{4}(mv^{2}) {/eq}

Now, from the energy conservation

{eq}E_{1} = E_{2} \\ mgH = \dfrac{3}{4}(mv^{2}) \\ 9.8 \times 2.51 = \dfrac{3}{4}(v^{2}) \\ v = 5.73\ m/s {/eq}

where

• v is the velocity of the cylinder