Question

A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill, starting from rest. if the angular speed of the disk is 4.27 rad/s at the bottom, how high did it start on the hill

Answer 1

Answer:

Height reached by the ball, h = 3.57 meters

Explanation:

It is given that,

Mass of the disk, m = 42 kg

Diameter of the disk, d = 3.2 m

Radius, r = 1.6 m

Angular speed of the disk, $\omega=4.27\ rad/s$

The kinetic energy of the disk is equal to its potential energy. Using the conservation of energy as :

$\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2=mgh$

$\dfrac{1}{2}mv^2+\dfrac{1}{2}(mr^2/2)\omega^2=mgh$

$\dfrac{1}{2}v^2+\dfrac{1}{2}(r^2/2)\omega^2=gh$

$\dfrac{1}{2}(r\omega)^2+\dfrac{1}{2}(r^2/2)\omega^2=gh$

$\dfrac{1}{2}(1.6\times 4.27)^2+\dfrac{1}{2}(1.6^2/2)\times 4.27^2=gh$

$h=\dfrac{35.0071}{9.8}$

h = 3.57 meters

So, the solid disk will reach to a height of 3.57 meters.