Question

A solid ball of mass M = 2.0 kg and radius R = 0.25 m starts from rest at a height h = 3.0 m above the bottom of the path. It rolls without slipping down the left side of the path. The right side of the path is frictionless. Moment of inertia of a hollow sphere is � = ! ! ��!. v What is the linear speed

Answer 1

Answer:

6.48 m/s

Explanation:

We are given that

Mass,M=2 kg

Radius,R=0.25 m

Height,h=3 m

Moment of inertia of solid sphere=$I=\frac{2}{5}MR^2$

We have to find the linear speed.

$\omega=\frac{v}{R}$

By law of conservation of energy

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

$mgh=\frac{1}{2}I(\frac{v}{R})^2+\frac{1}{2}mv^2=\frac{1}{2}v^2(\frac{I}{R^2}+m)$

Where $g=9.8m/s^2$

Substitute the values

$2\times 9.8\times 3=\frac{1}{2}(\frac{2}{5R^2}MR^2+M)=\frac{7}{10}Mv^2=\frac{7}{10}(2)v^2$

$v^2=\frac{2\times 9.8\times 3\times 10}{7\times 2}$

$v=\sqrt{\frac{2\times 9.8\times 3\times 5}{7}}$

$v=6.48m/s$