Question

A solid, uniform sphere of mass 2.0 kg and radius 1.8 m rolls from rest without slipping down an inclined plane of height 7.5 m. What is the angular velocity of the sphere at the bottom of the inclined plane

Answer 1

Answer:

$w^2=5.5rads/s$

Explanation:

From the question we are told that:

Mass $m=2.0kg$

Radius $r=1.8m$

Height $h=7.5m$

Generally the equation for Potential energy is mathematically given by

Potential energy=Kinetic energy+Rotational energy

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

Since there is no slipping

 $v=rw$

Therefore

 $mgh=\frac{1}{2}mr^2w^2+\frac{1}{2}Iw^2$

Where

      $I=\frac{1}{2}mr^2$

      $l=3.24m$

 $2*9.81*7.5=\frac{1}{2}(2)(1.8)^2w^2+\frac{1}{2}(3.24)w^2$

 $147.15=3.24w^2+1.62w^2$

 $w^2=\frac{147.15}{4.86}$

 $w^2=\sqrt{\frac{147.15}{4.86}}$

 $w^2=5.5rads/s$