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

`w^2=5.5rads/s`

Step-by-step 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=(1)/(2)mv^2+(1)/(2)Iw^2`

Since there is no slipping

`v=rw`

Therefore

`mgh=(1)/(2)mr^2w^2+(1)/(2)Iw^2`

Where

`I=(1)/(2)mr^2`

`l=3.24m`

`2*9.81*7.5=(1)/(2)(2)(1.8)^2w^2+(1)/(2)(3.24)w^2\\\\`

`147.15=3.24w^2+1.62w^2`

`w^2=(147.15)/(4.86)`

`w^2=\sqrt{(147.15)/(4.86)}`

`w^2=5.5rads/s`