# A sphere of radius `20 cm` and mass `45 g` is placed atop a ramp of height `0.75 m` and inclination angle `30` deg. If the ramp were frictionless the sphere would slide down the ramp in a time `t` . With friction, the sphere rolls without slipping down the ramp and reaches the bottom at time `t'` . What is `(t')/t` ? First use conservation of energy to find `v` at the bottom of the ramp when there is no friction.

`E_i=E_f`

`mgh=1/2mv^2`

`v=sqrt(2gh)`

Now plug into a kinematic equations to find `t` after the sphere has rolled down the ramp a distance we will call `Delta x` .

know `v^2=2a Delta x`

and

`v=t*a`

Then `v*v=2a Delta x`

`v=(2a Delta x)/v`

`t*a=(2a Delta x)/v`

`t=(2Delta x)/v`

`t=(2 Delta x)/sqrt(2gh)` ` `

Now repeat the process with friction to find `t'` .

`E_i=E_f`

`mgh=1/2m(v')^2+1/2I(omega)^2`

`I=2/5mR^2` for a solid sphere.

`mgh=1/2m(v')^2+1/2(2/5mR^2)((v')/R)^2`

`gh=1/2(v')^2+1/5(v')^2`

`gh=(1/2+1/5)(v')^2`

`sqrt((10gh)/7)=v'`

`t'=(2Delta x)/(v')=2 Delta x*sqrt(7/(10gh))`

Finally, we have

`(t')/t=(2 Delta x sqrt(7/(10gh)))/((2 Delta x)/sqrt(2gh))`

`(t')/t=sqrt(7/(10gh))*sqrt(2gh)`

`(t')/t=sqrt(7/5)`