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` ?

Expert Answers

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First use conservation of energy to find `v` at the bottom of the ramp when there is no friction.




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`



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'` .



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





`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))`



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