circular motion- speed of rider, centripetal acceleration and coefficient of friction

azizlwl

Quote by dani123

and the normal force is just mv²/R... right? after I get the normal force and times that by the μ that I got from doing v²/gR; do then go on to do μ=Ff/Fn? Cause wouldnt that just end up giving me the same value for μ that I got in the first place? Thank you very much for your time by the way!

What you are doing μ=Ff/Fn=(mv²/R)/mg=v²/Rg
Which is wrong.

Ff is the value of frictional force which is perpendicular to the Fn which is equal to mg.
Fn is the pressing force = centripetal force.

The orientation is turn 90° from normal horizontal plane. Now its a vertical plane.

dani123

I think I got it finally.. so I did:

μ=(9.8m/s²)*(2.6m)/ (7.10m/s)²
μ=0.505

does this seem like a reasonable answer? If someone could check for me, it would be greatly appreciated! Thank you!

Infinitum

Quote by dani123

I think I got it finally.. so I did:

μ=(9.8m/s²)*(2.6m)/ (7.10m/s)²
μ=0.505

does this seem like a reasonable answer? If someone could check for me, it would be greatly appreciated! Thank you!

Yep, that is correct!!

dani123

its about time hahah thank you so much for you help!

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dani123	I think I got it finally.. so I did: μ=(9.8m/s²)*(2.6m)/ (7.10m/s)² μ=0.505 does this seem like a reasonable answer? If someone could check for me, it would be greatly appreciated! Thank you!