Question

a 24kg metal ring with 24cm diameter rolls without stopping down a 30 degree incline from a height of 3.4 m

1) according to the law of conservation of energy what should be the linear speed of the ring at the bottom of the ramp (I'm guessing that i solve for Vfrictionless, which is square root of 2gh?)

2) if the ring has a moment of inertia of I=mr^2 what will its linear speed be at the base of the incline? For this part I got the math down to the square root of gh.

3) what is the avg linear arc of this ring down the incline. I have no idea how to even start this part

Answer 1

a)
PE -> KE = KE(translational) + KE (rotational)
PE = mgh
KE(translational) =mv²/2
KE (rotational) = Iω²/2 = mR²v²/2R² = mv²/2
(for ring (hoop) I= mR²; ω =v/R)
KE = mv²/2 + mv²/2 = mv²
mgh = mv² => v=sqrt(gh)
b) if the ring has I= mR²/2
KE (rotational) = Iω²/2 = mR²v²/2•2R² = mv²/4
KT = mv²/2 + mv²/4 =3 mv²/4
mgh =3 mv²/4 => v=sqrt(1.33gh)
c) ave v = total distance/total time
s= h/sinα
v₀=0 => v=v₀+at= at => a= v/t
s= v₀t+at²/2 = at²/2 = v t² /2t =vt/2 => t=2s/v
ave v = total distance/total time =
= s : (2s/v) = v/2
Substitute v from a)