A thin cylindrical ring starts from rest at a height h1 = 88 m. The ring has a radius R = 18 cm and a mass M = 4 kg.?
Part (a) Write an expression for the ring's initial energy at point 1, assuming that the gravitational potential energy at point 3 is zero.
Part (b) If the ring rolls (without slipping) all the way to point 2, what is the ring's energy at point 2 in terms of h2 and v2?
Part (c) Given h2 = 17 m, what is the velocity of the ring at point 2 in m/s?
Part (d) What is the ring's rotational velocity in rad/s?
Part (e) After passing point 2 the hill becomes frictionless and the ring's rotational velocity remains constant. What is the linear velocity of the ring at point 3 in m/s?
- Andrew SmithLv 71 month ago
a) The initial energy is potential energy Mgh.
b) For a ring the rotational energy is equal to the translational energy. ie it behaves as if it has a mass of 2M Ek = 1/2 * (2M) v^2If at point 2 there is still some potential energy then that must be added to get the total energy.
c) assuming that all the potential has gone to kinetice 1/2 * (2M)v^2 = M gh -> v= sqrt(gh) where h is the difference in height between point 1 and point 2.
d) w = v/r = sqrt( gh) / R
I have no idea of what point 3 is. I have assumed that point 3 is at the BOTTOM of a slope. If it is further down the slope then for the distance below point 2 the energy would ONLY go into translation.
ie 1/2*(2M) * v^2 = Mg(h2-h1) and 1/2 M (v3^2-v^2) = Mg ( h3-h2)
- alexLv 71 month ago
Where's the diagram ?
- NCSLv 71 month ago
mass M = 4 kg ♦♦ R = 18 cm = 0.18 m ♦♦ h₁ = 88 m
(a) E₁ = M*g*h₁
(b) "without slipping" means we have rotational and translational kinetic energy. The moment of inertia if the thin cylinder is
I = MR²
and so the total kinetic energy is
KE = ½mv² + ½I(v/R)² = ½mv² + ½*MR²*(v/R)² = mv²
E₂ = Mv₂² + M*g*h₂
and E₂ = E₁
in case you want to solve for v₂
(c) dropping units for ease (v₂ is in m/s)
4*9.8*88 = 4v₂² + 4*9.8*17
v₂ = 26.4 m/s
should be rounded to 2 digits
(d) ω₂ = v₂ / R = 26.4m/s / 0.18m = 147 rad/s
should be rounded to 150 rad/s owing to the data, IMO
(e) M*g*h₁ = ½Mv₃² + ½*MR²*ω₂²
since the rotational energy is unchanged from (c) and we can no longer relate v to ω.
mass M cancels -- the rest (sans units) is
9.8*88 = ½*v₃² + ½*0.18²*147²
v₃ = 32 m/s
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