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# ws14 - energy of E final = mgH ′ 1 2 Iω 2 = E initial =...

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Worksheet 14: Rotational Energy Questions 1. Consider a straight stick balancing on frictionless ice, what path does the center of mass fall? Since there is no net horizontal force, the center of mass cannot accelerate horizontally, hence must fall down in a straight vertical path. The distance it falls would be 2 , where is the stick’s length. 2. The marble will not make it up to its original height. The reason is this: at the bottom of the hill, the marble has turned all of its gravitational potential energy ( mgH ) into both linear and rotational kinetic energy ( 1 2 mv 2 + 1 2 2 ). It’s rotational kinetic energy is nonzero because the there was friction on that side of the hill and this caused the marble to roll without slipping. Once it travels up the other end, it’s able to turn its linear kinetic energy ( 1 2 mv 2 ) into potential energy, but since it cannot grip the frictionless surface, it keeps its rotational kinetic energy, for a Fnal

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Unformatted text preview: energy of E final = mgH ′ + 1 2 Iω 2 = E initial = mgH . We see that since some of the initial energy is ‘stuck’ in the rotational motion, it cannot go as high. Problems 1. A rod with a Fxed pivot falls down (a) KE final = 1 2 MgL (b) ω = r 3 g L (into the page) (c) v top = √ 3 gL 2. The bowling ball’s Fnal speed is: v 2 cm = 10 7 gL sin θ . 3. Various objects roll down a ramp: (a) The block reaches the bottom Frst. (b) The bowling ball and the marble reach the bottom at the same time. (c) The bowling ball reaches the bottom Frst. 4. A cylinder rolls down a track. 1 (a) v 2 f = 4 3 g ( H-h-R ) (b) Δ KE linear = 1 2 mv 2 f < Δ U = Mg ( H-h ) (c) D 2 = 8 3 h ( H-h-R ) 5. A sphere rolls down a circular ramp. (a) v 2 = 4 3 gR (1-cos θ ) (b) cos θ c = 4 7 ⇒ θ ≈ 55 . 2 o 2...
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ws14 - energy of E final = mgH ′ 1 2 Iω 2 = E initial =...

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