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Unformatted text preview: Quiet please ladies and gentlemen The lecture is about to begin n Linear world Rotational world x ∆ x t υ ∆ = ∆ [ ] m [ ] / m s a t υ ∆ = ∆ 2 / m s [ ] N W F x = ∆ [ ] J θ ∆ t θ ϖ ∆ = ∆ [ ] rad [ ] / rad s t ϖ α ∆ = ∆ 2 / rad s F l τ = [ ] N m W τ θ = ∆ [ ] J m [ ] kg 2 kgm 2 1 2 KE I ϖ = 2 1 2 KE m υ = [ ] J [ ] J R υ ϖ = a R α = 2 I mR = Point mass I τ α = ∑ F ma = ∑ [ ] / P m kgm s υ = 2 / L I kgm s ϖ = Conservation of Rotational momentum For an isolated rotating system angular momentum is conserved A skater with arms outstretched spins at 0.3 rev/sec. The body of the skater is a 50kg cylinder of radius 12cm the hands are each 0.5kg point masses on the end of 1m arms (neglect the mass of the arms) Q1. What is the angular momentum of the skater. If the skater folds their arms their body can be approximated as a cylinder of mass 51kg and radius 12.5cm? Q2. What is the new angular velocity? Q3. How much work was done by the skater? Given: Part 1 50 M kg = 1 2 m kg = 1 0.3 / rev s ϖ = Find I ϖ 0.12 R m = 1 r m = Conservation of Rotational momentum Find angular momentum Given: Part 1 50 M kg = 1 2 m kg = 1 0.3 / rev s ϖ = Find I ϖ I b h I I I = + 2 2 2 1 2 I I MR mr mr = + + 0.12 R m = 1 r m = ( 29 ( 29 ( 29 2 2 2 1 50 0.12 0.5 1 0.5 1 2 I I = + + 0.36 1 I I = + I I I L I ϖ = 2 1.36 kgm = I I I L I ϖ = ( 29 1.36 0.3 2 π = × 2 1 2.56 kgm s = Conservation of Rotational momentum For an isolated rotating system angular momentum is conserved A skater with arms outstretched spins at 0.3 rev/sec. The body of the skater is a 50kg cylinder of radius 12cm the hands are each 0.5kg point masses on the end of 1m arms (neglect the mass of the arms)...
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This note was uploaded on 04/07/2012 for the course PHYS 111 taught by Professor Dr.jackman during the Fall '07 term at Waterloo.
 Fall '07
 Dr.Jackman
 Physics

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