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# If mi is a mass element its angular momentum about

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Unformatted text preview: change the direction of p, i.e., no torques. ~ Rotation analog of the conservation of linear momentum ~ ˆ j r ri Look at the special case of motion in a circle ... angular momentum is conserved .... r vi ∆mi r L ˆ i O r r r p ... although a centripetal force acts on the object, it is Consider a planar object rotating about O. If ∆mi is a mass element, its angular momentum about the rotation axis is: But v i = riω . directed through the rotation axis; it produces zero torque and so there’s no change in angular momentum! r r r ˆ Li = ri × ( ∆mi ) v i = ∆mi ri v i k v ˆ ∴ Li = ( ∆mi ri2 )ω k . So, the total angular momentum is r r ˆ ˆ L = ∑i Li = ∑i ( ∆mi ri2 )ω k = Iω k ˆr But ω k = ω . r r ∴ L = Iω , r r i.e., in the direction of ω . cf: linear momentum p is r parallel to v . r r L = Iω No net torque ... ω I ∴1= 2 I2 ω1 RMM09VD1.MOV If there is a net external torque acting on an object then, from Newton’s 2nd Law of rotation, we find r r r r r dω d (Iω ) dL τ = Iα = I = =, dt dt dt DISCUSSION PROBLEM [10.1]: i.e., the rate of change of angular momentum of the object is equal to the net external torque. (Compare with r r dp Newton’s 2nd Law F = .) Note: if there is no net dt r r dL = 0 , as before. external torque, i.e., τ ⇒ 0, then dt A rotating cloud of gas begins to collapse under If a torque is applied for time ∆t , then r r (Units of L ⇒ J.s) τ.∆t = ∆L . r r Remember: F.∆t = ∆p, the impulse equation? gravitational forces within. As it collapses, what happens to its: • Moment of inertia? • Angular momentum? Let’s look some situations involving angular momentum ... • Angular velocity? ω1 l l ω2 ⇒ I2 I1 This is a conservation of angular momentum problem ... I1ω1 = I2ω 2 ... but we need to make some assumptions ... • body is a cylinder ( r ≈ 0.2 m, Mb ≈ 50 kg ). Question 3: An ice skater starts a spin with arms outstretched, rotating at 1.5rev/s. Estimate her rotational speed (in rev/s) when she brings her arms against her body. • each arm is a thin rod ( l ≈ 1m, Ma ≈ 5kg). 1 1 With arms “out” I1 ⇒ MbR12 + 2 × Ma (2l)2 2 12 1 1 = × 50 × (0.2)2 + 2 × × 5 × 22 = 4.33 kg ⋅ m2. 2 12 1 With arms “in” I2 ⇒ (Mb + 2Ma )R 22 2 1 = × (50 + 10) × (0.22)2 = 1.45 kg ⋅ m2 . 2 Iω 4.33 × 1.5 ∴ ω2 = 1...
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## This note was uploaded on 07/13/2011 for the course PHY 2048 taught by Professor Guzman during the Spring '08 term at FAU.

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