10-29-08LectPHY2048

10-29-08LectPHY2048 - Chapt. 11 examples Rolling (combined...

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Chapt. 11 examples Rolling (combined rotation & translation) The cylinder , of mass m , radius r , rolls from rest (without slipping) down the incline of angle θ What is its speed after travelling distance L along the incline? This means we can use conservation of mechanical energy without worry that there will be some loss of energy for which we failed to account. L θ m r ω v com Since the cylinder rolls there is friction in the problem (or the cylinder would slip rather than roll) but because it is static friction (no slipping) it is not associated with a displacement (recall that for rolling the point in contact with the surface is at rest) meaning there is no work and thus no thermal energy loss associated with this friction.
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L θ m r ω v com mech E0 Δ= UK 0 Δ+ h In travelling the distance L the cylinder falls through a height h , so, fiif UU K K0 −+− = Since started from rest K i = 0 . Taking U f = 0 at the the end of L , if 0 −= or in terms of the parameters given since, i Um g h = hL s i n = θ i g L s i n = θ We have,
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The kinetic energy at the end of L is split between translational and rotational terms, i.e. f Trans rot KK K = + 22 fc o m c o m 11 Km v I =+ ω But, com vr com v r ω= While for a cylinder about its central axis, 2 com 1 Im r 2 = So, 2 com o m v 1 v ( m r ) 2 r ⎛⎞ ⎜⎟ ⎝⎠ 2 f com com com 3 v m v m v 24 4 =+= L θ m r ω v com h
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Then, 2 if c o m 3 U K mgLsin mv 0 4 −= θ = 2 com 4 vg L s i n 3 4 L s i n 3 = θ Or, Alternatively, the textbook derives an expression for the acceleration of the com for a uniform round body rolling down a ramp due to gravity (pp. 279-280) as, L θ m r ω v com h 22 2 1 2 gsin 2 ag s i n 1 I /mr 1 mr /mr 3 θ θ =− θ ++ Then using, oo vv 2 a ( r r ) = +− Solving for the velocity,
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L θ m r ω v com h 2 com 2 v0 2 ( g s i n ) ( 0 L ) 3 =+ θ 2 com 4 vg L s i n 3 So that again, com 4 L s i n 3
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HITT#1-10/29/08 Suppose that in the previous problem that there is no friction so the cylinder slides down the ramp the distance L without rolling. L θ m r v com With respect to the value when rolling the v com will now be, A. Greater B. Smaller C. The same
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HITT#1-10/29/08 L θ m r v com A. Greater with if UK 0 −= i Um g h m g L s i n == θ 2 fc o m 1 Km v 2 = When rolling the gravitational potential energy was split between translational and rotational kinetic energy terms. Now there is no rotation so all the Δ U goes into the translation making the velocity there greater.
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10-29-08LectPHY2048 - Chapt. 11 examples Rolling (combined...

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