If rolling about axis through cm without slipping k 1

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If rolling about axis through CM without slipping K = 1 2 MV 2 CM + 1 2 I CM ω 2 and ω = V CM R so K = 1 2 M + I CM R 2 V 2 CM Put rolling without slipping + kinetic energy for rotation about CM together! :
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Phys 2A - Mechanics Example: Determine the minimum height for the ball to clear the loop-the-loop. ANS: As we have seen earlier in course the condition is that the centripetal acceleration at the top of the loop the loop equal the gravitational acceleration V 2 r loop = g Energy conservation: E i = E f K i + U i = E f + K f U = gravitational, K = translational plus rotational (without slipping) “i” = at height h, zero speed; “f” = at top of loop-the-loop R h = 1 + I MR 2 V 2 2 g + 2 r loop (should really be r loop -R, but will ignore for now) 0 + Mgh = 1 2 ( M + I/R 2 ) V 2 + MG (2 r loop )
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Phys 2A - Mechanics Moreover, for a uniform sphere Using the condition for the minimum height that V 2 r loop = g h min = 1 + I MR 2 r loop 2 + 2 r loop h min = 1 + 2 5 r loop 2 + 2 r loop I = 2 5 MR 2 h min - 2 r loop is the height above the loop-the-loop one should release from: h min - 2 r loop = 7 10 r loop Note that for all objects of circular cross section, I = f × MR 2 , where f is a pure number ( f = 2/5 for sphere). So for other objects h min - 2 r loop = 1 2 (1 + f ) r loop
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Phys 2A - Mechanics To be precise, the radius of the trajectory of the CM of the ball around the loop-the-loop is r loop -R so everywhere above one should use this for r loop h min = 7 10 ( r loop - R ) + 2( r loop - R ) For the DEMO, r loop = 8 . 5 in R = 0 . 75 in h min = 21 in so
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