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Physics
Classical Physics
Mechanics
Conservation of energy in rotating bodies
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[QUOTE="aclaret, post: 6462887, member: 687743"] as they above say your work is wrong, but u have some ok intuition. cylinder not rotate about its mass centre, however the mass centre point do rotate in circular arc about the contact point on right (well, so long as it is still contact!). this you can use to advantage! say, mass centre of cylinder has rotate by ##\theta## about right point of contact. let ##v = |\underline{v}|## be speed of the mass centre. can first use the energy conserve$$2gr(1-\cos{\theta}) = v^2$$now, remember that mass centre move as if it were particle which is act upon by resultant force, so you now pretend mass centre is a particle, and resolve resultant of force toward right contact point (i.e. you calculate centripetal component):$$-N + mg\cos{\theta} + N_2 \sin{2\theta} = \frac{mv^2}{r}$$where ##N## is force of contact from right point, and ##N_2## force of contact from left point. now you must do examine cases when ##N = 0## and ##N_2 = 0## respective, i.e. set each to ##0## in turn and use two equation to eliminate ##v##, and find ##\theta_{\text{max}}## for each. then you can tell, which lose contact first. see - much more easy than do use lot of messy coordinate and difficult equation, yes ;) [/QUOTE]
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Forums
Physics
Classical Physics
Mechanics
Conservation of energy in rotating bodies
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