Skip to main content
9 events
when toggle format what by license comment
Jun 11, 2020 at 9:33 history edited CommunityBot
Commonmark migration
May 16, 2016 at 22:17 comment added YoTengoUnLCD @Frobenius Thanks! After a while I understood that we were counting the energy of the system and not just that of the particle. It's all clear now.
May 16, 2016 at 22:16 vote accept YoTengoUnLCD
May 16, 2016 at 21:10 answer added sammy gerbil timeline score: 3
May 16, 2016 at 21:08 comment added Voulkos You don't count the same thing twice : The 2nd part $\frac {I}{2}\omega^2$ concerns the kinetic energy of the rotating disc, not of the small particle. Here $\:I = \textit{moment of inertia of the disc around the axis of rotation}\:$. The 1rst part $\:\frac m 2 v^2\:$ is the kinetic energy of the rotating particle, expressed also as $\:\frac m 2 r^2 \omega^2\:$, where $\:m r^2\:$ is the "moment of inertia" of the small particle.
May 16, 2016 at 21:07 answer added M. Enns timeline score: 4
May 16, 2016 at 19:52 history edited Qmechanic CC BY-SA 3.0
deleted 12 characters in body; edited tags
May 16, 2016 at 19:21 review First posts
May 16, 2016 at 19:25
May 16, 2016 at 19:19 history asked YoTengoUnLCD CC BY-SA 3.0