Saved Bookmarks
| 1. |
A ringrollsonhorizontalwithoutslipping. Provethat itstranslationalkineticenergyis equalto rotationalkineticenergy. |
|
Answer» SOLUTION :if themassof theringis mand itsradiusis rthenmoment of INERITA, `I = m r ^(2) ` LINEAR VELOCITY( v )= angularvelocity`( omega )xx `radius or` v= omgear ` ` therefore` translationalkineticenergy, ` E_(1)= (1)/(2)mv^(2)m ( omegar)^(2)= (1)/(2) momega ^(2)r^(2)` Rotaionalkinetic energy, `E_(2) =(1)/(2)I omega^(2)= (1)/(2) m r^(2). omega ^(2)= (1)/(2)m omega ^(2)r ^(2) ` `thereforeE_(1)= E_(2)` |
|