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)`


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