a body of radius r and mass m is rolling on a horizontal plane without sliping with speed v it then rolls up a hill of vertical height h. If h=3v^2/4g the body is
Answers
Answered by
3
Answer:The total kinetic energy of the body
K=KT=KR=12Mv2+12Iω2
=12Mv2[1+(MMr2)] [as v=rω]
when it rolls up an iregular inclined plane of height (h=3v24g), its KE is converted into PE, so by conservation of mechanical energy
12MV2[=1+1Mr2]=Mg[3v24g]
Which on simplification gives I=Mr22. This result clearly indicates that the body is either a disc or a cylinder.
Explanation:
Similar questions