Question

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

Answer 1

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: