The cylinder shown, of mass M and radius R, starts from rest and rolls without slipping down an incline of height H. At the bottom its translational speed is √(8gH/7). Which rotational inertia is consistent with this motion?

Question

G1 Physics · Accepted Answer

Energy at the bottom is MgH=½Mv²+½I(v²/R²). Given v²=8gH/7, solving gives I=3MR²/4. Hence B. Write translation ΣF=Ma and rotation Στ about CM=I about CMα, together with the no-slip relation v of CM=ωR and a of CM=αR whenever pure rolling applies. Solving the equations for the shown geometry gives 3MR²/4..