If a sphere is rolling, the ratio of translational energy to total kinetic energy is given by:

If a sphere is rolling, the ratio of translational energy to total kin

1.	If a sphere is rolling, the ratio of translational energy to total kinetic energy is given by:
Answer» Solution :The expression for total kinetic energy in pure ROLLING is, `KE=KE_("TRANS")+KE_(NOT)` For any object the total kinetic energy as per given equation `KE=(1)/(2)MV_(CM)^(2)+(1)/(2)Mv_(CM)^(2)((K^(2))/(R^(2)))` `KE=(1)/(2)Mv_(CM)^(2)(1+(K^(2))/(R^(2)))` Then `(1)/(2)Mv_(CM)^(2)(1+(K^(2))/(R^(2)))=(1)/(2)Mv_(CM)^(2)+(1)/(2)Mv_(CM)^(2)((K^(2))/(R^(2)))` The above equation suggests that in pure rolling the ratio of total kinetic energy, translational kinetic energy and rotational kinetic energy is given as, `KE:KE_("TRANS"):KE_("ROT"): : (1+(K^(2))/(R^(2))):1: ((K^(2))/(R^(2)))` Now, `KE_("TRANS"): KE_(ROT) : : 1: ((K^(2))/(R^(2)))` For a solid spere, `(K^(2))/(R^(2))=(2)/(5)` Then, `KE_("TRANS") : KE_("ROT") : : 1: (2)/(5) or KE_("TRANS") : KE_(NOT) : : 5:2`