A solid sphere is undergoing pure rolling. What is the ratio of its translational kinetic energy to rotational kinetic energy?

Explanation

The expression for total kinetic energy in pure rolling is,
`KE=KE_("TRANS")+KE_("ROT")`
For any object the total kinetic energy as per equation 5.58 and 5.59 is,
`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 sphere, `(K^(2))/(R^(2))=(2)/(5)`
Then, `KE_("TRANS"):KE_("ROT")::1:(2)/(5)` or
`KE_("TRANS"):KE_("ROT")::5:2`