Here's the question you clicked on:
deeprony7
A sphere of mass 50gm and radius 10cm rolls without slipping with a velocity of 5cm/s. Its total kinetic energy in ergs is?
For rolling w/o slipping, and choosing teh cm of the cm of teh sphere as the rotation point, \[K _{total}=K _{transl}+K _{rot}=1/2mv^{2}+1/2I \omega ^{2} = 1/2mv^{2}+1/2(2/5mr ^{2})(v/r)^{2}=7/10 mv^{2}\] You can also choose the point of contact as the rotation point and use teh parallel axis theorem to get the same answer
Choose the point of contact as the rotation point... thus the point of rotation has been moved a distance r from the center of mass of the sphere. Parallel axis theorem says \[I _{\parallel} = I _{cm} + mr ^{2}\] where, again, r is the distance the point of rotation has been displaced from the center of mass (cm) of the sphere. Now, all energy is rotational: \[K _{total} = K _{rot} = 1/2 I _{\parallel} \omega ^{2} \] \[= 1/2 (2/5 mr ^{2} + mr ^{2})*(v/r)^{2}\] \[=1/2(7/5 mr ^{2})*(v ^{2}/r ^{2}) = 7/10 mv ^{2}\]
You have a combination of linear and rotational KE. These guys are telling you true. P.S. Who uses ergs any more? I haven't heard the term since 1950s sci-fi.