A sphere of mass 50gm and radius 10cm rolls without slipping with a velocity of 5cm/s.
Its total kinetic energy in ergs is?

Question

anonymous · Answer

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

anonymous · Answer

how will i do that?

anonymous · 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: $$K _{total} = K _{rot} = 1/2 I _{\parallel} \omega ^{2} $$  $$= 1/2 (2/5 mr ^{2} + mr ^{2})*(v/r)^{2}$$ $$=1/2(7/5 mr ^{2})*(v ^{2}/r ^{2}) = 7/10 mv ^{2}$$

anonymous · Answer

interesting..=)

anonymous · Answer

8.75x10^(-5) Joule

anonymous · Answer

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.