What is the correct answer?

Question

I nearly quarrel v teacher bcox of this Q. The Q is as following:

A hollow cylinder of mass M and radius R
(a) slides with a speed v
(b) rolls with the same speed without slipping on a horizontal plane.
What is the ratio of the kinetic energies of the cylinder in the two cases? [ Moment of inertia of hollow cylinder = MR^2 ]
A 1:4
B 1:2
C 1:1
D 2:1
Which is the answer?

Can show me the workings or not?

My working is:
for situation (a),
kinetic energy= 1/2 mv^2
for situation (b),
total ke= rotational ke + translational ke
= 1/2 Iw^2 + 1/2 mv^2
= 1/2 (mR^2)(v/R)^2 + 1/2 mv^2
= mv^2
ke (a) : ke (b) = 1:2

But my teacher said his answer is the correct one:
translational ke/rotational ke
= (1/2 mv^2) / (1/2 Iw^2)
= (mv^2) / [mR^2 (v/R)^2]
= 1

Very headache!!!

Answer 1

when slipping ke=1/2mv^2
when its rolling 1/2mv^2*2/2=1/2mv^2
so its C ie 1:1

Answer 2

I'll guess B from choices A/B, as don't know the math formulas off top of my head. If you ignore the fact of both cylinders are moving, then the rolling cylinder has more energy than the resting cylinder due to its rotation. Since the rate of spin is related to radius and speed v, you should be able to mathematically be able to prove this.

Answer 3

C:

Obviously, friction will slow the sliding cylinder first, but, never the less -
Same velocity, same mass = same kinetic E.

I really don't think it matters if it's rolling, sliding or flying through the air.

Answer 4

b
beacuse m+ r is two
and v is one

Answer 5

b