A thin -walled pipe rolls along the floor. What is the ratio of its translational kinetic energy to its rotational kinetic energy about the central axis parallel to its length ?

Options

A. 1
B. `(3)/(2)`
C. `(2)/(3)`
D. `(4)/(3)`

Answer

A

Explanation

(a) The rotational kinetic energy is `K = (1)/(2) I omega^2`, where `I = mR^2` is its rotational inertia about the centre of mass The ratio is
`(K_(transl))/(K_(rot)) = ((1)/(2) mv_(com)^2)/(1/(2)(mR^2)(v_(com)//R)^2) = 1.00`.