A hula Hoop (I=1/2MR^2) rolls without sliding along the floor. The ratio of its translational kinetic energy to its rotational kinetic energy (about an axis through its center of mass) is:
A. 1
B. 2
C. 3
D. 1/2
KE*rolling)=1/2 I w^2=1/4 M r^2(v/r)^2=1/4 M v^2
KK(translational)=1/2 M v^2
Looks like translation is twice rolling
To determine the ratio of translational kinetic energy (K_t) to rotational kinetic energy (K_r), we need to calculate both values using the given formula for the moment of inertia (I) of a hula hoop, which is given as I = 1/2MR^2.
The translational kinetic energy (K_t) is given by the formula K_t = 1/2MV^2, where M represents the mass of the hula hoop and V represents its linear velocity.
To relate the translational velocity V to the angular velocity ω (rotation rate) and the radius R, we use the relation V = ωR. Therefore, V^2 = ω^2R^2.
Substituting this expression for V^2 into the formula for K_t, we get: K_t = 1/2Mω^2R^2.
The rotational kinetic energy (K_r) is given by the formula K_r = 1/2Iω^2.
Substituting the given moment of inertia (I) into the formula for K_r, we get: K_r = 1/2(1/2MR^2)ω^2.
The ratio of K_t to K_r is therefore: K_t/K_r = (1/2Mω^2R^2) / (1/2(1/2MR^2)ω^2) = (1/2Mω^2R^2) / (1/4Mω^2R^2).
Canceling out the common terms, we find that the ratio simplifies to K_t/K_r = 2.
So, the ratio of the translational kinetic energy to the rotational kinetic energy of the hula hoop is 2.
Therefore, the correct answer is B. 2.