A round body of mass M, radius R, and moment of inertia I= BMR^2 is rolling without slipping. Its kinetic energy can be decomposed into the energy of its center of mass motion (translation) and the enrgy of rotation about its center of mass. The fractions of the body's kinetc energy associated with these contributions depend only on B, Find these fractions...

To find the fractions of the body's kinetic energy associated with the contributions of translation and rotation, we need to consider the energy associated with each type of motion.

1. Translation Energy:
The translation energy of an object is given by the formula (1/2)mv^2, where m is the mass of the object and v is the velocity of the object's center of mass. In this case, the body's translation energy is the energy associated with its center of mass motion.

2. Rotation Energy:
The rotation energy of an object is given by the formula (1/2)Iω^2, where I is the moment of inertia of the object and ω is the angular velocity of the object about its center of mass. In this case, the body's rotation energy is the energy associated with its rotation about its center of mass.

Since the body is rolling without slipping, we can relate the linear velocity of the center of mass to the angular velocity. The linear velocity v is equal to the angular velocity ω multiplied by the radius R (v = ωR).

To find the fractions of the body's kinetic energy associated with translation and rotation, we need to determine their respective contributions relative to the total kinetic energy.

Total Kinetic Energy:
The total kinetic energy is the sum of the translation energy and rotation energy:
K_total = K_translation + K_rotation

K_translation = (1/2)Mv^2
K_rotation = (1/2)Iω^2

Substituting v = ωR:
K_translation = (1/2)M(ωR)^2 = (1/2)Mω^2R^2

Since I = BMR^2:
K_rotation = (1/2)(BMR^2)(ω^2) = (1/2)BMR^2ω^2

K_total = (1/2)Mω^2R^2 + (1/2)BMR^2ω^2

To find the fractions, we divide each term by K_total:
Fraction of Translation Energy:
K_translation / K_total = ((1/2)Mω^2R^2) / ((1/2)Mω^2R^2 + (1/2)BMR^2ω^2)
= Mω^2R^2 / (Mω^2R^2 + BMR^2ω^2)
= Mω^2R^2 / ω^2(R^2(M + BM))
= M / (M + BM)

Fraction of Rotation Energy:
K_rotation / K_total = ((1/2)BMR^2ω^2) / ((1/2)Mω^2R^2 + (1/2)BMR^2ω^2)
= BMR^2ω^2 / (Mω^2R^2 + BMR^2ω^2)
= (BM / (M + BM))

Therefore, the fraction of the body's kinetic energy associated with translation is M / (M + BM), and the fraction associated with rotation is BM / (M + BM), where B is a constant that depends on the body's distribution of mass.