Question

a uniform solid sphere of mass m and radius r rolls without sliding along a horizonal floor. the ratio of its translational kinetic energy to its rotational kinetic energy (about an axis through its center of mass) is:

Answer 1

The ratio of the translational kinetic energy to the rotational kinetic energy of a uniform solid sphere rolling without sliding is 5:2.

Step-by-step explanation:

When analyzing the motion of the uniform solid sphere, we can take into account that a rolling sphere has both translational and rotational motion.

The translational kinetic energy (KE_trans) can be represented by the equation KE_trans = 1/2mv^2, where m is the mass and v is the velocity of the sphere.

The rotational kinetic energy (KE_rot) is given by the equation KE_rot = 1/2Iω^2 where I is the moment of inertia for a sphere (I = 2/5mr^2 where r is the radius), and ω is the angular velocity.

Since the sphere is rolling without slipping, the relation between the translational velocity and the angular velocity is v = rω. Substituting this into the expression for KE_rot yields KE_rot = 1/2(2/5mr^2)(ω^2).

Simplifying this expression, we get KE_rot = 1/5mv^2.

Now, to find the ratio of translational kinetic energy to rotational kinetic energy, we divide KE_trans by KE_rot:

Ratio = KE_trans/KE_rot

= (1/2mv^2) / (1/5mv^2) =

(5/2).

Therefore, the ratio of the translational kinetic energy to the rotational kinetic energy of a uniform solid sphere rolling without sliding on a horizontal floor is 5:2.