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maw [93]
2 years ago
10

A solid sphere rolls along a horizontal, smooth surface at a constant linear speed without slipping. What is the ratio between t

he rotational kinetic energy about the center of the sphere and the sphere’s total kinetic energy?.
Physics
1 answer:
Pavlova-9 [17]2 years ago
3 0

The ratio of the rotational kinetic energy about the center of the sphere and the total kinetic energy of the sphere is 2/7.

<h3>What is the rotational kinetic energy of a sphere?</h3>

The rotational kinetic energy of the sphere is directly proportional to the square of angular acceleration.

K = \dfrac 12 I\omega ^2

Where,

I = rotational inertia of sphere = 2/5MR²

Where M is the mass of the sphere and R is the radius of the sphere,

ω = angular acceleration of sphere = V/R

Where V is the speed of the sphere.

So,

K = 1/2 × 2/5MR² × V²/R²

K = MV²/5

The translational kinetic energy of the sphere,

K' = \dfrac 12 MV^2

The total kinetic energy of the sphere,

K_t = K + K'

So,

K_t= \dfrac {MV^2}5 +\dfrac { MV^2}2\\\\K_t= \dfrac {7MV^2}{10}

Thus the Ratio of rotational kinetic energy to total kinetic energy,

\dfrac KK_t = \dfrac {\dfrac {MV^2}{5} }{ \dfrac {7MV^2}{10}}\\\\\dfrac KK_t = \dfrac 15 \times \dfrac { 10}7\\\\\dfrac KK_t= \dfrac 27

Therefore, the ratio of the rotational kinetic energy about the center of the sphere and the total kinetic energy of the sphere is 2/7.

Learn more about rotational kinetic energy:

brainly.com/question/14611300

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