contestada

A spherical object (with non-uniform density) of mass 38 kg and radius 0.33 m rolls along a horizontal surface at a constant linear speed without slipping. The moment of inertia of the object about a diameter is 0.43 M R2.
The object’s rotational kinetic energy about its own center is what fraction of the object’s total kinetic energy?

Respuesta :

princessesther2011

Answer:

0.104

Explanation:

Mass(m) = 3kg, radius(r) = 0.33m, moment of inertia (I) = 0.43kgm^2

Rotational Kinetic Energy (RTE) = w^2I ÷ 2

w = v/r w^2 = v^2/r^2

RTE = v^2/2r^2

Total Kinetic Energy (TKE) = mv^2/2

RTE/TKE = (v^2I/2r^2) ÷ (mv^2/2) = I/mr^2 = 0.43/(38×0.33^2) = 0.43/(38×0.1089) = 0.43/4.1382 = 0.104