A hoop of mass M and radius R rolls on a level surface without slipping. If the angular velocity of the wheel is w, what is its KINETIC ENERGY?

Respuesta :

Answer:

MR²w²

Explanation:

Given:

Mass of the hoop = M

Radius of the hoop = R

Angular velocity of the wheel = w

Since, the hoop is rolling without slipping

thus,

Total kinetic energy = Rotational Kinetic energy + Translation Kinetic Energy

or

The total kinetic energy = [tex]\frac{1}{2}Iw^2+\frac{1}{2}Mv^2[/tex]

here,

I is the moment of inertia of the hoop

I = MR²

v is the translational speed of the wheel

also,

v = Rw

therefore,

the total kinetic energy =  [tex]\frac{1}{2}MR^2w^2+\frac{1}{2}M(Rw)^2[/tex]

or

The total kinetic energy = MR²w²

Answer:.

Translational kinetic energy is basically normal KE, so 1/2(m)(v^2)

If you want total including rot. KE, it is:

1/2(I)(w^2) + 1/2(m)(v^2)