Question

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?

Answer 1

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)

Answer 2

MR²w²

Step-by-step 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 =
`(1)/(2)Iw^2+(1)/(2)Mv^2`

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 =
`(1)/(2)MR^2w^2+(1)/(2)M(Rw)^2`

or

The total kinetic energy = MR²w²