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)