Question

A hoop of mass 50 kg rolls without slipping. If the center-of-mass of the hoop has a translational speed of 4.0 m/s, the total kinetic energy of the hoop is

a. 0. 20 kJ
b. 0.40 kJ
c. 1.1 kJ
d. 3.9 kJ
e. None of these is correct.

Answer 1

total kinetic energy = 800 J

so correct option is e. None of these is correct

Step-by-step explanation:

given data

hoop mass = 50 kg

speed = 4.0 m/s

to find out

total kinetic energy of hoop

solution

we know that translational kinetic energy that is

translational kinetic energy = 0.5×m×v² ....................1

translational kinetic energy = 0.5×50×4²

translational kinetic energy = 400 J

and rotational kinetic energy will be

rotational kinetic energy = 0.5 × moment of inertia × (angular speed )² ............2

here

moment of inertia = mass × R²

and angular speed =
`(v)/(R)`

so rotational kinetic energy = 0.5×m×v²

rotational kinetic energy = 0.5×50×4²

rotational kinetic energy = 400 J

so total kinetic energy = 400 J + 400 J

total kinetic energy = 800 J

so correct option is e. None of these is correct