Question

A spherical object (with non-uniform density) of mass 16 kg and radius 0.22 m rolls along a horizontal surface at a constant linear speed without slipping. The moment of inertia of the object about a diameter is 0.59 M R2 . The object’s rotational kinetic energy about its own center is what fraction of the object’s total kinetic energy

Answer 1

Kr = 0.7618K

Step-by-step explanation:

Suppose that the object's velocity is V, then his kinetic energy is:

K =
`(mv^(2) )/(2)`

K =
`((16)v^(2) )/(2)`

K = 8
`v^(2)`

The rotational kinetic energy is

Kr =
`(Iw^(2) )/(2)`

where I: The moment of inertia

ω: angular velocity

Kr =
`((0.59)w^(2) )/(2)`

Kr = 0.295
`w^(2)`

How the movement is without slipping, then

ω =
`(v)/(r)`

ω =
`(v)/(0.22)`

Thus

Kr =
`(0.295v^(2) )/(0.22^(2) )`

Kr = 6.095
`v^(2)`

8
`v^(2)` ----> 1

6.095
`v^(2)`----->?

Kr = 0.7618K