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lord [1]
4 years ago
11

A spherical object (with non-uniform density) of mass 16 kg and radius 0.22 m rolls along a horizontal surface at a constant lin

ear 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
Physics
1 answer:
Lapatulllka [165]4 years ago
4 0

Answer:

Kr = 0.7618K

Explanation:

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

K = \frac{mv^{2} }{2}

K = \frac{(16)v^{2} }{2}

K = 8v^{2}

The rotational kinetic energy is

Kr = \frac{Iw^{2} }{2}

           where I: The moment of inertia

                      ω: angular velocity

Kr =\frac{(0.59)w^{2} }{2}

Kr = 0.295w^{2}

How the movement is without slipping, then  

ω = \frac{v}{r}

ω = \frac{v}{0.22}

Thus

Kr = \frac{0.295v^{2} }{0.22^{2} }

Kr = 6.095v^{2}

8v^{2}  ---->  1

6.095v^{2}----->?

Kr = 0.7618K

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