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3 years ago

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A wheel is rotating freely at angular speed 660 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initi

ally at rest and with 5 times the rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the angular speed of the resultant combination of the shaft and two wheels? (b) What fraction of the original rotational kinetic energy is lost?

1 answer:

sweet-ann [11.9K]3 years ago

6 0

Answer:

(a) 110 rev/ min

(b) 5/6

Explanation:

As per the conservation of linear momentum,

L ( initial ) = L ( final )

I' ω' = ( I' + I'' ) ωf

I' is the rotational inertia of first wheel and I'' is the rotational inertia of second wheel which is at rest.

(a)

So, ωf = I' ω' / ( I' + I'' )

As I'' = 5I'

ωf = I' ω' / ( I' + 5I' )

ωf = ω'/ 6

now we know ω' = 660 rev / min

therefore ωf = 660/6

= 110 rev/ min

(b)

Initial kinetic energy will be K'

K' = I'ω'² / 2

and final K.E. will be K'' = ( I' + I'' )ωf² / 2

K'' = ( I' + 5I' ) (ω'/ 6)²/ 2

K'' = 6I' ω'²/72

K'' = I' ω'²/ 12

therefore the fraction lost is

ΔK/K' = ( K' - K'' ) / K'

= {( I'ω'² / 2) - (I' ω'²/ 12)} / ( I'ω'² / 2)

= 5/6

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