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A wheel is rotating freely at angular speed 440 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with 9 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?

User TrongBang
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1 Answer

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Answer:

(a) Angular speed of resultant combination will be rev/min

(b) Fraction of total energy loss will be 0.9

Step-by-step explanation:

We have given angular speed of the first wheel
\omega _1=440rev/min

According to question moment of inertia of second wheel is 9 times the moment of inertia of first wheel

So
I_2=9I_1

(a) Now according to law of conservation of angular momentum


I_1\omega _1=(I_1+I_2)\omega _2


I_1* 440=(I_1+9I_1)\omega _2


\omega _2=44rev/min

(b) Change in rotational kinetic energy will be equal to
(KE_i-KE_f)/(KE_i).......eqn1


KE_I=(1)/(2)I_1* 440^2


KE_f=(1)/(2)10I_1* 44^2

Putting these values in eqn 1

Change in kinetic energy will be 0.9

User Clifford
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