Question

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?

Answer 1

Answer:

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

(b) Fraction of total energy loss will be 0.9

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

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

$I_1\times 440=(I_1+9I_1)\omega _2$

$\omega _2=44rev/min$

(b) Change in rotational kinetic energy will be equal to $\frac{KE_i-KE_f}{KE_i}$.......eqn1

$KE_I=\frac{1}{2}I_1\times 440^2$

$KE_f=\frac{1}{2}10I_1\times 44^2$

Putting these values in eqn 1

Change in kinetic energy will be 0.9