Question

A wheel is rotating freely at angular speed 800 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with twice the rotational inertia of the first, is suddenly coupled to the same shaft. The fraction of original Kinetic energy lost is

Answer 1

Given:

A wheel is rotating freely at angular speed 800 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with twice the rotational inertia of the first, is suddenly coupled to the same shaft.

To find:

Fraction of Kinetic Energy lost .

Calculation:

Applying Conservation of Angular Momentum :

$\therefore \: (I1)( \omega1) + (I2)( \omega2) = (I1 + I2)( \omega)$

$= > (I \times 800) + (2I \times 0) = (I + 2I)( \omega)$

$= > 800I = 3I \omega$

$= > \omega = \dfrac{800}{3}$

$= > \omega = 266.67 \: rev \: {m}^{ - 1}$

Original Kinetic Energy :

$KE1 = \dfrac{1}{2} I { \omega}^{2}$

$= > KE1 = \dfrac{1}{2} I { (800)}^{2}$

$= > KE1 = \dfrac{1}{2} I \times 640000$

$= > KE1 = 320000 I$

Final Kinetic Energy :

$KE2= \dfrac{1}{2} I { \omega}^{2}$

$= > KE2= \dfrac{1}{2} I { (266.67)}^{2}$

$= > KE2= \dfrac{1}{2} I \times 71112.89$

$= > KE2= 35556.44 I$

Kinetic energy lost :

$\Delta KE = (320000 - 35556.45)I$

$= > \Delta KE = 284443.56I$

Fraction of KE lost :

$\dfrac{ \Delta KE}{ KE} = \dfrac{284443.56I }{320000I} = 0.89$

Hence final answer is :

Fraction of KE lost is 0.89