Question

Angular momentum of body charges by 80 kg.$m^2$/s when its angular velocity changes from 20 rad/s to 40 rad/s. The change in rotational K.E. is

• A. 1200 J
• B. 1600 J
• C. 400 J
• D. 2400 J

Answer 1

Answer: (D)

2400 J

$I{w}_f-I{\omega }_i=80$
$I(40-20)=80$
$I=4$
Change in rotational K.E.

$\Rightarrow \Delta K.E=\frac{1}{2}I{\omega }_f^2-\frac{1}{2}I{\omega }_i^2$

$\Rightarrow \Delta K.E=\frac{1}{2}I({\omega }_f^2-w_i^2)$

$\Rightarrow \Delta K.E=\frac{1}{2}\times 4\times (1600-400)$

$\Rightarrow \Delta K.E=2\times 1200=2400J$