Question

If the angular velocity of a body rotating about an axis is doubled and its moment of inertia halved, the rotational kinetic energy will hanged by a factor pf :

• A. 4
• B. 2
• C. 1
• D. 1/2

Answer 1

The correct option is B 2

Initially

$W_1=$angular velocity

$I_1=$Moment of inertia

Rotational kinetic energy

$K_1=1/2I_1{W}_1^2$

Finally

$W_2/I_2$

$W_2=2W_1/I_2=\frac{I_1}{2}$

K.E.

$K_2=1/2I_2{W}_2^2$

$=\frac{1}{2}\times \frac{I_1}{2}\times 4W_1^2$

$k_2=2k_1$.