Question

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

• A. $4$
• B. $2$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $2$

Initial rotational kinetic energy$K=\frac{1}{2}I{\omega }^2$
Final rotational kinetic energy$K^{\prime }=\frac{1}{2}I{}^{\prime }\omega {{}^{\prime }}^2=\frac{1}{2}\left(\frac{I}{2}\right)(2\omega {)}^2=2K$