0 of 28 Concepts completed Due in less than 8 houns Multiple Choice Question An electron is moving with uniform circular motion around a circle of radius 6 cm , its initial rotational kinetic energy is 24 . J. If its (Iransiational) speed doubled and it kept moving around the same circle, then how much rotational kinetic energy would it have? 24 J 48 J 96J Need help? Review these concept resources. (1) Rend About the Concept

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Let the rotational kinetic energy be given by

$$
K = \frac{1}{2} m v^2.
$$

Initially, the electron’s kinetic energy is

$$
K_1 = \frac{1}{2} m v_1^2 = 24 \text{ J}.
$$

If the speed is doubled, namely $$\displaystyle v_2 = 2 v_1$$, then the new kinetic energy will be

$$
K_2 = \frac{1}{2} m (2v_1)^2.
$$

Simplify the expression:

$$
K_2 = \frac{1}{2} m \cdot 4v_1^2 = 4 \left(\frac{1}{2} m v_1^2\right) = 4K_1.
$$

Substituting the initial kinetic energy:

$$
K_2 = 4 \times 24 \text{ J} = 96 \text{ J}.
$$

Thus, the rotational kinetic energy when the speed is doubled is

$$
\boxed{96 \text{ J}}.
$$
When the electron's speed doubles, its rotational kinetic energy becomes 96 J.

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