# A 12-g CD with a radius of 6.0 cm rotates with an angular speed of 35 rad/s. A) What is its...

## Question:

A 12-g CD with a radius of 6.0 cm rotates with an angular speed of 35 rad/s.

A) What is its kinetic energy? Answer in Joules.

B) What angular speed must the CD have if its kinetic energy is to be doubled? Answer in rad/s.

## Rotational Energy:

The kinetic energy of a rotating object also known as the rotational energy is a component of the object's total kinetic energy. This can be determined using the rotational inertia and angular speed of the object.

Given:

• Mass of the CD, {eq}(m_1) = 12 \ \rm g = 0.012 \ \rm kg {/eq}
• Radius of the CD, {eq}(r) = 6.0 \ \rm cm = 0.06 \ \rm m {/eq}
• Angular speed of the CD, {eq}(\omega) = 35\ \dfrac{\rm rad}{\rm s} {/eq}

Part A) The formula for the angular kinetic energy of the CD is given by

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

The angular speed is given while the rotational inertia is solved as follows:

\begin{align} I &= \frac {1}{2} m_1 r^2 \\[0.3cm] &= \frac {1}{2} (0.012 \ \rm kg)(0.06 \ \rm m)^2 \\[0.3cm] &= 2.16 \times 10^{-5} \ \rm kg \cdot \rm m^2 \end{align}

Hence, solving for the rotational kinetic energy,

\begin{align} RKE &= \frac {1}{2} I \omega^2 \\[0.3cm] &= \frac {1}{2} (2.16 \times 10^{-5} \ \rm kg \cdot \rm m^2)\left(35 \frac {rad}{s} \right)^2 \\[0.3cm] &= 0.01323 \ \rm J \\[0.3cm] &\approx \boxed {0.013 \ \rm J} \end{align}

Part B) Doubling the rotational kinetic energy, that is

$$RKE = 0.01323 \times 2 = 0.02646 \ \rm J$$

Back solving the angular speed of the CD,

\begin{align} \omega &= \frac {2RKE}{I} \\[0.3cm] &= \frac {2 (0.02646 \ \rm J)}{2.16 \times 10^{-5} \ \rm kg \cdot \rm m^2} \\[0.3cm] &= \boxed {2450\ \frac {\rm rad}{\rm s}} \end{align} 