Rotational Kinetic Energy

Rotational kinetic energy equations

• A body moving with linear velocity has an associated linear kinetic energy: $$E_k = \frac{1}{2}mv^2$$
• This can also be written using linear momentum $p = mv$: $$E_k = \frac{p^2}{2m}$$
• Similarly, a rotating body with angular velocity has an associated rotational kinetic energy: $$E_k = \frac{1}{2}I\omega^2$$
• This can also be written using angular momentum $L = I\omega$: $$E_k = \frac{L^2}{2I}$$
• Where: $E_k$ = rotational kinetic energy (J), $I$ = moment of inertia (kg m$^2$), $\omega$ = angular velocity (rad s$^{-1}$), $L$ = angular momentum (kg m$^2$ s$^{-1}$).

Rolling without slipping

• Circular objects, such as wheels, are made to move with both linear and rotational motion. The wheels of a car or bicycle rotate, causing the vehicle to move forward.
• Rolling without slippingA combination of rotating and sliding (translational) motion where the point in contact with the surface has a velocity of zero. There is enough friction present to initiate rotational motion. is a combination of rotating and translational motion.
• When a disc rotates, each point on the disc has a different linear velocity depending on its distance from the centre ($v \propto r$). The linear velocity is the same at all points on the circumference.
• When a disc slips or slides, there is not enough friction present to allow the object to roll. Each point on the object has the same linear velocity, and the angular velocity is zero.
• When a disc rolls without slipping, there is enough friction present to initiate rotational motion. The point in contact with the surface has a velocity of zero. The centre of mass has a velocity of $v = \omega r$. The top point has a velocity of $2v$ or $2\omega r$.

Rolling down a slope

• A common scenario involving rotational and translational motion is an object (usually a ball or a disc) rolling down a slope.
• At the top of the slope, a stationary object has gravitational potential energy: $$E_p = mg\Delta h$$
• As the object rolls down, the gravitational potential energy is transferred to both translational (linear) and rotational kinetic energy.
• At the bottom of the slope, the total kinetic energy is: $$E_{K\text{ total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$
• The linear or angular velocity can then be determined by equating $E_p$ and $E_{K\text{ total}}$, substituting the moment of inertia for the object, and using the relationship $v = \omega r$.
• For example, for a solid sphere of mass $m$ and radius $r$, with $I = \frac{2}{5}mr^2$: $$mg\Delta h = \frac{1}{2}m(\omega r)^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\omega^2$$ $$mg\Delta h = \frac{1}{2}m\omega^2 r^2 + \frac{1}{5}m\omega^2 r^2$$ $$mg\Delta h = \frac{7}{10}m\omega^2 r^2$$

Worked example: turntable and gum

• A turntable (mass $M = 550$ g, radius $R = 120$ mm) is modelled as a solid disc with $I = \frac{1}{2}MR^2$. It rotates freely at $\omega_1 = 3.28$ rad s$^{-1}$. A lump of gum (mass $m = 9.0$ g, modelled as a solid sphere with $I = \frac{2}{5}mr^2$) is dropped from a small height and sticks at $r = 117$ mm. The new angular velocity is $\omega_2 = 3.11$ rad s$^{-1}$.
• Before the gum is dropped, only the turntable rotates: $$E_{k\text{ before}} = \frac{1}{2}I_T\omega_1^2 = \frac{1}{4}MR^2\omega_1^2$$
• After the gum sticks, the total moment of inertia increases. The gum now contributes: $$E_{k\text{ after}} = \frac{1}{2}\left(\frac{1}{2}MR^2 + \frac{2}{5}mr^2\right)\omega_2^2$$
• The change in rotational kinetic energy: $$\Delta E_k = \frac{1}{4}MR^2\omega_1^2 - \frac{1}{2}\left(\frac{1}{2}MR^2 + \frac{2}{5}mr^2\right)\omega_2^2$$
• Substituting values ($M = 550 \times 10^{-3}$ kg, $R = 120 \times 10^{-3}$ m, $m = 9 \times 10^{-3}$ kg, $r = 117 \times 10^{-3}$ m): $$\Delta E_k = 0.0213 - 0.01938 = 1.92 \times 10^{-3} \text{ J}$$