10-29-25 Chapter 9 Rotational Kinetic Energy

Rotational Kinetic Energy

• Limit our discussion to rigid objects: objects that don’t change size or shape as they move. 

• For a rigid object where the center of mass is stationary:

$K_{sys} = K_{trans} + K_{rot} + L_{vib} \Rightarrow K_{sys} = K_{rot}$

        $K_{trans}$ & $K_{vib}$ are crossed out

• Calculating this seems very difficult as velocity is different everywhere:

$K_{rot} = \frac 12 m_1 |\overrightarrow v_1 |² + \frac 12 m_2 |\overrightarrow v_2|² + \frac 12 m_3|\overrightarrow v_3|² + …$

Angular Speed

• Thankfully, a rigid object’s rotation can be completely described by the

  • Change in angle around its axis of rotation 

  • A single number describing how mass is distributed around that axis

• The rate of rotation is measured by the angular speed, $\omega_{rot}$:

  • $\omega_{rot} = \frac {\Delta \theta}{\Delta t}$

• For each atom in a rigid object:

  • $|\overrightarrow v_t|=|\overrightarrow r_1|\omega_{rot} : |\overrightarrow v_2| = |\overrightarrow r_2| \omega_{rot}$

Moment of Inertia

• Rotational kinetic energy can be simplified using angular speed:

$K_{rot} = \frac 12 m_1 |\overrightarrow v_1|² + \frac 12 m_2 |\overrightarrow v_2|² + \frac 12 m_3 |\overrightarrow v_3|² + …$

$\Rightarrow K_{rot} = \frac 12 m_1(|\overrightarrow r_1|\omega_{rot})² + \frac 12 m_2(|\overrightarrow r_2|\omega_{rot})² + \frac 12 m_3(|\overrightarrow r_3|\omega_{rot})² + …$

$\Rightarrow K_{rot} = \frac 12 (m_1|\overrightarrow r_1|² + m_2|\overrightarrow r_2|² + m_3 |\overrightarrow r_3|² + …)(\omega_{rot})²$

$\Rightarrow K_{rot} = \frac 12 I (\omega_{rot})²$

• Moment of inertia, !:

$I = m_1 |\overrightarrow r_1 |² + m_2 |\overrightarrow r_2|² + m_3 |\overrightarrow r_3|² + …$

$\Rightarrow I = \sum m |\overrightarrow r|²$

Q9.2.a

A diatomic molecule such as molecular nitrogen $(N_2)$ consists of two atoms each of mass m whose nuclei are a distance d apart.

What is the moment of inertia of the molecule around the z-axis (where the origin is located at its center of mass)? Assume that each atom’s mass is concentrated in its nucleus.

$A)$ $md²$

$B)$ $2md²$

$C)$ $\frac 12 md²$

$D)$ $\frac 14 md²$

$E)$  $4md²$

$m_1 = m_2 = m$

$CM = \frac {r_1 + r_2}{2}$

$|\overrightarrow r_1| = |\overrightarrow r_2| = \frac d2$

$I = \sum m|\overrightarrow r|²$

$= m(\frac d2)² + m(\frac d2)²$

$= 2m \frac {d²}4$

$= \frac 12 md²$

The answer is C

Q9.2.c

Two masses of 0.7 kg each are connected by a low mass rigid rod of length 0.4 m. The object rotates around its center with angular speed 13 radians/s, as shown. What is the rotational kinetic energy of this object around its center?

$A)$ 484 J

$B)$ 4.73 J

$C)$ 2.37 J

$D)$ 0.056 J

$E)$ 0 J

Gather) 

• $m_1 = 0.7 kg$

• $m_2 = 0.7 kg$

• $L = 0.4 m$

• $\omega_{rot} = 13 \frac {rad}s$

• $K_{rot} = ??$

• $m_{rod} = 0$

Organize)

• $m_1 = m_2 = m$

• $|\overrightarrow r_1| = |\overrightarrow r_2| = \frac L2$

• $I_{sys} = I_1 I_2 + I_{rod} = m_1|\overrightarrow r_1|² + m_2|\overrightarrow r_2|²$

  • $I_{rod}$ is crossed out

• $= 2 \cdot m \cdot (\frac L2)² = \frac 12 mL²$

Analyze)

• $K_{rot} = \frac 12 I_{sys}(\omega_{rot})² = \frac 12(\frac 12 mL²)(\omega_{rot})²$

• $= \frac 14 \cdot 0.7 kg \cdot (0.4 m)² \cdot (13 \frac {rad}s)²$

• $= 4.732 kg \frac {m²rad²}{s²} = 4.732 J$

The answer is B.

Moment of Inertia for Continuous Solids

• For a continuous solid object:

I= m_1r_1² + m_2r_2² + m_3r_3² + … = \sum_i(m_ir_i²) \Rightarrow I = \int\int\int \rho ( r ) \cdot r² dV

• A table of the most commonly used moments of inertia is provided for you on the sample equation sheet

• WARNING: Moment of inertia is always calculated relative to a specific axis of rotation. The value of moment of inertia will change if you consider rotation around a different axis.

Cylinder: $I_{cylinder} = \frac 1{12} ML² + \frac 14 MR²$

Disk: $I_{disk} = \frac 12 MR²$

Sphere: $I_{sphere} = \frac 25 MR²$

Simultaneous Rotation and Circular Translation

• When an object is moving in a circle and rotating at the same time, there will be two independent angular speeds:

$\omega_{trans}$ and $\omega_{rot}$

• Sometimes these angular speeds are the same, as is the case with the Moon’s motion. Rotation is still happening even though an observer on Earth only ever sees the same side of the Moon.