9.3 Rotational Kinetic Energy

9.3.1 Period and Angular Speed

• \omega = Angular Speed

• T = Period

\omega = \frac {2\pi}T

• The period T is the time it takes for a rotating object to make one complete revolution. The angular speed \omega has units of radians per second

v = \frac {2\pi r}T = (\frac {2\pi}T)r = \omega r

• An atom rotating at a constant rate around an axis at distance r from the axis goes a distance of the circumference 2\pi r in the period T, so the speed v of the atom can be expressed in terms of the angular speed

9.3.2 Moment of Inertia

Using the above definition, we have a compact expression for rotational kinetic energy:

• K_{rot} = \frac 12 I\omega ²

Example 9.3.A | The Moment of Inertia of a Diatomic Molecule

What is the moment of inertia of a diatomic nitrogen molecule N_2 around its center of mass? The mass of a nitrogen atom is 2.3 × 10^{-26} kg and the average distance between nuclei is 1.5 × 10^{-10} m. Apply the definition of a moment of inertia carefully.

m=2.3×10^{-26}

d = 1.5 × 10^{-10}

I = m_1 r²_{\pm 1} + m_2 r²_{\pm 2} + …

I = M(\frac d2)² + M(\frac d2)² = 2M(\frac d2)²

I = 2 (2.3 × 10^{-26} kg) (\frac {1.5 × 10^{-10}m}2)²

I = 2.59 × 10^{-46} kg \cdot m²

Example 9.3.B | The Moment of Inertia of a Bicycle Wheel

A bicycle wheel has almost all its mass M located in the outer rim at radius R. What is the moment of inertia of the bicycle wheel about its center of mass?

Let m represent the mass of one atom in the rim.

The moment of inertia is 

I = m_1r²_{\pm 1} + m_2r²_{\pm 2} + m_3r²_{\pm 3} + m_4 r²_{\pm 4}+ …

I = m_1R² + m_2 R² + m_3 R² + m_4 R² + …

I = [m_1+m_2+m_3+m_4+…]R²

I=MR²

We’ve assumed that the mass of the spokes is negligible compared to the mass of the rim, so that the total mass is just the mass of the atoms in the rim.

Example 9.3.C | Rotational Kinetic Energy and Work

A wheel is mounted on a stationary axle, which is nearly frictionless so that the wheel turns freely. The wheel has an inner ring with mass 5 kg and radius 10 cm and an outer ring with mass 2 kg and radius 25 cm; the spokes have negligible mass. A string with negligible mass is wrapped around the outer ring and you pull on it, increasing the rotational speed of the wheel. During the time that the wheel’s rotation changes from 4 revolutions per second to 7 revolutions per second, how much work do you do?

m_1 = 5 kg

r_1 = 10 cm = 0.1 m

m_2 = 2 kg

r_2 = 25cm = 0.25 m

\omega_i = (4 \frac {rev}s) (\frac {2\pi rad}{rev}) = 25.1  rad/s

w_f = (7 \frac {rev}s)(\frac {2\pi rad}{rev}) = 44.0 rad/s

W = ??

System: Wheel, String

Surroundings: Your hand, axle, Earth

Principle: Energy Principle

E_f = E_i + W

\frac 12 I \omega ²_f = \frac 12 I \omega²_i + W

W = \frac 12 I (\omega²_f - \omega²_i)

Analyze:

I = m_1 r²_{\pm 1} + m_2 r²_{\pm 2} + …

I = 5kg \cdot (0.1m)² + 2kg \cdot (0.25 m)² = 0.175 kg\cdot m²

W = \frac 12(0.175 kg\cdot m²)((44.0 rad/s)²_f - (25.1 rad/s)²_f)

W = 0.0875 kg\cdot m²(1936 rad²/s²_f - 630.01 rad²/s_i²)

W = 0.0875 kg\cdot m²(1305 rad²/s²)

W = 114.1875 J

Checkpoint 9.3-C-01

A barbell spins around a pivot at its center. The barbell consists of two small balls, each with mass 800 g, at the end of a very low mass rod whose length is 35 cm. The barbell spins with angular speed 40 radians/s. Calculate K_{rot}

m_1 = 0.8 kg

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