§10.3 — Rotational Dynamics

Newton's second law tells us how force changes linear motion:

\[ \vec F = m\vec a \]

Now the question: what changes rotational motion? The answer: torque — the rotational version of force. And just like force, torque—combined with moment of inertia—determines angular acceleration:

\[ \vec \tau = I\vec \alpha \]

Torque (\( \tau \))

Torque measures the rotational effect of a force.

Definition

For a force \( F \) at distance \( r \) from the axis, with angle \( \theta \) between the lever arm and the force:

\[ \tau = rF\sin\theta \]

Or as a vector:

\[ \vec \tau = \vec r \times \vec F \]

Unit: newton–meter (N·m)

Why the Lever Arm Matters

Only the perpendicular component of force produces torque. If force is parallel to the lever arm (\( \theta = 0° \)), torque is zero—no rotation:

\[ \tau = rF\sin(0°) = 0 \]

If force is perpendicular (\( \theta = 90° \)), torque is maximum:

\[ \tau = rF\sin(90°) = rF \]

Practical example: A door. Push perpendicular to the door at its edge, and it rotates easily. Push parallel to the door (sideways), and nothing happens.

Effective Lever Arm

For a perpendicular force, the effective lever arm is the perpendicular distance from the axis to the line of force:

\[ r_{\perp} = r\sin\theta \]

\[ \tau = F \cdot r_{\perp} \]

Numerical example: A 50 N force perpendicular to a door at distance 1.2 m from the hinge:

\[ \tau = 50 \times 1.2 = 60 \text{ N·m} \]

Same force applied at distance 0.5 m:

\[ \tau = 50 \times 0.5 = 25 \text{ N·m} \]

2.4 times less! This is why doorknobs are placed far from the hinges.

Newton's Second Law for Rotation

Newton's second law for rotational motion:

\[ \tau = I\alpha \]

where:

• \( \tau \) = net torque (N·m)
• \( I \) = moment of inertia (kg·m²)
• \( \alpha \) = angular acceleration (rad/s²)

Parallel with linear motion:

Linear	Rotational
\( F = ma \)	\( \tau = I\alpha \)
Mass = resistance to acceleration	Moment = resistance to angular acceleration

Example

A wheel with moment of inertia 0.4 kg·m² experiences a net torque of 8 N·m. What is the angular acceleration?

\[ \alpha = \frac{\tau}{I} = \frac{8}{0.4} = 20 \text{ rad/s}^2 \]

Fast! For comparison, typical linear acceleration on Earth is ~10 m/s².

Multiple Torques

If several forces act on an object, each produces its own torque. The net torque is the sum of all torques:

\[ \tau_{\text{net}} = \tau_1 + \tau_2 + \tau_3 + \cdots \]

Sign convention: counterclockwise torques are positive; clockwise negative.

Example

A disk on a frictionless surface. Two forces:

• \( F_1 = 10 \) N perpendicular to disk at distance 0.5 m (counterclockwise): \( \tau_1 = +5 \) N·m
• \( F_2 = 6 \) N perpendicular to disk at distance 0.5 m (clockwise): \( \tau_2 = -3 \) N·m

Net torque:

\[ \tau_{\text{net}} = 5 - 3 = 2 \text{ N·m} \]

If \( I = 1 \) kg·m²:

\[ \alpha = \frac{2}{1} = 2 \text{ rad/s}^2 \text{ (counterclockwise)} \]

Rotational Equilibrium

If an object is not spinning and should not start spinning, the net torque must be zero:

\[ \tau_{\text{net}} = 0 \implies \alpha = 0 \]

This matters for suspended objects (like a balanced beam) or objects spinning at constant angular velocity.

Worked Example — Atwood Machine With Pulley

Two masses \( m_1 = 4 \) kg and \( m_2 = 6 \) kg hang from a pulley of radius \( r = 0.1 \) m and moment of inertia \( I = 0.02 \) kg·m².

Find: linear acceleration and string tension.

Solution:

For the masses:

• Heavier mass falling: \( m_2 g - T = m_2 a \) (linear)
• Lighter mass rising: \( T - m_1 g = m_1 a \)

For the pulley (rotational): \[ \tau = (T_2 - T_1) r = I\alpha \]

If string doesn't slip: \( a = \alpha r \)

Combining equations:

\[ (m_2 - m_1)g - T(r) = (m_1 + m_2)a + \frac{I\alpha}{r} \]

Simplifying (final formula):

\[ a = \frac{(m_2 - m_1)g}{m_1 + m_2 + I/r^2} \]

Substituting:

\[ a = \frac{(6-4) \times 9.8}{4 + 6 + 0.02/0.01} = \frac{19.6}{10 + 2} = \frac{19.6}{12} \approx 1.63 \text{ m/s}^2 \]

Torque Direction and the Right-Hand Rule

Right-hand rule:

• Fingers point in rotation direction
• Thumb points along \( \vec{\tau} \)

Result: If \( \vec{\tau} \) points upward, rotation is counterclockwise (viewed from above).

What You Should Know

• Torque definition: \( \tau = rF\sin\theta \) or \( \vec\tau = \vec r \times \vec F \)
• Second law (rotation): \( \tau = I\alpha \)
• Net torque: sum of all torques
• Sign convention: counterclockwise (+), clockwise (−)
• Equilibrium: \( \tau_{\text{net}} = 0 \)
• Right-hand rule: for direction

Preview of §10.4

Even if an object spins at constant rate (no angular acceleration), it possesses rotational kinetic energy. How does this energy relate to angular velocity and moment of inertia? Even better: how is work done on a spinning object?

📚 See also: Halliday Vol 1, Ch 10, §10.3 — Rotational dynamics. 🔗 Reference: §9.5 (Angular momentum) — definition of torque and its relation to angular momentum change.

⇧ Back to chapter