Physics

Chapter 10a Fixed-Axis Rotation

Imron Rosyadi

Learning Objectives

By the end of this section, you will be able to:

• Describe the physical meaning of rotational variables as applied to fixed-axis rotation
• Explain how angular velocity is related to tangential speed
• Calculate the instantaneous angular velocity given the angular position function
• Find the angular velocity and angular acceleration in a rotating system
• Calculate the average angular acceleration when the angular velocity is changing
• Calculate the instantaneous angular acceleration given the angular velocity function

Angular Velocity

We expand our description of motion to rotation—specifically, rotational motion about a fixed axis.

• Uniform circular motion: motion in a circle at constant speed.
• Angular position (\(\theta\)): angle swept out by the position vector from the origin to the particle.
  • Increases counterclockwise.
  • Units: radians (rad). \(2\pi\) radians = \(360^{\circ}\).
  • Dimensionless quantity.

• Arc length (\(s\)): distance traced by the particle along its circular path.
• Relationship between \(\theta\), \(r\), and \(s\): \[\theta = \frac{s}{r}\] Where \(r\) is the radius of the circle.
• Vector form: \[\vec{s} = \vec{\theta} \times \vec{r}\] \(\vec{\theta}\) points out of the page for counterclockwise rotation.

Figure 10.2: A particle follows a circular path.

Figure 10.3: Vector relationships for angular position.

Instantaneous Angular Velocity

• Magnitude of angular velocity (\(\omega\)): time rate of change of the angle \(\theta\). \[\omega = \lim_{\Delta t \to 0} \frac{\Delta \theta}{\Delta t} = \frac{d\theta}{dt}\]
• Units: radians per second (rad/s).
• Conversion: multiply revolutions/s by \(2\pi\) to get rad/s.
• Direction:
  • Counterclockwise rotations are positive.
  • Clockwise rotations are negative.

The average angular velocity is defined as \(\bar{\omega} = \frac{\Delta \theta}{\Delta t}\). The instantaneous angular velocity is the limit of the average angular velocity as \(\Delta t\) approaches zero, which is the derivative of angular position with respect to time. This is analogous to how instantaneous linear velocity is defined from average linear velocity.

Angular Velocity and Tangential Speed

• Tangential speed (\(v_t\)) of a particle in circular motion: \[v_t = r\omega\]
  • Tangential speed increases linearly with the radius \(r\) from the axis of rotation for a constant angular velocity.

Figure 10.4: Different tangential speeds on a rotating disk.

• Angular velocity vector (\(\vec{\omega}\)): points along the axis of rotation.
  • Determined by the right-hand rule:
    • Curl fingers in the direction of rotation (counterclockwise).
    • Thumb points in the direction of \(\vec{\omega}\) (positive \(z\)-axis for counterclockwise).
• Vector relationship for tangential velocity: \[\vec{v_t} = \vec{\omega} \times \vec{r}\]

Example: Rotation of a Flywheel

A flywheel sweeps out an angle at \(\theta = (45.0 \text{ rad/s})t\). It rotates counterclockwise.

1. Angular velocity: \(\omega = \frac{d\theta}{dt} = 45.0 \text{ rad/s}\). (Constant)
2. Direction of angular velocity: By the right-hand rule, with counterclockwise rotation, the thumb points out of the page.
3. Radians rotated in 30 s: \(\Delta \theta = \theta(30 \text{ s}) - \theta(0 \text{ s}) = (45.0 \text{ rad/s})(30.0 \text{ s}) - 0 = 1350.0 \text{ rad}\).
4. Tangential speed at 10 cm from axis: \(v_t = r\omega = (0.1 \text{ m})(45.0 \text{ rad/s}) = 4.5 \text{ m/s}\).

Note

In 30 seconds, the flywheel rotates approximately 215 revolutions (\(1350.0 \text{ rad} / (2\pi \text{ rad/rev})\)).

This highlights the energy storage capacity of flywheels.

Angular Acceleration

• Instantaneous angular acceleration (\(\alpha\)): derivative of angular velocity with respect to time. \[\alpha = \lim_{\Delta t \to 0} \frac{\Delta \omega}{\Delta t} = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}\]

• Units: radians per second squared (rad/s\(^2\)).

• Angular acceleration vector (\(\vec{\alpha}\)):

  • If \(\vec{\omega}\) is along the positive \(z\)-axis and $

  $ is positive, \(\vec{\alpha}\) points along the \(+z\)-axis (increasing rotation rate).

  • If \(\vec{\omega}\) is along the positive \(z\)-axis and $

  $ is negative, \(\vec{\alpha}\) points along the \(-z\)-axis (decreasing rotation rate).

Figure 10.7: Angular acceleration direction relative to angular velocity.

• Tangential acceleration (\(a_t\)): related to angular acceleration.

  \[a_t = r\alpha\]

  This is due to the change in magnitude of tangential velocity.

• Vector form of tangential acceleration: \[\vec{a_t} = \vec{\alpha} \times \vec{r}\]

Problem-Solving Strategy: Rotational Kinematics

1. Identify Rotational Motion: Confirm the problem involves fixed-axis rotation.
2. Identify Unknowns: Determine what quantities need to be found. A sketch helps.
3. Identify Knowns: List all given information or what can be inferred.
4. Select Equations: Choose the appropriate kinematic equations (Table 10.1 in notes) to solve for unknowns. Consider translational analogs if helpful.
5. Solve Numerically: Substitute known values with units. Use radians for angles.
6. Check Reasonableness: Does the answer make physical sense?

Table 10.1 Kinematic Equations:

\(\theta_f = \theta_0 + \bar{\omega}t\)

\(\omega_f = \omega_0 + \alpha t\)

\(\theta_f = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2\)

\(\omega_f^2 = \omega_0^2 + 2\alpha(\Delta\theta)\)

These equations are valid for constant angular acceleration.

Example: Spinning Bicycle Wheel

A bicycle wheel accelerates from rest to 250 rpm in 5.00 s.

1. Average angular acceleration (\(\alpha\)): \(\Delta\omega = 250 \text{ rev/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 26.2 \text{ rad/s}\)

   \(\alpha = \frac{\Delta\omega}{\Delta t} = \frac{26.2 \text{ rad/s}}{5.00 \text{ s}} = 5.24 \text{ rad/s}^2\)

2. Time to stop with \(\alpha = -87.3 \text{ rad/s}^2\):

   Initial \(\omega_0 = 26.2 \text{ rad/s}\), final \(\omega_f = 0 \text{ rad/s}\).

   \(\omega_f = \omega_0 + \alpha t \Rightarrow t = \frac{\omega_f - \omega_0}{\alpha} = \frac{0 - 26.2 \text{ rad/s}}{-87.3 \text{ rad/s}^2} = 0.300 \text{ s}\)

Note

The acceleration to spin up the wheel is small and positive, taking 5 seconds. The acceleration to stop the wheel (braking) is large and negative, stopping it quickly in 0.3 seconds.

Example: Wind Turbine

A wind turbine stops in 30 s with \(\omega(t) = [(t \text{ s}^{-1} - 30.0)^2 / 100.0] \text{ rad/s}\). It rotates counterclockwise.

1. Directions of \(\vec{\omega}\) and \(\vec{\alpha}\):

   • Rotation is counterclockwise, so \(\vec{\omega}\) points out of the page.
   • Since it’s slowing down, \(\vec{\alpha}\) is opposite \(\vec{\omega}\), so

   \(\vec{\alpha}\) points into the page.

2. Average angular acceleration (\(\bar{\alpha}\)):

   \(\omega_0 = \omega(0) = [(0-30.0)^2/100.0] \text{ rad/s} = 9.0 \text{ rad/s}\)

   \(\omega_f = \omega(30) = [(30-30.0)^2/100.0] \text{ rad/s} = 0 \text{ rad/s}\)

   \(\bar{\alpha} = \frac{\Delta\omega}{\Delta t} = \frac{0 - 9.0 \text{ rad/s}}{30.0 \text{ s} - 0 \text{ s}} = -0.3 \text{ rad/s}^2\). (Negative, consistent with direction.)

3. Instantaneous angular acceleration at \(t=0, 15, 30 \text{ s}\):

   \(\alpha(t) = \frac{d\omega}{dt} = \frac{d}{dt} \left[ \frac{(t-30)^2}{100} \right] = \frac{2(t-30)}{100} = \frac{t-30}{50} \text{ rad/s}^2\)

   \(\alpha(0 \text{ s}) = \frac{0-30}{50} = -0.6 \text{ rad/s}^2\)

   \(\alpha(15 \text{ s}) = \frac{15-30}{50} = -0.3 \text{ rad/s}^2\)

   \(\alpha(30 \text{ s}) = \frac{30-30}{50} = 0 \text{ rad/s}^2\)

Learning Objectives

By the end of this section, you will be able to:

• Derive the kinematic equations for rotational motion with constant angular acceleration
• Select from the kinematic equations for rotational motion with constant angular acceleration the appropriate equations to solve for unknowns in the analysis of systems undergoing fixed-axis rotation
• Use solutions found with the kinematic equations to verify the graphical analysis of fixed-axis rotation with constant angular acceleration

Kinematics of Rotational Motion

• Under constant angular acceleration, rotational variables are related.

• Average angular velocity (\(\bar{\omega}\)): \[\bar{\omega} = \frac{\omega_0 + \omega_f}{2}\]

• Angular position from average angular velocity: \[\theta_f = \theta_0 + \bar{\omega}t\]

• Angular velocity from angular acceleration:

  Derived from \(\alpha = \frac{d\omega}{dt}\). Integrating gives: \[\omega_f = \omega_0 + \alpha t\]

The derivation of \(\omega_f = \omega_0 + \alpha t\) involves integrating \(\alpha dt = d\omega\) from \(t_0\) to \(t\) and \(\omega_0\) to \(\omega_f\). Since \(\alpha\) is constant, it pulls out of the integral.

Kinematics of Rotational Motion (Cont.)

• Angular displacement from angular velocity and acceleration: Derived from \(\omega = \frac{d\theta}{dt}\) and substituting \(\omega(t) = \omega_0 + \alpha t\). Integrating gives: \[\theta_f = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2\]
• Angular velocity from angular displacement and acceleration: Derived by combining the previous equations to eliminate \(t\). \[\omega_f^2 = \omega_0^2 + 2\alpha(\Delta\theta)\]
• These four equations (summarized in the next slide) are the kinematic equations for rotational motion with constant angular acceleration. They are direct analogs to linear kinematic equations.

The derivation of \(\theta_f = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2\) involves integrating \(\omega dt = d\theta\). Here, \(\omega\) is not constant, so we substitute \(\omega = \omega_0 + \alpha t\) before integrating.

Rotational Kinematic Equations (Constant \(\alpha\))

Rotational	Translational
\(\theta_f = \theta_0 + \bar{\omega}t\)	\(x_f = x_0 + \bar{v}t\)
\(\omega_f = \omega_0 + \alpha t\)	\(v_f = v_0 + at\)
\(\theta_f = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2\)	\(x_f = x_0 + v_0 t + \frac{1}{2}at^2\)
\(\omega_f^2 = \omega_0^2 + 2\alpha(\Delta\theta)\)	\(v_f^2 = v_0^2 + 2a(\Delta x)\)

Note

These equations are fundamental for analyzing fixed-axis rotation when angular acceleration is constant. Notice the direct analogy between rotational and translational variables.

Example: Acceleration of a Fishing Reel

Fishing line unwinds from a reel (radius \(r = 4.50 \text{ cm}\)) from rest with \(\alpha = 110 \text{ rad/s}^2\) for 2.00 s.

1. Final angular velocity (\(\omega_f\)) after 2 s:

   Knowns: \(\omega_0 = 0 \text{ rad/s}\) (starts from rest), \(\alpha = 110 \text{ rad/s}^2\), \(t = 2.00 \text{ s}\).

   Using \(\omega_f = \omega_0 + \alpha t\):

   \(\omega_f = 0 + (110 \text{ rad/s}^2)(2.00 \text{ s}) = 220 \text{ rad/s}\).

2. Number of revolutions the reel makes:

   Knowns: \(\theta_0 = 0 \text{ rad}\), \(\omega_0 = 0 \text{ rad/s}\), \(\alpha = 110 \text{ rad/s}^2\), \(t = 2.00 \text{ s}\).

   Using \(\theta_f = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2\):

   \(\theta_f = 0 + 0 + \frac{1}{2}(110 \text{ rad/s}^2)(2.00 \text{ s})^2 = 220 \text{ rad}\).

   Converting to revolutions: \(220 \text{ rad} \times \frac{1 \text{ rev}}{2\pi \text{ rad}} \approx 35.0 \text{ rev}\).

Important

Units are crucial! Ensure angles are in radians for calculations involving angular velocity and acceleration.

The reel spins very fast (220 rad/s \(\approx\) 2100 rpm) after just 2 seconds.

Example: Reel Slows Down and Stops

The fisherman applies a brake, causing an angular acceleration of \(\alpha = -300 \text{ rad/s}^2\). How long does it take the reel to stop?

• Knowns:
  • Initial angular velocity \(\omega_0 = 220 \text{ rad/s}\) (from previous example).
  • Final angular velocity \(\omega_f = 0 \text{ rad/s}\) (comes to a stop).
  • Angular acceleration \(\alpha = -300 \text{ rad/s}^2\).
• Using \(\omega_f = \omega_0 + \alpha t\): \(0 = 220 \text{ rad/s} + (-300 \text{ rad/s}^2)t\) \(t = \frac{-220 \text{ rad/s}}{-300 \text{ rad/s}^2} = 0.733 \text{ s}\).

Warning

Pay attention to the signs of angular acceleration: negative indicates slowing down if \(\omega\) is positive.

The large negative acceleration causes a quick stop, which can snap fishing lines.

Example: Angular Acceleration of a Propeller

Propeller’s angular velocity starts at 30 rad/s and drops linearly to 0 rad/s over 5 seconds.

1. Angular acceleration (\(\alpha\)):

   The slope of the angular velocity-time graph is \(\alpha\).

   \(\alpha = \frac{\Delta\omega}{\Delta t} = \frac{0 \text{ rad/s} - 30 \text{ rad/s}}{5 \text{ s} - 0 \text{ s}} = -6.0 \text{ rad/s}^2\).

   Verification using \(\omega_f = \omega_0 + \alpha t\):

   \(0 = 30 + (-6.0)(5) \Rightarrow 0 = 30 - 30\), which is correct.

2. Angle rotated (\(\Delta\theta\)) during these 5 seconds:

   The area under the \(\omega\)-t graph is the angular displacement. This is a triangle.

   \(\Delta\theta = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (5 \text{ s})(30 \text{ rad/s}) = 75 \text{ rad}\).

   Verification using \(\theta_f = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2\) (assuming \(\theta_0=0\)):

   \(\theta_f = (30 \text{ rad/s})(5 \text{ s}) + \frac{1}{2}(-6.0 \text{ rad/s}^2)(5 \text{ s})^2 = 150 - 75 = 75 \text{ rad}\).

Tip

Graphical analysis (slope, area under curve) is a powerful tool to complement kinematic equations.

Learning Objectives

By the end of this section, you will be able to:

• Given the linear kinematic equation, write the corresponding rotational kinematic equation
• Calculate the linear distances, velocities, and accelerations of points on a rotating system given the angular velocities and accelerations

Angular vs. Linear Variables

	Linear Variable	Rotational Variable
Position	\(x\)	\(\theta\)
Velocity	\(v = \frac{dx}{dt}\)	\(\omega = \frac{d\theta}{dt}\)
Acceleration	\(a = \frac{dv}{dt}\)	\(\alpha = \frac{d\omega}{dt}\)

• Units:
  • Linear position: meters (m)
  • Angular position: radians (rad) (dimensionless)
  • Linear velocity: m/s
  • Angular velocity: rad/s
  • Linear acceleration: m/s\(^2\)
  • Angular acceleration: rad/s\(^2\)

Note

The relationships between angular and linear variables are key to understanding how rotational motion affects points on a rotating object.

Relating Linear and Rotational Variables (Circular Motion)

• Tangential Speed (\(v_t\)): The linear speed of a point at radius \(r\). \[v_t = r\omega\]
• Centripetal Acceleration (\(a_c\)): Always points inward, towards the center of rotation. \[a_c = \frac{v_t^2}{r} = r\omega^2\]

Figure 10.14: Centripetal and tangential accelerations.

• Tangential Acceleration (\(a_t\)): Due to change in magnitude of \(v_t\). \[a_t = r\alpha\]

• Total Linear Acceleration (\(\vec{a}\)): Vector sum of centripetal and tangential accelerations. \[\vec{a} = \vec{a_c} + \vec{a_t}\]

  Magnitude: \(|\vec{a}| = \sqrt{a_c^2 + a_t^2}\)

Summary of Relationships (Circular Motion)

Rotational	Translational	Relationship (\(r\) = radius)
\(\theta\)	\(s\)	\(\theta = \frac{s}{r}\)
\(\omega\)	\(v_t\)	\(\omega = \frac{v_t}{r}\)
\(\alpha\)	\(a_t\)	\(\alpha = \frac{a_t}{r}\)
	\(a_c\)	\(a_c = \frac{v_t^2}{r} = r\omega^2\)

Important

These relationships are critical for translating between angular and linear descriptions of motion for any point on a rotating rigid body.

Example: Linear Acceleration of a Centrifuge

Centrifuge (\(r = 20 \text{ cm}\)) accelerates from \(10,000 \text{ rpm}\) to rest in 30 seconds. Counterclockwise rotation. Find total acceleration at \(t=29.0 \text{ s}\).

1. Angular acceleration (\(\alpha\)):

   \(\omega_0 = 10,000 \text{ rpm} = 10,000 \times \frac{2\pi}{60} \text{ rad/s} = 1047.2 \text{ rad/s}\).

   \(\alpha = \frac{\omega_f - \omega_0}{t} = \frac{0 - 1047.2 \text{ rad/s}}{30.0 \text{ s}} = -34.9 \text{ rad/s}^2\).

2. Tangential acceleration (\(a_t\)):

   \(a_t = r\alpha = (0.2 \text{ m})(-34.9 \text{ rad/s}^2) = -7.0 \text{ m/s}^2\).

3. Angular velocity (\(\omega\)) at \(t=29.0 \text{ s}\):

   \(\omega = \omega_0 + \alpha t = 1047.2 \text{ rad/s} + (-34.9 \text{ rad/s}^2)(29.0 \text{ s}) = 35.1 \text{ rad/s}\).

4. Tangential speed (\(v_t\)) at \(t=29.0 \text{ s}\):

   \(v_t = r\omega = (0.2 \text{ m})(35.1 \text{ rad/s}) = 7.0 \text{ m/s}\).

5. Centripetal acceleration (\(a_c\)) at \(t=29.0 \text{ s}\):

   \(a_c = \frac{v_t^2}{r} = \frac{(7.0 \text{ m/s})^2}{0.2 \text{ m}} = 245.0 \text{ m/s}^2\).

6. Magnitude of total acceleration (\(|\vec{a}|\)):

   \(|\vec{a}| = \sqrt{a_c^2 + a_t^2} = \sqrt{(245.0)^2 + (-7.0)^2} = 245.1 \text{ m/s}^2\).

7. Direction of total acceleration:

   The angle \(\theta = \tan^{-1}\left(\frac{a_t}{a_c}\right) = \tan^{-1}\left(\frac{-7.0}{245.0}\right) = -1.6^\circ\).

   This means the vector is slightly angled clockwise from the purely inward centripetal direction.

Learning Objectives

By the end of this section, you will be able to:

• Describe the differences between rotational and translational kinetic energy
• Define the physical concept of moment of inertia in terms of the mass distribution from the rotational axis
• Explain how the moment of inertia of rigid bodies affects their rotational kinetic energy
• Use conservation of mechanical energy to analyze systems undergoing both rotation and translation
• Calculate the angular velocity of a rotating system when there are energy losses due to nonconservative forces

Rotational Kinetic Energy

• Any moving object has kinetic energy. For translational motion, \(K = \frac{1}{2}mv^2\).

• For a rotating rigid body, each point has a different tangential velocity.

• We express kinetic energy using angular velocity (\(\omega\)), which is the same for all points.

• For a single particle rotating around a fixed axis:

  \(K = \frac{1}{2}mv_t^2 = \frac{1}{2}m(\omega r)^2 = \frac{1}{2}(mr^2)\omega^2\)

• For a rigid rotating body (sum of many small masses \(m_j\) at radii \(r_j\)): \[K = \sum_j \frac{1}{2}m_j v_j^2 = \sum_j \frac{1}{2}m_j (r_j \omega)^2 = \frac{1}{2} \left( \sum_j m_j r_j^2 \right) \omega^2\]

• Units: Joules (J).

Moment of Inertia (\(I\))

• The quantity \(\sum_j m_j r_j^2\) is called the moment of inertia (\(I\)). \[I = \sum_j m_j r_j^2\]
• Units: kg \(\cdot\) m\(^2\).
• Definition: The quantitative measure of rotational inertia.
  • Analogous to mass in translational motion.
  • Greater \(I\) means greater resistance to change in angular velocity.
• Dependence on mass distribution:
  • Objects with more mass concentrated farther from the axis of rotation have larger moments of inertia.
  • Example: A hollow cylinder has greater rotational inertia than a solid cylinder of the same mass and radius about its central axis.

Rotational Kinetic Energy and Moment of Inertia

• The expression for the kinetic energy of a rotating rigid body: \[K = \frac{1}{2}I\omega^2\]
• This equation shows rotational kinetic energy is directly proportional to moment of inertia and the square of angular velocity.
• Applications: Flywheel energy-storage devices (e.g., in cars).

Rotational	Translational
\(I = \sum_j m_j r_j^2\)	\(m\)
\(K = \frac{1}{2}I\omega^2\)	\(K = \frac{1}{2}mv^2\)

Tip

Understanding the analogy between \(I\) and \(m\) is crucial. \(I\) represents how difficult it is to rotate an object, while \(m\) represents how difficult it is to translate an object.

Example: Moment of Inertia of a System of Particles

Six 20-g washers are spaced 10 cm apart on a 0.5 m rod of negligible mass. Axis of rotation at 25 cm (center).

Figure 10.19: Six washers on a rotating rod.

1. Moment of inertia (\(I\)) of the system:

   Mass of each washer \(m = 0.02 \text{ kg}\).

   Distances from center: \(0.05 \text{ m}, 0.15 \text{ m}, 0.25 \text{ m}\) (for two washers each).

   \(I = \sum m_j r_j^2 = (0.02 \text{ kg})[2(0.25 \text{ m})^2 + 2(0.15 \text{ m})^2 + 2(0.05 \text{ m})^2]\)

   \(I = 0.02 \text{ kg} [2(0.0625) + 2(0.0225) + 2(0.0025)] \text{ m}^2\)

   \(I = 0.02 \text{ kg} [0.125 + 0.045 + 0.005] \text{ m}^2 = 0.02 \text{ kg} (0.175) \text{ m}^2 = 0.0035 \text{ kg} \cdot \text{m}^2\).

2. Moment of inertia if two closest washers are removed:

   \(I' = (0.02 \text{ kg})[2(0.25 \text{ m})^2 + 2(0.15 \text{ m})^2] = 0.02 \text{ kg} [0.125 + 0.045] \text{ m}^2 = 0.02 \text{ kg} (0.170) \text{ m}^2 = 0.0034 \text{ kg} \cdot \text{m}^2\).

3. Rotational kinetic energy if system rotates at 5 rev/s:

   \(\omega = 5 \text{ rev/s} \times 2\pi \text{ rad/rev} = 10\pi \text{ rad/s} \approx 31.4 \text{ rad/s}\).

   \(K = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.0035 \text{ kg} \cdot \text{m}^2)(10\pi \text{ rad/s})^2 \approx 1.73 \text{ J}\).

The washers close to the axis contribute minimally to the moment of inertia. This emphasizes the \(r^2\) dependence in the formula.

Problem-Solving Strategy: Rotational Energy

1. Identify Energy/Work: Determine if the problem involves energy or work in rotation.
2. Define System: Clearly identify the system of interest. A sketch helps.
3. Analyze Energy Types: Identify all forms of work and energy involved (kinetic, potential, rotational, translational).
4. Conservation of Mechanical Energy: If no nonconservative forces (like friction) are present, use \(K_i + U_i = K_f + U_f\).
5. Nonconservative Forces: If nonconservative forces are present, account for energy losses/gains.
6. Simplify: Eliminate terms that are zero or cancel out.
7. Evaluate: Check if the numerical answer is reasonable for the physical situation.

Example: Calculating Helicopter Energies

Helicopter has four blades, each \(4.00 \text{ m}\) long, \(50.0 \text{ kg}\) (thin rods rotating about one end). Total loaded mass \(1000 \text{ kg}\).

1. Rotational kinetic energy (\(K_R\)) in blades at 300 rpm:

   \(\omega = 300 \text{ rpm} = 300 \times \frac{2\pi}{60} \text{ rad/s} = 10\pi \text{ rad/s} \approx 31.4 \text{ rad/s}\).

   Moment of inertia for one blade (thin rod, axis at end) \(I_{blade} = \frac{1}{3}ML^2\).

   \(I_{total} = 4 \times \frac{1}{3}(50.0 \text{ kg})(4.00 \text{ m})^2 = 4 \times \frac{1}{3}(50.0 \text{ kg})(16.0 \text{ m}^2) = 1067.0 \text{ kg} \cdot \text{m}^2\).

   \(K_R = \frac{1}{2}I_{total}\omega^2 = \frac{1}{2}(1067.0 \text{ kg} \cdot \text{m}^2)(31.4 \text{ rad/s})^2 \approx 5.26 \times 10^5 \text{ J}\).

2. Translational kinetic energy (\(K_T\)) at \(20.0 \text{ m/s}\), compare with \(K_R\):

   \(K_T = \frac{1}{2}mv^2 = \frac{1}{2}(1000.0 \text{ kg})(20.0 \text{ m/s})^2 = 2.00 \times 10^5 \text{ J}\).

   Ratio \(K_T/K_R = \frac{2.00 \times 10^5 \text{ J}}{5.26 \times 10^5 \text{ J}} \approx 0.380\).

Important

Most of the helicopter’s kinetic energy is in its spinning blades. This highlights the importance of rotational kinetic energy in such systems.

Example: Energy in a Boomerang

Boomerang (mass \(1.0 \text{ kg}\)) hurled at \(30.0 \text{ m/s}\) at \(40.0^\circ\) above horizontal, rotating at \(10.0 \text{ rev/s}\).

Moment of inertia \(I = \frac{1}{12}mL^2\), where \(L = 0.7 \text{ m}\).

1. Total energy when it leaves the hand:

   \(I = \frac{1}{12}(1.0 \text{ kg})(0.7 \text{ m})^2 = 0.0408 \text{ kg} \cdot \text{m}^2\).

   \(\omega = 10.0 \text{ rev/s} \times 2\pi \text{ rad/rev} = 20\pi \text{ rad/s} \approx 62.83 \text{ rad/s}\).

   \(K_R = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.0408 \text{ kg} \cdot \text{m}^2)(62.83 \text{ rad/s})^2 \approx 80.9 \text{ J}\).

   \(K_T = \frac{1}{2}mv^2 = \frac{1}{2}(1.0 \text{ kg})(30.0 \text{ m/s})^2 = 450.0 \text{ J}\).

   \(K_{Total} = K_R + K_T = 80.9 \text{ J} + 450.0 \text{ J} = 530.9 \text{ J}\).

2. Maximum height (\(h\)) from hand (neglecting air resistance):

   Use conservation of mechanical energy (\(E_{Before} = E_{Final}\)).

   \(E_{Before} = \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2 + \frac{1}{2}I\omega^2\).

   At max height, \(v_y = 0\), so \(E_{Final} = \frac{1}{2}mv_x^2 + \frac{1}{2}I\omega^2 + mgh\).

   Since \(v_x\) and \(\omega\) are constant (no air resistance), these terms cancel:

   \(\frac{1}{2}mv_y^2 = mgh\).

   \(v_y = (30.0 \text{ m/s})\sin(40^\circ) = 19.28 \text{ m/s}\).

   \(h = \frac{v_y^2}{2g} = \frac{(19.28 \text{ m/s})^2}{2(9.8 \text{ m/s}^2)} = 18.97 \text{ m}\).

Note

In the absence of air resistance, the rotational kinetic energy does not affect the maximum height because it is conserved throughout the projectile motion.

Learning Objectives

By the end of this section, you will be able to:

• Calculate the moment of inertia for uniformly shaped, rigid bodies
• Apply the parallel axis theorem to find the moment of inertia about any axis parallel to one already known
• Calculate the moment of inertia for compound objects

Moment of Inertia (\(I\)) for Continuous Objects

• For an object made of discrete point masses, \(I = \sum_i m_i r_i^2\).
• For continuously distributed mass, we use integration: \[I = \int r^2 dm\]
  • Here \(dm\) is an infinitesimal piece of mass at distance \(r\) from the axis.
• The moment of inertia depends on the axis of rotation chosen.

Figure 10.23: Barbell rotating about different axes.

Example: Barbell with masses \(m\) at each end, length \(2R\).

• Axis through center: \(I_1 = mR^2 + mR^2 = 2mR^2\).

• Axis through one end: \(I_2 = m(0)^2 + m(2R)^2 = 4mR^2\).

  It is twice as hard to rotate about the end.

Calculating \(I\): Uniform Thin Rod (Axis through center)

• Mass \(M\), length \(L\). Axis perpendicular to rod, through its midpoint.

• Linear mass density \(\lambda = M/L\). So \(dm = \lambda dx\).

• Integration limits: \(x = -L/2\) to \(x = L/2\).

  \[I = \int r^2 dm = \int_{-L/2}^{L/2} x^2 \lambda dx\]

  \[I = \lambda \left[ \frac{x^3}{3} \right]_{-L/2}^{L/2} = \frac{\lambda}{3} \left[ \left(\frac{L}{2}\right)^3 - \left(-\frac{L}{2}\right)^3 \right]\]

  \[I = \frac{\lambda}{3} \left[ \frac{L^3}{8} - \left(-\frac{L^3}{8}\right) \right] = \frac{\lambda}{3} \left( \frac{2L^3}{8} \right) = \frac{\lambda L^3}{12}\]

  Substitute \(\lambda = M/L\):

  \[I = \frac{M}{L} \frac{L^3}{12} = \frac{1}{12}ML^2\]

Figure 10.25: Thin rod rotating about its center.

Calculating \(I\): Uniform Thin Rod (Axis at end)

• Mass \(M\), length \(L\). Axis perpendicular to rod, through one end.

• Linear mass density \(\lambda = M/L\). So \(dm = \lambda dx\).

• Integration limits: \(x = 0\) to \(x = L\).

  \[I = \int r^2 dm = \int_{0}^{L} x^2 \lambda dx\]

  \[I = \lambda \left[ \frac{x^3}{3} \right]_{0}^{L} = \frac{\lambda}{3} [L^3 - 0^3] = \frac{\lambda L^3}{3}\]

  Substitute \(\lambda = M/L\):

  \[I = \frac{M}{L} \frac{L^3}{3} = \frac{1}{3}ML^2\]

Figure 10.26: Thin rod rotating about its end.

Note

The moment of inertia is larger when the axis is at the end (\(\frac{1}{3}ML^2\)) compared to the center (\(\frac{1}{12}ML^2\)). This is because more mass is distributed further from the axis of rotation.

The Parallel-Axis Theorem

• Purpose: To find the moment of inertia about any axis parallel to an axis through the object’s center of mass (CM).
• Theorem: If \(I_{CM}\) is the moment of inertia about an axis through the center of mass, and \(d\) is the distance from this axis to a new parallel axis, then the moment of inertia about the new axis is: \[I_{parallel-axis} = I_{CM} + md^2\] Where \(m\) is the total mass of the object.

Tip

Applying to the thin rod example: For a rod rotating about its end:

\(I_{CM} = \frac{1}{12}ML^2\).

The distance from the center of mass (midpoint) to the end is \(d = L/2\).

\(I_{end} = I_{CM} + md^2 = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1}{12}ML^2 + \frac{3}{12}ML^2 = \frac{4}{12}ML^2 = \frac{1}{3}ML^2\).

This matches the direct integration result.

Calculating \(I\): Uniform Thin Disk (Axis through center)

• Mass \(M\), radius \(R\). Axis perpendicular to disk, through its center.

• Surface mass density \(\sigma = M/A = M/(\pi R^2)\). So \(dm = \sigma dA\).

• Consider infinitesimal rings of radius \(r\) and width \(dr\). Area of ring \(dA = 2\pi r dr\).

• Integration limits: \(r = 0\) to \(r = R\).

  \[I = \int r^2 dm = \int_{0}^{R} r^2 \sigma (2\pi r dr)\]

  \[I = 2\pi\sigma \int_{0}^{R} r^3 dr = 2\pi\sigma \left[ \frac{r^4}{4} \right]_{0}^{R} = 2\pi\sigma \frac{R^4}{4}\]

  Substitute \(\sigma = M/(\pi R^2)\):

  \[I = 2\pi \frac{M}{\pi R^2} \frac{R^4}{4} = \frac{2M R^2}{4} = \frac{1}{2}MR^2\]

Figure 10.27: Thin disk rotating about its center.

Moment of Inertia for Compound Objects

• The moment of inertia of a compound object is the sum of the moments of inertia of its individual parts about a common axis.

  \[I_{total} = \sum_i I_i\]

Figure 10.28: Disk at the end of a rod.

Example: Thin disk (mass \(m_d\), radius \(R\)) at the end of a thin rod (mass \(m_r\), length \(L\)). Axis of rotation at point A (end of rod).

1. Rod: Axis at its end.

   \(I_{rod} = \frac{1}{3}m_r L^2\).

2. Disk: Axis is parallel to the disk’s CM axis. Distance from disk’s CM to axis A is \(d = L+R\).

   \(I_{disk, CM} = \frac{1}{2}m_d R^2\).

   Using parallel-axis theorem: \(I_{disk} = I_{disk, CM} + m_d d^2 = \frac{1}{2}m_d R^2 + m_d(L+R)^2\).

3. Total Moment of Inertia:

   \(I_{total} = I_{rod} + I_{disk} = \frac{1}{3}m_r L^2 + \frac{1}{2}m_d R^2 + m_d(L+R)^2\).

Example: Person on a Merry-Go-Round

A \(25 \text{ kg}\) child (\(m_c\)) stands at \(r_c = 1.0 \text{ m}\) from the axis. Merry-go-round (\(m_m = 500 \text{ kg}\), radius \(r_m = 2.0 \text{ m}\)) is a uniform solid disk.

Figure 10.29: Child on a merry-go-round.

• Moment of inertia of the child (\(I_c\)):

  Child approximated as a point mass.

  \(I_c = m_c r_c^2 = (25 \text{ kg})(1.0 \text{ m})^2 = 25 \text{ kg} \cdot \text{m}^2\).

• Moment of inertia of the merry-go-round (\(I_m\)):

  Solid disk rotating about its center.

  \(I_m = \frac{1}{2}m_m r_m^2 = \frac{1}{2}(500 \text{ kg})(2.0 \text{ m})^2 = \frac{1}{2}(500 \text{ kg})(4.0 \text{ m}^2) = 1000 \text{ kg} \cdot \text{m}^2\).

• Total moment of inertia (\(I_{total}\)):

  \(I_{total} = I_c + I_m = 25 \text{ kg} \cdot \text{m}^2 + 1000 \text{ kg} \cdot \text{m}^2 = 1025 \text{ kg} \cdot \text{m}^2\).

Note

The merry-go-round’s moment of inertia is significantly larger, as its mass is more distributed further from the axis than the child’s mass.

Example: Angular Velocity of a Pendulum

A rod-shaped pendulum (\(L = 30 \text{ cm}\), \(M = 300 \text{ g}\)) released from rest at \(30^\circ\). Find angular velocity at its lowest point.

Figure 10.30: Rod pendulum.

• Strategy: Use conservation of mechanical energy (\(U_i + K_i = U_f + K_f\)).

• Initial State (top of swing):

  • \(K_i = 0\) (released from rest).
  • Potential energy \(U_i = Mgh_{cm}\). The center of mass (CM) of the rod is at \(L/2\).
  • The vertical height \(h_{cm}\) of the CM above its lowest point: \(h_{cm} = \frac{L}{2} - \frac{L}{2}\cos\theta = \frac{L}{2}(1 - \cos\theta)\).
  • \(U_i = Mg \frac{L}{2}(1 - \cos\theta)\).

• Final State (lowest point):

  • \(U_f = 0\) (set reference point at lowest CM position).
  • Rotational kinetic energy \(K_f = \frac{1}{2}I\omega^2\).
  • Moment of inertia for a rod rotating about its end: \(I = \frac{1}{3}ML^2\).

• Conservation of Energy:

  \(Mg \frac{L}{2}(1 - \cos\theta) + 0 = 0 + \frac{1}{2} \left(\frac{1}{3}ML^2\right)\omega^2\)

  \(Mg \frac{L}{2}(1 - \cos\theta) = \frac{1}{6}ML^2\omega^2\)

  Cancel \(M\) and \(L\):

  \(g \frac{1}{2}(1 - \cos\theta) = \frac{1}{6}L\omega^2\)

  \(\omega^2 = \frac{3g}{L}(1 - \cos\theta)\)

  \(\omega = \sqrt{\frac{3g}{L}(1 - \cos\theta)}\).

• Numerical Calculation:

  \(L = 0.3 \text{ m}\), \(\theta = 30^\circ\).

  \(\omega = \sqrt{\frac{3(9.8 \text{ m/s}^2)}{0.3 \text{ m}}(1 - \cos 30^\circ)} = \sqrt{98(1 - 0.866)} = \sqrt{98(0.134)} = \sqrt{13.132} \approx 3.62 \text{ rad/s}\).

Key Takeaways

• Rotational Variables: Angular position (\(\theta\)), angular velocity (\(\omega\)), and angular acceleration (\(\alpha\)) describe rotational motion, analogous to linear displacement, velocity, and acceleration.
• Relationships:
  • Tangential speed \(v_t = r\omega\).
  • Tangential acceleration \(a_t = r\alpha\).
  • Centripetal acceleration \(a_c = r\omega^2 = v_t^2/r\).
  • Total linear acceleration \(\vec{a} = \vec{a_c} + \vec{a_t}\).
• Kinematic Equations: For constant angular acceleration, these equations are direct analogs of linear kinematic equations.
• Rotational Kinetic Energy: \(K = \frac{1}{2}I\omega^2\). This is the energy associated with rotation.
• Moment of Inertia (\(I\)):
  • The rotational analog of mass. It measures an object’s resistance to changes in angular velocity.
  • Depends on mass distribution relative to the axis of rotation: \(I = \sum m_j r_j^2\) (for point masses) or \(I = \int r^2 dm\) (for continuous objects).
  • Parallel-Axis Theorem: \(I_{parallel-axis} = I_{CM} + md^2\). This simplifies finding \(I\) for parallel axes.
• Conservation of Energy: Can be applied to systems involving both translational and rotational kinetic energy, as well as potential energy.

Key Equations

Equation	Description
\(\theta = \frac{s}{r}\)	Angular position from arc length
\(\omega = \frac{d\theta}{dt}\)	Instantaneous angular velocity
\(\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}\)	Instantaneous angular acceleration
\(v_t = r\omega\)	Tangential speed
\(a_t = r\alpha\)	Tangential acceleration
\(a_c = r\omega^2 = \frac{v_t^2}{r}\)	Centripetal acceleration
\(\omega_f = \omega_0 + \alpha t\)	Rotational kinematic equation (constant \(\alpha\))
\(\theta_f = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2\)	Rotational kinematic equation (constant \(\alpha\))
\(\omega_f^2 = \omega_0^2 + 2\alpha(\Delta\theta)\)	Rotational kinematic equation (constant \(\alpha\))
\(I = \sum_j m_j r_j^2\) (discrete)	Moment of inertia
\(I = \int r^2 dm\) (continuous)	Moment of inertia (integral form)
\(K = \frac{1}{2}I\omega^2\)	Rotational kinetic energy
\(I_{parallel-axis} = I_{CM} + md^2\)	Parallel-axis theorem

Key Terms

Term	Definition
Angular Position (\(\theta\))	The angle swept out by a particle’s position vector from a fixed reference direction. Measured in radians.
Angular Velocity (\(\omega\))	The time rate of change of angular position. Measured in rad/s.
Angular Acceleration (\(\alpha\))	The time rate of change of angular velocity. Measured in rad/s\(^2\).
Tangential Speed (\(v_t\))	The linear speed of a point on a rotating object, tangent to its circular path.
Tangential Acceleration (\(a_t\))	The linear acceleration component tangent to the circular path, due to a change in tangential speed.
Centripetal Acceleration (\(a_c\))	The linear acceleration component directed towards the center of rotation, due to a change in direction of velocity.
Moment of Inertia (\(I\))	A measure of an object’s resistance to changes in its angular velocity; the rotational analog of mass.
Rotational Kinetic Energy (\(K_R\))	The kinetic energy associated with the rotation of an object.
Parallel-Axis Theorem	A theorem that relates the moment of inertia about an axis through the center of mass to that about any parallel axis.