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Torque and Rotational Dynamics

AP Physics 1 · Topic 5

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5.1

Rotational Kinematics
Syllabus
Learning ObjectiveEssential Knowledge

5.1.A
Describe the rotation of a system with respect to time using angular displacement, angular velocity, and angular acceleration.

  • 5.1.A.1 Angular displacement is the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis.
    • Equation: $\Delta\theta = \theta - \theta_0$
    • 5.1.A.1.i A rigid system is one that holds its shape but in which different points on the system move in different directions during rotation. A rigid system cannot be modeled as an object.
    • 5.1.A.1.ii One direction of angular displacement about an axis of rotation—clockwise or counterclockwise—is typically indicated as mathematically positive, with the other direction becoming mathematically negative.
    • 5.1.A.1.iii If the rotation of a system about an axis may be well described using the motion of the system's center of mass, the system may be treated as a single object. For example, the rotation of Earth about its axis may be considered negligible when considering the revolution of Earth about the center of mass of the Earth–Sun system.
  • 5.1.A.2 Average angular velocity is the average rate at which angular position changes with respect to time.
    • Equation: $\omega_{\text{avg}} = \dfrac{\Delta\theta}{\Delta t}$
  • 5.1.A.3 Average angular acceleration is the average rate at which the angular velocity changes with respect to time.
    • Equation: $\alpha_{\text{avg}} = \dfrac{\Delta\omega}{\Delta t}$
  • 5.1.A.4 Angular displacement, angular velocity, and angular acceleration around one axis are analogous to linear displacement, velocity, and acceleration in one dimension and demonstrate the same mathematical relationships.
    • 5.1.A.4.i For constant angular acceleration, the mathematical relationships between angular displacement, angular velocity, and angular acceleration can be described with the following equations:
      • Equation: $\omega = \omega_0 + \alpha t$
      • Equation: $\theta = \theta_0 + \omega_0 t + \dfrac{1}{2}\alpha t^2$
      • Equation: $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$
    • 5.1.A.4.ii Graphs of angular displacement, angular velocity, and angular acceleration as functions of time can be used to find the relationships between those quantities.

Boundary statement: Descriptions of the directions of rotation for a point or object are limited to clockwise and counterclockwise with respect to a given axis of rotation.

Source: College Board AP Course and Exam Description

Rotation is described by angular quantities that mirror the linear ones:

One radian is the angle whose arc length equals the radius One radian is the angle whose arc length equals the radius

  • angular displacement 角位移 $\theta$ (in radians 弧度),
  • angular velocity 角速度 $\omega=\dfrac{\Delta\theta}{\Delta t}$,
  • angular acceleration 角加速度 $\alpha=\dfrac{\Delta\omega}{\Delta t}$.

For constant $\alpha$, the rotational kinematic equations have the same form as the linear ones, with $\theta,\omega,\alpha$ replacing $x,v,a$: $\omega=\omega_0+\alpha t$, $\theta=\omega_0 t+\tfrac12\alpha t^2$, and $\omega^2=\omega_0^2+2\alpha\theta$.

Worked example. A wheel starts from rest and speeds up uniformly to $30\ \text{rad/s}$ in $6.0\ \text{s}$. Find its angular acceleration and the total angle turned:

$$\alpha=\frac{\Delta\omega}{\Delta t}=\frac{30}{6.0}=5.0\ \text{rad/s}^2,\qquad \theta=\tfrac12\alpha t^2=\tfrac12\times5.0\times6.0^2=90\ \text{rad}.$$

Vocabulary Train
English Chinese Pinyin
angular displacement 角位移 jiǎo wèi yí
radians 弧度 hú dù
angular velocity 角速度 jiǎo sù dù
angular acceleration 角加速度 jiǎo jiā sù dù
Exercise sheet
5.2

Connecting Linear and Rotational Motion
Syllabus
Learning ObjectiveEssential Knowledge

5.2.A
Describe the linear motion of a point on a rotating rigid system that corresponds to the rotational motion of that point, and vice versa.

  • 5.2.A.1 For a point at a distance $r$ from a fixed axis of rotation, the linear distance $s$ traveled by the point as the system rotates through an angle $\Delta\theta$ is given by the equation $\Delta s = r\Delta\theta$.
  • 5.2.A.2 Derived relationships of linear velocity and of the tangential component of acceleration to their respective angular quantities are given by the following equations:
    • Equation: $s = r\theta$
    • Equation: $v = r\omega$
    • Equation: $a_T = r\alpha$
  • 5.2.A.3 For a rigid system, all points within that system have the same angular velocity and angular acceleration.

Boundary statement: Descriptions of the directions of rotation for a point or object are limited to clockwise and counterclockwise with respect to a given axis of rotation.

Source: College Board AP Course and Exam Description

A point at radius $r$ from the axis has linear quantities tied to the angular ones:

$$s=r\theta,\qquad v=r\omega,\qquad a_t=r\alpha.$$
Points farther from the axis move faster. This link lets you switch between "how fast the wheel spins" and "how fast a point on its rim moves."

As the radius turns through an angle, a point moves along an arc at speed v As the radius turns through an angle, a point moves along an arc at speed v

Worked example. A bicycle wheel of radius $0.35\ \text{m}$ spins at $12\ \text{rad/s}$. A point on the rim (and so the bike) moves at $v=r\omega=0.35\times12=4.2\ \text{m/s}$. A point halfway to the axis moves at half that speed.

Exercise sheet
5.3

Torque
Syllabus
Learning ObjectiveEssential Knowledge

5.3.A
Identify the torques exerted on a rigid system.

  • 5.3.A.1 Torque results only from the force component perpendicular to the position vector from the axis of rotation to the point of application of the force.
  • 5.3.A.2 The lever arm is the perpendicular distance from the axis of rotation to the line of action of the exerted force.

5.3.B
Describe the torques exerted on a rigid system.

  • 5.3.B.1 Torques can be described using force diagrams.
    • 5.3.B.1.i Force diagrams are similar to free-body diagrams and are used to analyze the torques exerted on a rigid system.
    • 5.3.B.1.ii Similar to free-body diagrams, force diagrams represent the relative magnitude and direction of the forces exerted on a rigid system. Force diagrams also depict the location at which those forces are exerted relative to the axis of rotation.
  • 5.3.B.2 The magnitude of the torque exerted on a rigid system by a force is described by the following equation, where $\theta$ is the angle between the force vector and the position vector from the axis of rotation to the point of application of the force.
    • Equation: $\tau = rF_{\perp} = rF\sin\theta$

Boundary statement: While AP Physics 1 expects students to mathematically manipulate the magnitude of torque using vector conventions, the direction of torque is beyond the scope of the course.

Source: College Board AP Course and Exam Description

The principle of moments (torque)

Torque 力矩 is the rotational effect of a force – how effectively it turns an object about an axis:

$$\tau=r F\sin\theta = F\cdot r_\perp,$$
where $r_\perp$ is the moment arm 力臂 (the perpendicular distance from the axis to the force's line of action). A force applied farther out, or more perpendicular, produces more torque. Torque has a sign (clockwise vs counterclockwise).

The moment of a force depends on the perpendicular distance from the pivot The moment of a force depends on the perpendicular distance from the pivot

Worked example. You push with $20\ \text{N}$ at the end of a $0.30\ \text{m}$ wrench. Perpendicular to the wrench the torque is $\tau=rF=0.30\times20=6.0\ \text{N m}$. If you push at $60^{\circ}$ to the wrench instead, only the perpendicular part counts: $\tau=rF\sin 60^{\circ}=0.30\times20\times0.87=5.2\ \text{N m}$ – which is why you push at a right angle for the most turning effect.

Explore

Balance torques on a beam

Torque is force times perpendicular distance, $\tau=Fd$. The beam is in rotational equilibrium when the torques on each side are equal.

Vocabulary Train
English Chinese Pinyin
Torque 力矩 lì jǔ
moment arm 力臂 lì bì
Exercise sheet
5.4

Rotational Inertia
Syllabus
Learning ObjectiveEssential Knowledge

5.4.A
Describe the rotational inertia of a rigid system relative to a given axis of rotation.

  • 5.4.A.1 Rotational inertia measures a rigid system's resistance to changes in rotation and is related to the mass of the system and the distribution of that mass relative to the axis of rotation.
  • 5.4.A.2 The rotational inertia of an object rotating a perpendicular distance $r$ from an axis is described by the equation
    • Equation: $I = mr^2$
  • 5.4.A.3 The total rotational inertia of a collection of objects about an axis is the sum of the rotational inertias of each object about that axis:
    • Equation: $I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$

5.4.B
Describe the rotational inertia of a rigid system rotating about an axis that does not pass through the system's center of mass.

  • 5.4.B.1 A rigid system's rotational inertia in a given plane is at a minimum when the rotational axis passes through the system's center of mass.
  • 5.4.B.2 The parallel axis theorem uses the following equation to relate the rotational inertia of a rigid system about any axis that is parallel to an axis through its center of mass:
    • Equation: $I' = I_{\text{cm}} + Md^2$

Boundary statement: AP Physics 1 only expects students to calculate the rotational inertia for systems of five or fewer objects arranged in a two-dimensional configuration.

Boundary statement: Students do not need to know the rotational inertia of extended rigid systems, as these will be provided within the exam. Students should have a qualitative understanding of the factors that affect rotational inertia; for example, how rotational inertia is greater when mass is farther from the axis of rotation, which is why a hoop has more rotational inertia than a solid disk of the same mass and radius.

Source: College Board AP Course and Exam Description

Rotational inertia 转动惯量 (moment of inertia) $I$ measures how hard it is to change an object's rotation – the rotational version of mass. It depends on both the mass and how far that mass sits from the axis: mass spread farther out gives a larger $I$. For a point mass, $I=mr^2$; for extended bodies, standard formulas are provided (a hoop is $mR^2$, a solid disk $\tfrac12 mR^2$). This is why a figure skater spins faster when she pulls her arms in – she reduces $I$.

Vocabulary Train
English Chinese Pinyin
Rotational inertia 转动惯量 zhuǎn dòng guàn liàng
Exercise sheet
5.5

Rotational Equilibrium
Syllabus
Learning ObjectiveEssential Knowledge

5.5.A
Describe the conditions under which a system's angular velocity remains constant.

  • 5.5.A.1 A system may exhibit rotational equilibrium (constant angular velocity) without being in translational equilibrium, and vice versa.
    • 5.5.A.1.i Free-body and force diagrams describe the nature of the forces and torques exerted on an object or rigid system.
    • 5.5.A.1.ii Rotational equilibrium is a configuration of torques such that the net torque exerted on the system is zero.
      • Equation: $\sum \tau_i = 0$
    • 5.5.A.1.iii The rotational analog of Newton's first law is that a system will have a constant angular velocity only if the net torque exerted on the system is zero.
  • 5.5.A.2 A rotational corollary to Newton's second law states that if the torques exerted on a rigid system are not balanced, the system's angular velocity must be changing.

Boundary statement: AP Physics 1 does not expect students to simultaneously analyze rotation in multiple planes.

Source: College Board AP Course and Exam Description

An object is in rotational equilibrium 转动平衡 when the net torque is zero, so its angular velocity stays constant. For a balanced (static) object, both the net force and the net torque are zero. Choosing the axis at an unknown force's location removes it from the torque equation – a useful trick for beam and ladder problems.

At balance the clockwise and anticlockwise moments about the pivot are equal At balance the clockwise and anticlockwise moments about the pivot are equal

Worked example. A $30\ \text{kg}$ child sits $2.0\ \text{m}$ from the pivot of a seesaw. Where must a $40\ \text{kg}$ child sit on the other side to balance it? Set the clockwise torque equal to the anticlockwise torque (the $g$'s cancel):

$$30\times2.0=40\times d\;\Rightarrow\;d=\frac{60}{40}=1.5\ \text{m}.$$
The heavier child sits closer to the pivot – less distance, same torque.

Explore

Find the balance point

For rotational equilibrium the total clockwise torque equals the total anticlockwise torque. Move the forces and distances until the beam balances.

Vocabulary Train
English Chinese Pinyin
rotational equilibrium 转动平衡 zhuǎn dòng píng héng
Exercise sheet
5.6

Newton's Second Law in Rotational Form
Syllabus
Learning ObjectiveEssential Knowledge

5.6.A
Describe the conditions under which a system's angular velocity changes.

  • 5.6.A.1 Angular velocity changes when the net torque exerted on the object or system is not equal to zero.
  • 5.6.A.2 The rate at which the angular velocity of a rigid system changes is directly proportional to the net torque exerted on the rigid system and is in the same direction. The angular acceleration of the rigid system is inversely proportional to the rotational inertia of the rigid system.
    • Equation: $\alpha_{\text{sys}} = \dfrac{\sum \tau}{I_{\text{sys}}} = \dfrac{\tau_{\text{net}}}{I_{\text{sys}}}$
  • 5.6.A.3 To fully describe a rotating rigid system, linear and rotational analyses may need to be performed independently.

Source: College Board AP Course and Exam Description

Net torque produces angular acceleration, in direct analogy with $F=ma$:

$$\sum\tau = I\alpha.$$
So a larger net torque, or a smaller rotational inertia, gives a larger angular acceleration. Solve rotation problems just like translation problems, with $\tau\leftrightarrow F$, $I\leftrightarrow m$, and $\alpha\leftrightarrow a$.

Worked example. A net torque of $12\ \text{N m}$ acts on a wheel with rotational inertia $I=3.0\ \text{kg m}^2$. Its angular acceleration is $\alpha=\tau/I=12/3.0=4.0\ \text{rad/s}^2$ – the exact rotational twin of $a=F/m$.

Exercise sheet
5.6

Exam tips
  • Torque $\tau=Fr_\perp$ uses the perpendicular distance from the pivot; a force through the pivot gives zero torque.
  • For balance, set clockwise torque = anticlockwise torque; choosing the pivot at an unknown force removes it from the equation.
  • Use the rotational analogues: $\tau\leftrightarrow F$, $I\leftrightarrow m$, $\alpha\leftrightarrow a$, so $\sum\tau=I\alpha$ mirrors $\sum F=ma$.
  • Rotational inertia depends on how far the mass sits from the axis, not just its amount — a hoop resists spinning more than a disc of equal mass.
  • Convert angles to radians and link linear to angular with $v=r\omega$, $a_t=r\alpha$.

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