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Torque and Rotational Dynamics

AP Physics C: Mechanics · 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 Angular velocity is the rate at which angular position changes with respect to time.
    • Equation: $\omega = \dfrac{d\theta}{dt}$
  • 5.1.A.3 Angular acceleration is the rate at which angular velocity changes with respect to time.
    • Equation: $\alpha = \dfrac{d\omega}{dt}$
  • 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:
      • $\omega = \omega_0 + \alpha t$
      • $\theta = \theta_0 + \omega_0 t + \dfrac{1}{2}\alpha t^2$
      • $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$
    • 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: AP Physics C: Mechanics expects students to be able to mathematically manipulate the magnitudes of angular displacement, angular velocity, and angular acceleration using vector conventions. However, the directions of said vectors will not be assessed on the exam.

Descriptions of the directions of rotational kinematics quantities for a point or rigid body are limited to clockwise and counterclockwise with respect to a given axis of rotation.

Source: College Board AP Course and Exam Description

Rotation mirrors linear motion with angular quantities, defined as derivatives:

$$\omega=\frac{d\theta}{dt},\qquad \alpha=\frac{d\omega}{dt}=\frac{d^2\theta}{dt^2}.$$

Here $\theta$ is angular displacement 角位移 in radians 弧度, $\omega$ the angular velocity 角速度, and $\alpha$ the angular acceleration 角加速度. (AP works with their magnitudes and signs; the 3D vector directions are not assessed.) For constant $\alpha$, the kinematic equations are the linear ones re-lettered:

$$\omega=\omega_0+\alpha t,\qquad \Delta\theta=\omega_0t+\tfrac12\alpha t^2,\qquad \omega^2=\omega_0^2+2\alpha\,\Delta\theta.$$

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

Worked example. A fan blade slows from $30\ \text{rad/s}$ to rest with $\alpha=-6.0\ \text{rad/s}^2$: it takes $t=5.0\ \text{s}$ and turns through $\Delta\theta=\dfrac{0-30^2}{2(-6.0)}=75\ \text{rad}$ – about $12$ revolutions.

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:
    • $s = r\theta$
    • $v = r\omega$
    • $a_T = r\alpha$
  • 5.2.A.3 For a rigid system, all points within that system have the same angular velocity and angular acceleration.

Boundary statement: AP Physics C: Mechanics expects students to be able to mathematically manipulate the magnitudes of angular displacement, angular velocity, and angular acceleration using vector conventions. However, the directions of the vectors will not be assessed on the exam.

Descriptions of the directions of rotational kinematics quantities for a point or rigid body 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 moves along its arc with

$$s=r\theta,\qquad v=r\omega,\qquad a_t=r\alpha,$$

so points farther out move faster. A rotating point generally has two acceleration components at once: the tangential acceleration 切向加速度 $a_t=r\alpha$ (speeding up along the arc) and the centripetal acceleration 向心加速度 $a_c=\dfrac{v^2}{r}=r\omega^2$ (turning, towards the axis). They are perpendicular, so $a=\sqrt{a_t^2+a_c^2}$.

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

Vocabulary Train
English Chinese Pinyin
tangential acceleration 切向加速度 qiè xiàng jiā sù dù
centripetal acceleration 向心加速度 xiàng xīn jiā sù dù
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 torque exerted on a rigid system about a chosen pivot point by a given force is described by $\vec{\tau} = \vec{r} \times \vec{F}$.
    • 5.3.B.2.i The cross-product between two vectors, $\vec{A}$ and $\vec{B}$, results in a vector quantity of magnitude $\vec{A} \times \vec{B} = AB\sin\theta$.
    • 5.3.B.2.ii The direction of the vector resulting from the cross-product of vectors $\vec{A}$ and $\vec{B}$ is perpendicular to both vectors $\vec{A}$ and $\vec{B}$ and therefore is normal to the plane defined by vectors $\vec{A}$ and $\vec{B}$.
    • 5.3.B.2.iii The direction of the vector resulting from the cross-product of vectors $\vec{A}$ and $\vec{B}$ can be qualitatively determined by applying the appropriate right-hand rule.

Source: College Board AP Course and Exam Description

The principle of moments (torque)

Torque 力矩 is the turning effect of a force – the rotational cause, as force is the translational cause. It is a vector product 矢量积:

$$\vec{\tau}=\vec{r}\times\vec{F},\qquad \tau=rF\sin\theta=F\,r_\perp,$$

where $r_\perp$, the moment arm 力臂, is the perpendicular distance from the axis to the force's line of action. More force, applied farther from the axis, more perpendicular: more torque. A force pointing straight at (or away from) the axis has zero torque. In a plane, give counterclockwise and clockwise torques opposite signs and add.

As a vector product, $\vec\tau=\vec r\times\vec F$ points perpendicular to the plane of $\vec r$ and $\vec F$, with its direction fixed by the right-hand rule 右手定则: curl your right fingers from $\vec r$ toward $\vec F$ and your thumb points along $\vec\tau$ – out of the page for a counterclockwise turn, into it for a clockwise one. The same rule gives the direction of angular momentum $\vec L=\vec r\times\vec p$.

The torque of a force depends on the perpendicular distance from the axis The torque of a force depends on the perpendicular distance from the axis

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
vector product 矢量积 shǐ liàng jī
moment arm 力臂 lì bì
right-hand rule 右手定则 yòu shǒu dìng zé
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 $I = mr^2$.
  • 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.A.4 For a solid that can be considered as a collection of differential masses, $dm$, the solid's rotational inertia can be calculated using the equation $I = \int r^2\, dm$, where $r$ is the perpendicular distance from $dm$ to the axis of rotation.

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 C: Mechanics only expects students to use calculus in the derivations of the rotational inertia of thin rods of uniform or nonuniform density about an arbitrary axis perpendicular to the rod, as well as derivations of the rotational inertia of a thin cylindrical shell, disk, or rigid bodies that can be considered to be made up of coaxial rings or shells about an axis that passes through their centers (e.g., annular rings).

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 puck of the same mass and radius.

Source: College Board AP Course and Exam Description

Rotational inertia 转动惯量 $I$ measures resistance to angular acceleration – the rotational analog of mass, but it depends on where the mass sits. For point masses, $I=\sum m_ir_i^2$; for a continuous body,

$$I=\int r^2\,dm.$$

Mass far from the axis dominates, because of the $r^2$.

Worked example (rod). A uniform rod (mass $M$, length $L$) about its centre: with linear density 线密度 $\lambda=M/L$, $dm=\lambda\,dx$, so $I=\displaystyle\int_{-L/2}^{L/2}x^2\,\frac{M}{L}\,dx=\frac{1}{12}ML^2$. AP also expects nonuniform rods: if $\lambda(x)=cx$ on $[0,L]$, first find $M=\int_0^L cx\,dx=\tfrac12cL^2$, then $I=\int_0^L x^2(cx)\,dx=\tfrac14cL^4=\tfrac12ML^2$ about the light end.

Worked example (disk from rings). A solid disk is a nest of rings: a ring at radius $r$ of width $dr$ has $dm=\dfrac{M}{\pi R^2}\,2\pi r\,dr$, so

$$I=\int_0^R r^2\,dm=\frac{2M}{R^2}\int_0^R r^3\,dr=\tfrac12MR^2.$$

The same coaxial-shell method handles cylindrical shells and annular rings.

The parallel-axis theorem 平行轴定理 shifts any known $I$ to a parallel axis a distance $d$ away:

$$I'=I_{\text{cm}}+Md^2.$$

For the rod about one end: $I=\tfrac1{12}ML^2+M\big(\tfrac{L}{2}\big)^2=\tfrac13ML^2$.

Vocabulary Train
English Chinese Pinyin
Rotational inertia 转动惯量 zhuǎn dòng guàn liàng
linear density 线密度 xiàn mì dù
parallel-axis theorem 平行轴定理 píng xíng zhóu dìng lǐ
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 C: Mechanics does not expect students to simultaneously analyze rotation in multiple planes.

Source: College Board AP Course and Exam Description

Newton's first law has a rotational form: with zero net torque, angular velocity is constant – an object in rotational equilibrium 转动平衡 either does not rotate or rotates steadily. Full static equilibrium 静平衡 needs both $\sum F=0$ and $\sum\tau=0$ (about any axis). The standard move for beams and ladders: put the axis at the point where an unknown force acts, so that force drops out of the torque equation.

At balance the clockwise and anticlockwise torques about the axis are equal At balance the clockwise and anticlockwise torques about the axis are equal

Worked example. A uniform $20\ \text{kg}$ beam, $4.0\ \text{m}$ long, is pivoted at its left end and held horizontal by a vertical rope at its right end. A $40\ \text{kg}$ child sits $1.0\ \text{m}$ from the pivot. Torques about the pivot: $T(4.0)=20g(2.0)+40g(1.0)$, so $T=\dfrac{392+392}{4.0}=196\ \text{N}$. Then vertical force balance gives the pivot's upward push: $F_p=(60)(9.8)-196=392\ \text{N}$. Choosing the pivot as the axis removed $F_p$ from the torque equation entirely.

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
static equilibrium 静平衡 jìng 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: $\alpha_{\text{sys}} = \dfrac{\Sigma\tau}{I_{\text{sys}}} = \dfrac{\tau_{\text{net}}}{I_{\text{sys}}}$
  • 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

A net torque produces angular acceleration in proportion to the rotational inertia:

$$\alpha=\frac{\sum\tau}{I}\qquad\Big(\text{more generally }\sum\vec{\tau}=\frac{d\vec{L}}{dt}\Big).$$

Solve rotational dynamics exactly like translational dynamics, with $\tau\leftrightarrow F$, $I\leftrightarrow m$, $\alpha\leftrightarrow a$ – free-body diagram, equations, solve.

A massive pulley: the two tensions differ because the pulley needs net torque A massive pulley: the two tensions differ because the pulley needs net torque

Worked example (massive pulley). Masses $m_1=4.0\ \text{kg}$ and $m_2=2.0\ \text{kg}$ hang from a rope over a pulley 滑轮 of rotational inertia $I=0.50\ \text{kg}\cdot\text{m}^2$ and radius $R=0.20\ \text{m}$. Three Newton's-law equations, one per body:

$$m_1g-T_1=m_1a,\qquad T_2-m_2g=m_2a,\qquad (T_1-T_2)R=I\alpha=\frac{Ia}{R}.$$

Adding them (with $I/R^2=12.5\ \text{kg}$): $a=\dfrac{(m_1-m_2)g}{m_1+m_2+I/R^2}=\dfrac{19.6}{18.5}=1.1\ \text{m/s}^2$. Then $T_1=m_1(g-a)=35\ \text{N}$ and $T_2=m_2(g+a)=22\ \text{N}$. The tensions must differ – otherwise nothing would spin the pulley. (If a problem says the pulley is light, then $I\approx0$ and the tensions become equal again.)

Exam skill. Rotational FRQs score the setup: separate free-body diagrams for each mass and the pulley, the string constraint $a=R\alpha$ stated, and the sign convention consistent. The single most common error is assuming one tension throughout a rope that passes over a massive pulley.

Vocabulary Train
English Chinese Pinyin
pulley 滑轮 huá lún
Exercise sheet
5.6

Exam tips
  • Use the rotational analogues: $\theta,\omega=\tfrac{d\theta}{dt},\alpha=\tfrac{d\omega}{dt}$, with the constant-$\alpha$ equations mirroring linear kinematics.
  • Convert between linear and angular with $v=r\omega$ and $a_t=r\alpha$ (plus centripetal $a_c=\tfrac{v^2}{r}=\omega^2 r$).
  • Keep radians throughout and fix a positive sense of rotation.
  • Relate torque to angular acceleration through $\tau=I\alpha$.
  • Draw the rotation axis — the moment arm is the perpendicular distance to it.

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