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AP Physics C: Mechanics Unit 5 Review: Torque and Rotational Motion

Review AP Physics C: Mechanics Unit 5 to build fluency with torque, rotational inertia, and Newton's second law in rotational form. This unit translates the force-and-motion framework from Units 1 and 2 into its rotational counterparts, connecting angular kinematics to dynamics through calculus-based derivations.

Use the topic guides, practice questions, FRQ practice, and AP score calculator available on this page to work through every concept before exam day.

What is AP Physics C: Mechanics unit 5?

Unit 5 is the rotational counterpart to the translational dynamics you studied in Units 1 and 2. Every linear quantity gets a rotational analog: displacement becomes angular displacement, force becomes torque, mass becomes rotational inertia, and F = ma becomes sum-tau = I-alpha.

Unit 5 asks you to describe how rigid systems rotate, calculate torques using the cross product, find rotational inertia through integration and the parallel axis theorem, and apply Newton's first and second laws in rotational form to predict angular motion.

Rotational kinematics and the linear connection

Topics 5.1 and 5.2 establish angular displacement (delta-theta), angular velocity (omega = d-theta/dt), and angular acceleration (alpha = d-omega/dt), then link them to linear quantities at a point a distance r from the axis: s = r-theta, v = r-omega, and a_T = r-alpha. All points on a rigid body share the same omega and alpha, but their linear speeds differ by radius.

Torque and rotational inertia

Topic 5.3 defines torque as the cross product tau = r x F, with magnitude rF sin-theta and direction from the right-hand rule. Topic 5.4 defines rotational inertia I as the resistance to rotational change: I = mr^2 for a point mass, I = integral r^2 dm for a continuous solid, and I' = I_cm + Md^2 via the parallel axis theorem.

Rotational Newton's laws

Topic 5.5 states that when net torque is zero, angular velocity is constant (rotational Newton's first law). Topic 5.6 states that when net torque is nonzero, alpha = sum-tau / I_sys. Combined linear and rotational analyses are often needed for systems like a mass on a string wrapped around a pulley.

Rotation mirrors translation

Every concept in Unit 5 has a direct translational analog. Recognizing those pairings (theta with x, omega with v, alpha with a, tau with F, I with m, sum-tau = I-alpha with F = ma) lets you transfer problem-solving strategies you already know from Units 1 and 2 directly into rotational contexts, and it prepares you for the rotational energy and angular momentum work in Unit 6.

AP Physics C: Mechanics unit 5 topics

5.1

Rotational Kinematics

Defines angular displacement (delta-theta in radians), angular velocity (omega = d-theta/dt), and angular acceleration (alpha = d-omega/dt). For constant alpha, the four kinematic equations apply directly. Graph interpretation follows the same slope-and-area rules as linear kinematics.

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5.2

Connecting Linear and Rotational Motion

Links angular quantities to the linear motion of a point at radius r: s = r*theta, v = r*omega, a_T = r*alpha. All points on a rigid body share omega and alpha, but linear speeds differ by radius. The no-slip condition v_cm = R*omega applies to rolling objects.

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5.3

Torque

Torque is the cross product tau = r x F with magnitude rF sin-theta. Only the force component perpendicular to r produces torque. The lever arm is the perpendicular distance from the axis to the line of action of the force. Direction is determined by the right-hand rule.

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5.4

Rotational Inertia

Rotational inertia I = mr^2 for a point mass, I = integral r^2 dm for continuous solids, and I' = I_cm + Md^2 via the parallel axis theorem. Mass farther from the axis contributes more. AP Physics C requires calculus-based derivations for rods, disks, cylindrical shells, and annular rings.

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5.5

Rotational Equilibrium and Newton's First Law in Rotational Form

When sum-tau = 0, angular velocity is constant: this is Newton's first law for rotation. Rotational and translational equilibrium are independent conditions. Choosing a strategic pivot eliminates unknown forces from the torque equation.

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5.6

Newton's Second Law in Rotational Form

When net torque is nonzero, alpha = sum-tau / I_sys. For combined translational and rotational systems, write separate sum-F = ma and sum-tau = I*alpha equations, then connect them with a constraint equation such as a = alpha*R.

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practice snapshot

Hardest AP Physics C: Mechanics unit 5 topics

This snapshot uses Fiveable practice activity to show where students tend to miss questions and which review moves are worth prioritizing first.

63%average MCQ accuracy

Across 2.3k multiple-choice practice attempts for this unit.

2.3kMCQ attempts

Practice activity included in this snapshot.

3%average FRQ score

Across 4 scored free-response attempts for this unit.

Hardest topics in unit 5

MCQ miss rate
5.4

Review Rotational Inertia with attention to how the concept appears in AP-style source and evidence questions.

45%442 tries
5.1

Review Rotational Kinematics with attention to how the concept appears in AP-style source and evidence questions.

36%563 tries
5.3

Review Torque with attention to how the concept appears in AP-style source and evidence questions.

36%410 tries
5.5

Review Rotational Equilibrium and Newton's First Law in Rotational Form with attention to how the concept appears in AP-style source and evidence questions.

35%307 tries

Unit 5 review notes

5.1

Rotational Kinematics

A rigid body rotating about a fixed axis is described by angular displacement delta-theta (in radians), angular velocity omega = d-theta/dt, and angular acceleration alpha = d-omega/dt. For constant alpha, the same four kinematic equations from linear motion apply with theta, omega, and alpha replacing x, v, and a. Graphs of theta(t), omega(t), and alpha(t) follow the same slope-and-area logic as their linear counterparts.

  • Angular displacement delta-theta: Angle in radians through which a rigid body rotates; delta-theta = theta - theta_0. Counterclockwise is typically positive.
  • Angular velocity omega: omega = d-theta/dt in rad/s; instantaneous rate of change of angular position.
  • Angular acceleration alpha: alpha = d-omega/dt in rad/s^2; for constant alpha, omega = omega_0 + alpha*t and theta = theta_0 + omega_0*t + (1/2)*alpha*t^2.
  • Rigid body: A system that holds its shape during rotation; different points move in different directions but the body cannot be modeled as a single point object.
  • Graph interpretation: The slope of a theta(t) graph gives omega; the slope of an omega(t) graph gives alpha; the area under an alpha(t) graph gives the change in omega.
Write the four constant-alpha kinematic equations from memory, then identify which graph relationship gives you alpha from an omega(t) plot.
Linear quantitySymbolRotational analogSymbol
DisplacementxAngular displacementtheta
VelocityvAngular velocityomega
AccelerationaAngular accelerationalpha
MassmRotational inertiaI
ForceFTorquetau
5.2

Connecting Linear and Rotational Motion

For any point at distance r from a fixed axis, the arc length is s = r*theta, the tangential speed is v = r*omega, and the tangential acceleration is a_T = r*alpha. All points on a rigid body share the same omega and alpha, but their linear speeds scale with r. The centripetal (radial) acceleration at that point is a_r = r*omega^2, directed toward the axis.

  • s = r*theta: Arc length traveled by a point at radius r when the body rotates through angle theta (in radians).
  • v = r*omega: Tangential speed of a point at radius r; points farther from the axis move faster even though omega is the same for all points.
  • a_T = r*alpha: Tangential component of linear acceleration; changes the speed of the point along its circular path.
  • a_r = r*omega^2: Centripetal (radial) acceleration directed toward the axis; does not change the point's speed, only its direction.
  • No-slip condition: For rolling without slipping, v_cm = R*omega and a_cm = R*alpha, linking the translational and rotational motions of the rolling object.
A disk of radius 0.4 m spins at omega = 5 rad/s. Find the tangential speed and centripetal acceleration of a point on the rim.
5.3

Torque

Torque is the rotational effect of a force. Only the component of force perpendicular to the position vector from the axis to the point of application produces torque. The magnitude is tau = r*F*sin-theta, where theta is the angle between r and F. The lever arm is the perpendicular distance from the axis to the line of action of the force, giving the equivalent form tau = F * (lever arm). Direction follows the right-hand rule from the cross product tau = r x F.

  • tau = r x F: Torque as a cross product; magnitude is rF sin-theta, direction is perpendicular to both r and F by the right-hand rule.
  • Lever arm: Perpendicular distance from the axis of rotation to the line of action of the force; maximizing the lever arm maximizes torque for a given force.
  • Perpendicular force component: Only F_perp (the component of force perpendicular to r) contributes to torque; the radial component along r produces zero torque.
  • Sign convention: Counterclockwise torques are typically positive and clockwise torques negative; consistency within a problem is required.
  • Force diagram: Like a free-body diagram but also shows where each force is applied relative to the axis, which is essential for calculating each torque.
A 10 N force is applied at 0.5 m from a pivot at 30 degrees from the radial direction. Calculate the torque and identify whether it is clockwise or counterclockwise.
5.4

Rotational Inertia

Rotational inertia I measures a rigid system's resistance to changes in rotation. It depends on both total mass and how that mass is distributed relative to the axis. For a point mass, I = mr^2. For a collection of point masses, I_total = sum(m_i * r_i^2). For a continuous solid, I = integral r^2 dm, where r is the perpendicular distance from each mass element to the axis. The parallel axis theorem I' = I_cm + Md^2 shifts the axis from the center of mass to any parallel axis a distance d away.

  • I = mr^2: Rotational inertia of a single point mass m at perpendicular distance r from the axis.
  • I = integral r^2 dm: Calculus-based formula for continuous solids; requires expressing dm in terms of a spatial variable (dx, dr, etc.) using the object's geometry and density.
  • Parallel axis theorem: I' = I_cm + Md^2; rotational inertia about any axis equals I about the parallel center-of-mass axis plus Md^2, where d is the distance between the axes.
  • Mass distribution: Mass farther from the axis contributes more to I (scales as r^2); a hollow cylinder has greater I than a solid disk of the same mass and radius.
  • Common results: Solid disk: I = (1/2)MR^2; thin cylindrical shell: I = MR^2; thin rod about center: I = (1/12)ML^2; thin rod about end: I = (1/3)ML^2.
Derive the rotational inertia of a uniform thin rod of mass M and length L about one end using I = integral r^2 dm, then verify with the parallel axis theorem from the center result.
ObjectAxisI
Point massDistance r from massmr^2
Solid diskThrough center, perpendicular(1/2)MR^2
Thin cylindrical shellThrough center, perpendicularMR^2
Thin rodThrough center, perpendicular(1/12)ML^2
Thin rodThrough one end, perpendicular(1/3)ML^2
5.5

Rotational Equilibrium and Newton's First Law in Rotational Form

A system is in rotational equilibrium when the net torque about any chosen axis is zero (sum-tau = 0), which means its angular velocity is constant. This is the rotational analog of Newton's first law. A system can be in rotational equilibrium without being in translational equilibrium, and vice versa. When solving equilibrium problems, choose the pivot point strategically to eliminate unknown torques from the equation.

  • sum-tau = 0: Condition for rotational equilibrium; the angular velocity is constant (including zero) when net torque is zero.
  • Independent equilibrium conditions: Rotational equilibrium (sum-tau = 0) and translational equilibrium (sum-F = 0) are separate conditions; a system can satisfy one without the other.
  • Pivot selection: Any point can serve as the pivot for a torque sum; choosing a point where an unknown force acts eliminates that force from the torque equation.
  • Torque due to gravity: For a uniform object, gravity acts at the center of mass; the torque from gravity equals Mg times the horizontal distance from the pivot to the center of mass.
  • Unbalanced torque: If sum-tau is not zero, angular velocity must be changing; the system is not in rotational equilibrium and alpha is nonzero.
A uniform 2 m beam of mass 10 kg is hinged at one end and held horizontal by a vertical rope at the other end. Find the rope tension and hinge force using both sum-F = 0 and sum-tau = 0.
5.6

Newton's Second Law in Rotational Form

When net torque on a rigid system is not zero, the system has angular acceleration given by alpha = sum-tau / I_sys, the rotational form of Newton's second law. This is the direct analog of F = ma. For systems that both translate and rotate (such as a mass hanging from a string wrapped around a pulley), you must write a separate sum-F = ma equation for the translational motion and a separate sum-tau = I*alpha equation for the rotational motion, then connect them with a constraint equation such as a = alpha*R.

  • sum-tau = I*alpha: Newton's second law in rotational form; net torque equals rotational inertia times angular acceleration, directly analogous to F = ma.
  • Combined analysis: For a pulley-mass system, write sum-F = ma for the hanging mass and sum-tau = I*alpha for the pulley, then use a = alpha*R to solve the system.
  • Constraint equation: A kinematic relationship linking linear and angular motion, such as a_cm = alpha*R for rolling or a = alpha*R for a string on a pulley, reducing unknowns.
  • Direction consistency: Choose a positive direction for both linear and rotational motion at the start and apply it consistently across all equations in a combined problem.
  • Rotational inertia in dynamics: A larger I means a smaller alpha for the same net torque; mass distribution determines how quickly a system responds to an applied torque.
A disk of mass M and radius R has a string wrapped around it with a hanging mass m. Write and solve the system of equations to find the angular acceleration of the disk.

Practice AP Physics C: Mechanics unit 5 questions

Try AP-style multiple-choice questions and written prompts after you review the notes.

Example AP-style MCQs

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MCQ

AP-style practice question

Question

System A is a point mass MM attached to a massless rod of length RR. System B is a uniform thin ring of mass MM and radius RR rotating about an axis perpendicular to the ring at its edge. Which claim correctly compares their rotational inertias?

IB=2IAI_B = 2I_A because the ring has both center-of-mass inertia and parallel-axis shift.

IB=IAI_B = I_A because both systems have mass MM distributed at distance RR from the axis.

IB=4IAI_B = 4I_A because the ring's mass is distributed along a circumference of length 2πR2\pi R.

IA=2IBI_A = 2I_B because the point mass concentrates all matter at the maximum distance RR.

MCQ

AP-style practice question

Question

A cylinder of mass MM and radius RR is pulled by a horizontal force FF applied at its center of mass. The cylinder rolls without slipping on a rough horizontal surface. Which claim correctly identifies the direction of the friction force ff exerted by the surface?

Opposite to FF, because a net torque is required to produce angular acceleration.

In the direction of FF, because it assists the translational motion of the center.

Opposite to FF, because it must oppose the translational velocity to prevent slip.

Zero, because the cylinder rolls without slipping and does not slide on the surface.

Example FRQs

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FRQ

Rotating disk with falling mass and string tension

1. A uniform disk of mass M and radius R is mounted on a frictionless horizontal axle through its center, as shown in Figure 1. A string is wrapped around the disk, and a block of mass m is attached to the free end of the string. The block is released from rest at time t = 0, and as it falls, the string unwinds without slipping, causing the disk to rotate. The rotational inertia of the disk about its center is I=12MR2I = \frac{1}{2}MR^2. As the system moves, the tension in the string varies with time.

Figure 1. Disk on a frictionless axle with a string wrapped around the rim and a hanging block.

A clean black-and-white side-view apparatus diagram with no grid.

Overall layout:
- The circular disk is centered in the left half of the figure.
- A vertical string hangs in the right half of the figure, descending from the disk’s rim to a rectangular hanging block.
- The drawing clearly indicates the disk rotates as the string unwinds.

Disk and axle:
- Draw a perfect circle for the disk.
- Mark the disk’s exact center with a small filled dot.
- Through the center, draw a short horizontal axle line (a thin straight line) extending equally left and right from the center dot.
- Add the text label "frictionless axle" above the axle line near the center.
- Place the label "M" inside the disk, slightly above the center dot.

Radius marking:
- Draw a straight radius line from the center dot to the rim pointing directly to the right (a horizontal radius).
- At the rim endpoint of this radius line, place a short tick mark indicating the rim.
- Label this radius line "R" with the text placed just above the radius line near its middle.

String wrapping and exit point:
- Depict the string as a thin line wrapped around the disk’s rim over a short arc near the right side of the disk (show 2–3 closely spaced curved turns hugging the rim to indicate wrapping).
- Define the string’s leaving point unambiguously: the string leaves the disk at the rightmost point on the rim (the point where the horizontal radius meets the rim).
- From that rightmost rim point, draw the string as a straight vertical line downward (perfectly vertical, aligned with the disk’s rightmost point).
- At the leaving point on the rim, show the string tangent to the disk by making the outgoing segment vertical and the radius to that point horizontal (explicitly perpendicular).

Block:
- Draw the block as a medium-sized rectangle hanging at the bottom end of the vertical string.
- The block is entirely below the disk and does not touch the disk.
- Attach the top center of the block to the bottom end of the string (string connects to the midpoint of the block’s top edge).
- Label the block with "m" centered inside the rectangle.

Rotation direction indicator:
- Draw a curved arrow around the disk near the upper-right quadrant showing clockwise rotation.
- Place the arrowhead on the curved arrow so that it clearly indicates clockwise motion.
- Add the text label "clockwise" next to the curved arrow.

No other forces are shown in Figure 1 (it is an apparatus diagram only).

Figure 2. Free-body diagrams for the disk and the block (forces only).

IMPORTANT: This figure is a BLANK FREE-BODY DIAGRAM TEMPLATE for students to complete. Do NOT draw any force arrows, vector arrows, or directional indicators of any kind. The rendered image must show ONLY the geometric objects with their labels and reference points — no forces, no arrows, no coordinate axes, no axis arrows, and no coordinate grid are to appear anywhere in the image.

Two separate blank free-body diagram template panels placed side-by-side on a plain white background. No axes, no coordinate grid, no axis arrows, and no force arrows appear anywhere in the figure.

LEFT PANEL (Disk template): Draw a medium-large circle outline representing the disk, centered in the left panel. IMPORTANT: Write the label 'Disk' just above the circle (outside it). Write the letter 'M' inside the circle near its center. Place a small filled black dot at the exact center of the circle to indicate the axle location. Draw a short straight line segment from the center dot to the rightmost point on the rim, pointing directly to the right, and label it 'R' just above that line segment. Place a small solid filled black dot at the rightmost point on the rim (the endpoint of the radius line) and label it 'P' just outside the rim next to that dot. These reference markings (center dot, radius line labeled R, and rim point P) are geometric setup marks, not force arrows, and must be drawn. Draw nothing else on the left panel — no arrows of any kind.

RIGHT PANEL (Block template): Draw a medium-sized rectangle outline representing the block, centered in the right panel, smaller than the disk circle. IMPORTANT: Write the label 'Block' just above the rectangle (outside it). Write the letter 'm' inside the rectangle at its center. Draw nothing else on the right panel — no arrows of any kind.

IMPORTANT FINAL CHECK: The completed image must contain zero force arrows, zero vector arrows, and zero coordinate axis arrows. The only lines in the image are the circle outline, the rectangle outline, the short radius line segment from center to point P, and the label text strings ('Disk', 'M', 'R', 'P', 'Block', 'm'). Any arrow of any kind would be incorrect. All arrow styling rules are intentionally omitted because no arrows should be rendered.
A.
i.

On Figure 2, draw and label arrows to represent all forces acting on the disk and on the block. The tails of the arrows should be at the points where the forces are exerted. Draw arrows of lengths proportional to the magnitudes of the forces.

ii.

Derive an expression for the magnitude of the angular acceleration α\alpha of the disk as the system moves. Express your answer in terms of M, m, R, g, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

B.

Derive an expression for the angular velocity of the disk at time t=π2ωt = \frac{\pi}{2\omega}. Express your answer in terms of M, R, τ0\tau_0, ω\omega, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information. In a different scenario, the disk is initially at rest and a time-varying torque τ(t)=τ0cos(ωt)\tau(t) = \tau_0\cos(\omega t) is applied to the disk about its axis of rotation, where τ0\tau_0 is a positive constant and ω\omega is a positive constant angular frequency. The torque is applied from time t = 0 to time t=π2ωt = \frac{\pi}{2\omega}. There is no string or hanging mass in this scenario.

FRQ

Rotating disk with hanging masses and string tension

4. A uniform solid disk of mass MD=2.0M_D = 2.0 kg and radius R=0.30R = 0.30 m is mounted on a frictionless horizontal axle through its center, as shown in Figure 1. A light string is wrapped around the disk. One end of the string is attached to a block of mass m1=1.5m_1 = 1.5 kg that hangs vertically. The other end of the string passes over the disk and is attached to a second block of mass m2=0.50m_2 = 0.50 kg that also hangs vertically on the opposite side of the disk. The rotational inertia of a uniform solid disk about its central axis is I=12MDR2I = \frac{1}{2}M_D R^2. The system is released from rest at time t=0t = 0.

Figure 1. Two-block string wrapped on a uniform solid disk (MD = 2.0 kg, R = 0.30 m) mounted on a frictionless axle.

Single-panel black-and-white physics setup diagram (no graph axes, no grid). The drawing fills the page width.

Central rotating object:
- A large circle representing a uniform solid disk is centered horizontally on the page and placed in the upper half of the panel.
- The disk is labeled inside the circle with the text: "Disk" and directly beneath it "M_D = 2.0 kg".
- The disk’s center is marked with a small solid dot.
- A horizontal axle passes through the disk’s center: draw a straight horizontal line segment through the center dot, extending a short distance beyond the left and right edges of the disk.
- The axle is labeled above the axle line near the disk center: "frictionless axle".
- A fixed support is shown holding the axle: draw a simple rigid support bracket above the axle, centered over the disk, consisting of two vertical posts rising from the axle ends to a short horizontal bar. Label this support "fixed support".

Radius labeling (must show exact value):
- Draw a radius arrow from the center dot to the disk rim pointing directly to the right (3 o’clock direction).
- Put the text "R = 0.30 m" next to this radius arrow, with the text parallel to the arrow.

String path (continuous single light string, wrapped on the rim):
- Draw a thin string that is in contact with the disk rim on the left side and on the right side, with two vertical hanging segments.
- The string must be drawn tangent to the disk at the leftmost point (9 o’clock) and at the rightmost point (3 o’clock):
  - Left segment: a perfectly vertical line descending straight down from the left tangent point of the rim.
  - Right segment: a perfectly vertical line descending straight down from the right tangent point of the rim.
- Show that the string is wrapped around the disk across the top half of the rim: connect the left tangent point to the right tangent point with a thin arc that follows the disk’s top semicircle, lying exactly on the rim. (The arc indicates contact/wrap; it should coincide with the disk boundary.)
- Place the label "light string" above the top arc, centered over the disk.

Hanging blocks (positions and labels):
- Block 1 (left side):
  - At the bottom end of the left vertical string segment, attach a rectangular block.
  - The block’s top face touches the string end (string attaches at the block’s top center).
  - The block is positioned in the lower left quadrant of the panel, directly below the left disk tangent point so the string is perfectly vertical.
  - Inside the block write: "m_1". Next to the block (to the left of it) write: "m_1 = 1.5 kg".
- Block 2 (right side):
  - At the bottom end of the right vertical string segment, attach a rectangular block.
  - The block’s top face touches the string end (string attaches at the block’s top center).
  - The block is positioned in the lower right quadrant of the panel, directly below the right disk tangent point so the string is perfectly vertical.
  - Inside the block write: "m_2". Next to the block (to the right of it) write: "m_2 = 0.50 kg".

Relative heights and symmetry requirements (to remove ambiguity):
- The two blocks are drawn at the same vertical level (their bottom faces aligned horizontally), indicating the initial configuration before release.
- The left and right vertical string segments have equal drawn lengths from rim to block, reinforcing symmetry of the starting position.

Direction-of-motion indicators for Scenario 1 (as described in the prompt):
- Next to the left block, draw a bold downward arrow labeled "block 1 motion".
- Next to the right block, draw a bold upward arrow labeled "block 2 motion".
- On the disk, draw a curved arrow on the front face indicating rotation consistent with left side moving down and right side moving up (counterclockwise). Label the curved arrow "\u03c9, \u03b1".

Text-only constraints:
- Only the following numerical values appear anywhere in the figure: "2.0 kg", "0.30 m", "1.5 kg", "0.50 kg".
- No other numbers, no coordinate markings.

Line styling:
- Disk outline: thick black line.
- String: thin black line.
- Arrows (motion and rotation): medium-thick black with filled arrowheads.
- All labels are printed clearly in a sans-serif font.
A.

While the system is accelerating, the magnitudes of the tension forces in the string on each side of the disk are T1T_1 (on the side with block 1) and T2T_2 (on the side with block 2).

Indicate whether T1T_1 is greater than, less than, or equal to T2T_2 by writing one of the following.

  • T1>T2T_1 > T_2
  • T1<T2T_1 < T_2
  • T1=T2T_1 = T_2

Justify your answer using qualitative reasoning beyond referencing equations.

B.

Derive an expression for the magnitude of the linear acceleration aa of block 1. Express your answer in terms of m1m_1, m2m_2, MDM_D, RR, gg, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Figure 2. Same disk-and-string setup as Figure 1, but the right-hand block is replaced with m3 = 1.0 kg.

Single-panel black-and-white physics setup diagram with the same geometry and styling as Figure 1, occupying the same relative positions on the page.

Keep identical elements from Figure 1 (must match exactly):
- The uniform solid disk is centered horizontally in the upper half of the panel with a thick outline, center dot, and horizontal axle through the center.
- The axle is held by the same fixed support bracket above it and labeled "frictionless axle" and "fixed support".
- The radius arrow points directly to the right (3 o’clock) and is labeled exactly "R = 0.30 m".
- The disk interior label remains: "Disk" and "M_D = 2.0 kg".
- The same single light string is drawn tangent at the leftmost and rightmost rim points with two vertical hanging segments, and the string is wrapped along the top semicircle of the rim from left tangent point to right tangent point. The label "light string" sits above the top arc.
- The left block remains unchanged: a rectangle attached to the bottom of the left vertical string segment, positioned directly below the left tangent point, with "m_1" inside and the text "m_1 = 1.5 kg" printed just to its left.

Change on the right side (must be explicit):
- Replace the right-hand hanging block with a rectangular block labeled inside as "m_3".
- Next to this right block (to the right of it) print the exact text: "m_3 = 1.0 kg".
- The right block is attached at its top center to the bottom end of the right vertical string segment, directly below the right tangent point.

Relative heights and symmetry requirements:
- The left block (m1) and the right block (m3) are drawn with their bottom faces aligned horizontally (same initial height).
- The left and right vertical string segments from rim to blocks are drawn equal in length.

No additional scenario arrows unless shown in the original Figure 2:
- Do not include any up/down motion arrows or rotation arrows in Figure 2 unless they are explicitly labeled as part of this figure; this figure is only the modified-mass setup.

Text-only constraints:
- The only numerical values that appear are: "2.0 kg", "0.30 m", "1.5 kg", "1.0 kg".

Line styling:
- Match Figure 1: thick disk outline, thin string, medium label text, no shading, no grid, no axes.
C.

Indicate whether the magnitude of the angular acceleration α\alpha of the disk in this scenario is greater than, less than, or equal to the magnitude of the angular acceleration of the disk in the original scenario. In a different scenario, block 2 is replaced with a block of mass m3=1.0m_3 = 1.0 kg, as shown in Figure 2. The system is again released from rest. The disk has the same mass MD=2.0M_D = 2.0 kg and radius R=0.30R = 0.30 m as in the original scenario.

Briefly justify your answer.

FRQ

Rotational dynamics of disk-block system

2. A uniform solid disk of mass M = 2.0 kg and radius R = 0.40 m is mounted on a horizontal frictionless axle through its center, as shown in Figure 1. The disk is initially at rest. A string is wrapped around the outer edge of the disk and attached to a block of mass m = 0.50 kg that hangs vertically. The block is released from rest at time t = 0 s and falls a distance h = 1.5 m before hitting the ground at time t₁. After the block hits the ground, the string goes slack and the disk continues to rotate freely.

Figure 1. Disk–pulley (axle) system with hanging block and measured drop distance.

Monochrome line-art physics apparatus diagram (no shading), drawn in a wide rectangular frame.

Overall layout:
- The disk is the dominant object and is placed in the left half of the frame, centered vertically.
- A vertical string hangs from the right side of the disk down into the lower half of the frame.
- The block hangs on the string in the lower-right quadrant.
- A horizontal ground line spans the entire bottom edge of the frame.

Disk and axle:
- Draw a perfect circle for the disk.
- Mark the disk’s center with a small filled dot.
- Through the center dot, draw a short horizontal axle line segment that extends a small distance on both sides of the disk, indicating a horizontal axle pointing toward/away from the viewer.
- Label the axle immediately above it: "frictionless axle".
- Label the disk just above the top-left of the circle: "uniform solid disk".
- Add a mass label near the disk: "M = 2.0 kg".
- Add a radius label: draw a straight radius line from the center dot to the rightmost rim of the disk and label it "R = 0.40 m".

String and wrap direction:
- Draw a thin string tangent to the disk at the disk’s rightmost point (the 3 o’clock point), showing the string leaving the rim vertically downward.
- Show the string as contacting the rim only at that tangent point (no gaps between string and rim at the contact point).
- Indicate wrapping by drawing a short curved segment of string hugging the rim over a small arc near the tangent point (a subtle arc segment on the rim), making it visually clear the string is wrapped around the outer edge.
- Label the string near the vertical segment: "string".

Hanging block:
- Draw a small rectangle (the block) attached to the bottom end of the vertical string.
- Center the block directly below the tangent point so the string is perfectly vertical.
- Label the block to the right of it: "m = 0.50 kg".

Drop distance and ground:
- Draw a thick horizontal line as the ground across the full width at the bottom.
- The block is shown above the ground with a clear gap.
- Indicate the drop distance h as a vertical double-headed dimension arrow drawn to the right of the block:
  - The top of this dimension arrow aligns horizontally with the block’s initial position (use a short horizontal dashed reference line through the block’s center).
  - The bottom of the dimension arrow touches the ground line.
  - Label next to the dimension arrow: "h = 1.5 m".

Extra text constraints:
- Do NOT include any other numbers.
- Do NOT include any angular velocity arrows on the disk in this figure.
- All labels are horizontal, in clear print: "M = 2.0 kg", "R = 0.40 m", "m = 0.50 kg", "h = 1.5 m", "frictionless axle".

Figure 2. Force-vector templates for the disk and for the block (vectors drawn from the central dot).

Monochrome worksheet-style figure containing two separate, clearly boxed panels side-by-side with equal size.

Panel layout:
- Left panel title centered at the top inside the panel: "Disk".
- Right panel title centered at the top inside the panel: "Block".
- Each panel contains a single object outline and a single filled dot used as the required tail point for all force vectors.

Left panel (Disk template):
- Draw a circle (disk outline) centered in the panel.
- Place a small filled dot exactly at the geometric center of the circle.
- Through the dot, draw a very short horizontal line segment to indicate the axle location (purely visual cue).
- No force arrows are pre-drawn.
- Provide small, faint direction reference ticks (not arrows) outside the disk: one at the top, bottom, left, and right of the circle to help students orient directions, but with no letters on them.

Right panel (Block template):
- Draw a square/rectangle block outline centered in the panel (slightly taller than wide).
- Place a small filled dot exactly at the geometric center of the block.
- No force arrows are pre-drawn.

Global styling requirements:
- The central dots are the only filled elements.
- No other text besides the panel titles "Disk" and "Block".
- No background grid.
- The panels have thin rectangular borders to clearly separate them.
A.

Draw and label free-body diagrams showing all the forces exerted on the disk and on the block while the block is falling. Use Figure 2 to draw each force vector starting on the dot and pointing in the direction of the force. The length of each vector should be proportional to the magnitude of the force. Clearly label each force using standard notation (e.g., FTF_T for tension, FgF_g for gravitational force). While the block is falling and before it hits the ground, the string exerts forces on both the disk and the block.

B.

Derive an expression for the magnitude of the angular acceleration α\alpha of the disk while the block is falling. Express your answer in terms of M, m, R, g, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information. The block falls with acceleration aya_y in the downward direction. The rotational inertia of the disk about its central axis is I=12MR2I = \frac{1}{2}MR^2.

Figure 3. Axes for sketching angular velocity ω versus time t, with t₁ marked.

A blank set of Cartesian axes for a hand-drawn sketch.

Axes formatting (all required numbers shown as tick labels):
- Horizontal axis labeled "t (s)".
- Vertical axis labeled "ω (rad/s)".
- Origin at the lower-left corner of the plotting region is labeled "0".
- Arrows are drawn on the positive ends of both axes.

Horizontal axis scale:
- The time axis starts at 0 and ends at 6.
- Tick marks every 1 second with numeric labels: 0, 1, 2, 3, 4, 5, 6.
- The special time label "t₁" is marked by a vertical tick on the time axis located exactly at 3 seconds.
- Directly under that tick, write "t₁".

Vertical axis scale:
- The angular-velocity axis starts at 0 and ends at 12.
- Tick marks every 2 rad/s with numeric labels: 0, 2, 4, 6, 8, 10, 12.

No curve is drawn (axes only):
- The plotting region is empty except for the axes, ticks, tick labels, and the "t₁" marker.
- No grid lines.
- No title other than the caption outside the axes (if any).
C.

Sketch a graph of the angular velocity ω\omega of the disk as a function of time t from t = 0 to a time significantly after t1t_1 using Figure 3. Your graph should show the behavior both before and after the block hits the ground. Clearly indicate any discontinuities or changes in the relationship between ω\omega and t. The angular velocity of the disk increases as the block falls. At time t1t_1, the block hits the ground and the string goes slack. After t1t_1, the disk continues to rotate freely about the axle with no friction.

Figure 4. Rotational inertia options for three disks.

A clean, bordered table with 3 columns and 4 rows total (1 header row plus 3 data rows). All text is black on white.

Column headers (top row):
- Column 1 header: "Disk"
- Column 2 header: "Mass"
- Column 3 header: "Rotational inertia about center"

Data rows (each row is a single line in the table):
- Row 1: "Disk 1 (solid)" | "M" | "½MR²"
- Row 2: "Disk 2 (ring)"  | "M" | "MR²"
- Row 3: "Disk 3 (solid)" | "2M" | "MR²"

Formatting details:
- Use the squared symbol in R².
- Use the one-half symbol "½" (not a decimal).
- Center text vertically within each cell.
- Left-align the Disk names in column 1; center the symbols in columns 2 and 3.
D.

Indicate which disk (Disk 1, Disk 2, or Disk 3) will have the greatest angular velocity immediately after the block hits the ground. If two or more disks will have the same angular velocity, indicate this explicitly. The experiment is repeated three times using different disks, each with radius R = 0.40 m. The same block of mass m = 0.50 kg falls the same distance h = 1.5 m in each trial. The properties of the three disks are shown in Figure 4.

Disk 1 Disk 2 Disk 3 Two or more disks tie

Briefly justify your answer using physics principles, referencing relevant equations or relationships.

Key terms

TermDefinition
angular displacementThe angle in radians through which a rigid body rotates about a specified axis, calculated as delta-theta = theta - theta_0.
radianThe standard unit of angular measurement in rotational kinematics; arc length equals radius times angle in radians.
tangential accelerationThe component of linear acceleration directed tangent to the circular path of a point on a rotating body; a_T = r*alpha.
lever armThe perpendicular distance from the axis of rotation to the line of action of an applied force; multiplying force by lever arm gives torque magnitude.
Moment armThe perpendicular distance from the axis of rotation to the line of action of a force; equivalent to lever arm and used interchangeably in torque calculations.
cross productThe vector operation tau = r x F that defines torque; its magnitude is rF sin-theta and its direction is perpendicular to both r and F by the right-hand rule.
position vectorThe vector r from the axis of rotation to the point where a force is applied; its magnitude and the angle it makes with the force determine the torque.
mass distributionThe spatial arrangement of mass relative to the axis of rotation; mass farther from the axis contributes more to rotational inertia because I scales as r^2.
non-uniform mass distributionA mass distribution where density varies with position; requires I = integral r^2 dm with a position-dependent expression for dm.
translational equilibriumThe condition sum-F = 0; independent of rotational equilibrium and must be checked separately in static problems.
rolling motionCombined translational and rotational motion where the contact point has zero velocity; governed by the no-slip constraint v_cm = R*omega.
no-slip conditionThe constraint that the contact point between a rolling object and a surface has zero relative velocity, giving v_cm = R*omega and a_cm = R*alpha.
constraint equationA kinematic relationship linking linear and angular motion, such as a = alpha*R for a string on a pulley, used to connect the translational and rotational equations in a combined problem.

Common unit 5 mistakes

Confusing angular and linear quantities

Omega and alpha describe the whole rigid body; v and a_T describe a specific point at radius r. Plugging omega directly into a linear equation without multiplying by r is one of the most frequent errors in Unit 5 problems.

Using the wrong component of force for torque

Only the force component perpendicular to the position vector r produces torque. If a force is applied at an angle, you must use F*sin-theta (or equivalently the lever arm), not the full magnitude of F.

Forgetting to apply the parallel axis theorem correctly

I' = I_cm + Md^2 requires I_cm, the rotational inertia about the center of mass, not about some other convenient axis. Using the wrong base value of I leads to an incorrect result.

Treating rotational and translational equilibrium as the same condition

Sum-F = 0 and sum-tau = 0 are independent equations. A beam can have zero net torque but a nonzero net force (or vice versa). Both conditions must be checked separately in static equilibrium problems.

Inconsistent sign conventions in combined problems

In a pulley-mass system, if you define downward as positive for the hanging mass, you must define the corresponding rotation direction as positive for the pulley. Mixing sign conventions across the two equations produces incorrect constraint equations.

How this unit shows up on the AP exam

Combined translational and rotational analysis

AP Physics C: Mechanics frequently presents systems where a mass accelerates linearly while a pulley or spool rotates. You are expected to write separate sum-F = ma and sum-tau = I*alpha equations, identify the correct rotational inertia, and apply a constraint equation to solve for angular acceleration, linear acceleration, or tension. Showing each equation separately before substituting is important for earning full credit.

Calculus-based rotational inertia derivations

Free-response questions may ask you to derive I for a rod of uniform or nonuniform density, a disk, or an annular ring using I = integral r^2 dm. You are expected to set up the integral explicitly, express dm in terms of a spatial variable and the object's geometry, evaluate the integral, and apply the parallel axis theorem if the axis is not through the center of mass.

Rotational equilibrium with force and torque diagrams

Problems involving beams, hinges, or suspended rods require both a force diagram and a torque equation. You are expected to choose a pivot strategically, correctly identify each torque's sign and lever arm, and recognize that satisfying sum-tau = 0 does not automatically satisfy sum-F = 0. Justifying your pivot choice and showing the torque sum explicitly are standard expectations.

Final unit 5 review checklist

  • Final Unit 5 review checklistUse this list to confirm you can handle every major skill in Unit 5 before the exam.
  • Write and apply the constant-alpha kinematic equationsGiven initial angular velocity, angular acceleration, and time (or angle), solve for any unknown using omega = omega_0 + alpha*t, theta = theta_0 + omega_0*t + (1/2)*alpha*t^2, and omega^2 = omega_0^2 + 2*alpha*(delta-theta).
  • Convert between angular and linear quantitiesUse s = r*theta, v = r*omega, and a_T = r*alpha to find the linear motion of a specific point on a rotating rigid body given its distance from the axis.
  • Calculate torque using the cross product and lever armFind tau = rF sin-theta for a force applied at angle theta from the radial direction, or equivalently multiply the force by its lever arm. Assign correct sign based on clockwise or counterclockwise direction.
  • Derive rotational inertia using integration and the parallel axis theoremSet up I = integral r^2 dm for uniform and nonuniform rods, disks, cylindrical shells, and annular rings. Apply I' = I_cm + Md^2 to shift the axis off the center of mass.
  • Apply rotational equilibrium (sum-tau = 0)Choose a pivot strategically, write the torque equation, and solve for unknown forces or distances. Recognize that rotational equilibrium does not require translational equilibrium.
  • Solve combined translational and rotational dynamics problemsWrite sum-F = ma for linear motion and sum-tau = I*alpha for rotation, then use a constraint equation (a = alpha*R or v_cm = omega*R) to connect the two equations and solve for unknowns.

How to study unit 5

Step 1: Rotational kinematics and the linear connection (Topics 5.1-5.2)Read the Topic 5.1 and 5.2 guides. Practice writing the constant-alpha equations from memory and converting between angular and linear quantities using s = r*theta, v = r*omega, and a_T = r*alpha. Sketch theta(t) and omega(t) graphs and identify slopes and areas.
Step 2: Torque (Topic 5.3)Read the Topic 5.3 guide. Practice drawing force diagrams that show where each force is applied relative to the axis. Calculate torques using both tau = rF sin-theta and the lever arm method. Apply the right-hand rule to determine direction.
Step 3: Rotational inertia (Topic 5.4)Read the Topic 5.4 guide. Derive I for a uniform rod about its end and about its center using integration, then verify the end result with the parallel axis theorem. Memorize the standard results for disks and cylindrical shells. Practice setting up dm in terms of a spatial variable for nonuniform rods.
Step 4: Rotational equilibrium (Topic 5.5)Read the Topic 5.5 guide. Solve beam and lever problems by writing sum-tau = 0 about a strategically chosen pivot. Practice identifying which forces produce zero torque about a given pivot and why.
Step 5: Newton's second law in rotational form and combined systems (Topic 5.6)Read the Topic 5.6 guide. Set up and solve pulley-mass and rolling-object problems by writing separate sum-F = ma and sum-tau = I*alpha equations, then connecting them with a constraint equation. Use the AP score calculator to estimate your overall score as you complete practice problems.

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Topic study guides

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Frequently Asked Questions

What topics are covered in AP Physics Mech Unit 5?

AP Physics C: Mechanics Unit 5 covers 6 topics: Rotational Kinematics, Connecting Linear and Rotational Motion, Torque, Rotational Inertia, Rotational Equilibrium and Newton's First Law in Rotational Form, and Newton's Second Law in Rotational Form. Together they build the rotational counterparts to the linear dynamics you studied earlier in the course. See the full topic breakdown at AP Physics C: Mechanics Unit 5.

How much of the AP Physics Mech exam is Unit 5?

Unit 5 makes up 10-15% of the AP Physics C: Mechanics exam. That weight covers Torque and Rotational Dynamics, including Rotational Kinematics, Torque, Rotational Inertia, Rotational Equilibrium, and Newton's Second Law in Rotational Form. It's a meaningful chunk of the exam, so strong fluency with these concepts pays off on both the MCQ and FRQ sections.

What's on the AP Physics Mech Unit 5 progress check (MCQ and FRQ)?

The AP Physics C: Mechanics Unit 5 progress check in AP Classroom includes both MCQ and FRQ parts drawn from all six unit topics: Rotational Kinematics, Connecting Linear and Rotational Motion, Torque, Rotational Inertia, Rotational Equilibrium, and Newton's Second Law in Rotational Form. The MCQ part tests conceptual understanding and calculation, while the FRQ part typically asks you to derive expressions, apply Newton's Second Law in rotational form, or analyze rotational equilibrium scenarios. For matched practice before the progress check, visit AP Physics C: Mechanics Unit 5.

How do I practice AP Physics Mech Unit 5 FRQs?

The best way to practice Unit 5 FRQs is to work through problems that mirror the three most common question types: deriving rotational inertia expressions, applying Newton's Second Law in rotational form to systems with torque, and analyzing rotational equilibrium. These topics, especially 5.4 Rotational Inertia and 5.6 Newton's Second Law in Rotational Form, show up most often in free-response questions on the AP Physics C: Mechanics exam. Start by writing out full solutions with clear diagrams and labeled torque directions, then check your reasoning step by step. You'll find FRQ-aligned practice at AP Physics C: Mechanics Unit 5.

Where can I find AP Physics Mech Unit 5 practice questions?

For AP Physics C: Mechanics Unit 5 practice questions, including multiple-choice and practice test problems on Torque and Rotational Dynamics, head to AP Physics C: Mechanics Unit 5. You'll find MCQ-style questions covering Rotational Kinematics, Torque, Rotational Inertia, and Newton's Second Law in Rotational Form, plus FRQ practice that mirrors what College Board tests on the actual exam.

How should I study AP Physics Mech Unit 5?

Start with Rotational Kinematics (5.1) and Connecting Linear and Rotational Motion (5.2) to build the foundation, since every later topic depends on those analogies between linear and rotational quantities. Then work through Torque (5.3) and Rotational Inertia (5.4) together, practicing the integral definitions of inertia for common shapes. Once those feel solid, tackle Rotational Equilibrium (5.5) and Newton's Second Law in Rotational Form (5.6) with full free-body and torque diagrams on every problem. A few concrete steps that help: draw the rotational analog table (x vs. theta, v vs. omega, F vs. tau) and keep it visible while you practice. Do at least one FRQ per topic from past exams, writing out every step rather than just checking the answer. Visit AP Physics C: Mechanics Unit 5 for organized practice by topic.

Ready to review Unit 5?Start with the notes, check the topic cards, and use the practice or resource links when they are available for this course.