AP Physics 1 · Topic 5.6

Topic 5.6: Newton's Second Law in Rotational Form

Unit 5: Torque and Rotational Dynamics10-15% of the multiple-choice section

Newton's second law in rotational form says the angular acceleration of a system equals the net torque on it divided by its rotational inertia. It looks like F equals ma, but rotational inertia is not a fixed property the way mass is: it changes with the axis you choose.

AP Physics: Unit 5 (topics 5.6 Newton's Second Law in Rotational Form). AP Physics 1 Unit 5, Topic 5.6, covering learning objective 5.6.A, describe the conditions under which a system's angular velocity changes, and essential knowledge 5.6.A.1 through 5.6.A.3. The CED lists four suggested skills for this topic: 1.A, 2.A, 2.C, and 3.C. The CED attaches no boundary statement to Topic 5.6. Unit 5 is weighted at 10 to 15 percent of the multiple-choice section and about 15 to 20 class periods.

What the CED asks of Topic 5.6

Topic 5.6 has one learning objective and three essential knowledge statements, and no boundary statement of its own.

Objective 5.6.A asks you to describe the conditions under which a system's angular velocity changes. It is the exact complement of objective 5.5.A, which asks when the angular velocity remains constant, and the pair mirrors 2.4.A and 2.5.A in the translational half of the course.

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. The relevant equation is the one printed on the equation sheet:

\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}} = \frac{\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.

The CED lists four suggested skills for Topic 5.6: 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. In the CED's grouping, 1.A belongs to Practice 1, Creating Representations; 2.A and 2.C to Practice 2, Mathematical Routines; and 3.C to Practice 3, Scientific Questioning and Argumentation.

Skill 2.C shapes the questions. Comparing scenarios is why so many Topic 5.6 items give you the same object twice with the axis moved, or the same torque applied to two systems, and ask which spins up faster. Statement 5.6.A.2 is written in the language of proportionality for that reason: it is about how $\alpha$ scales, not just how to compute it once.

The equation as the sheet actually prints it

Most textbooks write the rotational second law as $\tau_{\text{net}} = I\alpha$ . The AP Physics 1 Table of Information does not. It prints

\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}

which is the same relationship solved for angular acceleration, printed in exactly the same shape as the translational one, $\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$ . Reading the two lines side by side is a faster way to learn the analogy than memorizing a table of it.

Rearranged, $\tau_{\text{net}} = I \alpha$ is fine to use and fine to write on a free-response question. Just know that the sheet gives the quotient form, so hunting for $\tau = I\alpha$ under exam pressure only costs time. The other rotational lines the sheet does print:

Printed on the AP Physics 1 sheet	What it gives you
$\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}$	The rotational second law
$\tau = r_{\perp} F = r F \sin\theta$	The magnitude of each torque
$I = \sum m_i r_i^2$	Rotational inertia of a set of point masses
$I' = I_{\text{cm}} + M d^2$	The parallel axis theorem
$\omega = \omega_0 + \alpha t$ and the other two angular kinematics equations	What happens after you have $\alpha$
$v = r\omega$ and $a_T = r\alpha$	The bridge to linear quantities

What is not printed is the rotational inertia of any extended shape. The Topic 5.4 boundary statement explains why: AP Physics 1 only expects students to calculate the rotational inertia for systems of five or fewer objects arranged in a two-dimensional configuration, and students do not need to know the rotational inertia of extended rigid systems, as these will be provided within the exam. So if a question involves a disk, a rod, a hoop, or a sphere, look for the given value of $I$ in the stem.

Also not printed is $\sum \tau = 0$ , the rotational equilibrium condition from Topic 5.5, which you derive by setting $\alpha_{\text{sys}} = 0$ . Several relationships students expect to look up on the AP Physics 1 formula sheet are one-step derivations like this.

The analogy with F equals ma, term by term

Every symbol in the translational second law has a rotational partner, and the algebra is identical once the swap is made.

Translational	Rotational	Relationship
Force $F$	Torque $\tau$	$\tau = r F \sin\theta$
Mass $m$	Rotational inertia $I$	$I = \sum m_i r_i^2$
Acceleration $a$ , in $\mathrm{m/s^2}$	Angular acceleration $\alpha$ , in $\mathrm{rad/s^2}$	$a_T = r\alpha$
Velocity $v$ , in $\mathrm{m/s}$	Angular velocity $\omega$ , in $\mathrm{rad/s}$	$v = r\omega$
$\vec{a}_{\text{sys}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$	$\alpha_{\text{sys}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}$	Both printed on the sheet

Three features of the translational law carry across cleanly, and they are worth stating because each is directly testable.

It is the net that matters. One torque tells you nothing on its own. Add the signed torques, then divide. A wheel with a 4.0 newton meter drive torque and a 4.0 newton meter friction torque has zero angular acceleration despite two large torques acting.
Direction is preserved. EK 5.6.A.2 says the angular acceleration is in the same direction as the net torque, so in a single plane with a sign convention declared as in Topic 5.5, $\alpha$ carries the sign of $\tau_{\text{net}}$ . A negative $\alpha$ on a system spinning in the positive direction means it is slowing, not reversing.
Constant torque means constant angular acceleration. That is what licenses the three angular kinematics equations from Topic 5.1. If the torque varies, they do not apply, exactly as the linear kinematic equations fail under varying force.

The proportionality language in EK 5.6.A.2 is the version the exam tends to test. Doubling the net torque doubles $\alpha$ with the system unchanged; doubling $I$ halves $\alpha$ with the torque unchanged. Those are questions with no arithmetic in them at all.

Where the analogy breaks: rotational inertia is not a fixed property

This is the difference that matters most, and it is the one that costs marks.

Mass is a property of an object. A 2.0 kg block has a mass of 2.0 kg on a table, in orbit, sideways, or spinning, and you never have to ask "2.0 kg about what?"

Rotational inertia is not like that. It is a property of an object and a chosen axis, jointly. The same rod has one rotational inertia about its center and a different, larger one about its end, and neither is more correct. EK 5.4.A.1 says rotational inertia is related to the mass of the system and the distribution of that mass relative to the axis of rotation, and that second clause has no translational counterpart.

Two CED statements pin the size of the effect.

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. So no axis choice ever gives you less than $I_{\text{cm}}$ .
5.4.B.2. The parallel axis theorem, $I' = I_{\text{cm}} + M d^2$ , tells you exactly how much more you get for moving the axis a distance $d$ away, parallel to the one through the center of mass. Since $Md^2$ grows as the square of the offset, moving the axis is a powerful lever.

The practical consequences for Topic 5.6:

Never quote a rotational inertia without naming its axis. "The rotational inertia of the rod is $0.180\ \mathrm{kg \cdot m^2}$ " is incomplete. About its center, it is.
**The $I$ in the second law must be about the same axis as the torques.** Mixing an $I$ about the center of mass with torques computed about an end produces a meaningless number. Choose the axis once, then compute both about it.
The free pivot choice from Topic 5.5 does not survive here. In rotational equilibrium the net torque is zero about every axis, so you may move the pivot for convenience. Once $\tau_{\text{net}} \neq 0$ , $\sum \tau$ depends on the axis, and you must use the physical axis the system rotates about, or the axis through the center of mass when it is also translating freely.
Same object, different answers. The Topic 5.4 boundary statement's own illustration: a hoop has more rotational inertia than a solid disk of the same mass and radius, because its mass sits farther from the axis. Given equal torques, the disk wins.

Topic 5.4 covers the calculation of $I$ in full, and the second law is only as good as the $I$ you feed it.

Degrees for the geometry, radians for the motion

The rotational second law has both kinds of angle in it, one on each side of the equals sign, so this is worth getting straight before the first calculation.

On the torque side, $\theta$ is a geometric angle measured in degrees. In $\tau = r F \sin\theta$ , that angle is the one between the force vector and the position vector from the axis of rotation to the point of application of the force. The AP Physics 1 Table of Information supplies a table of trigonometric values for common angles listed as $0^\circ$ , $30^\circ$ , $37^\circ$ , $45^\circ$ , $53^\circ$ , $60^\circ$ , and $90^\circ$ , so degree mode is the expected setting.

On the angular acceleration side, everything is in radians. EK 5.1.A.1 states that angular displacement is the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis, and angular velocity and angular acceleration inherit that unit as rad/s and $\mathrm{rad/s^2}$ .

Symbol	Unit	Never
$\theta$ inside $\sin\theta$ in the torque equation	Degrees	Radians, unless your calculator is in radian mode on purpose
$\Delta\theta$ , angular displacement	Radians	Degrees, or the arc length comes out wrong
$\omega$ , angular velocity	rad/s	rev/s, without converting first
$\alpha$ , angular acceleration	$\mathrm{rad/s^2}$	Degrees per second squared
$\tau$ , torque	$\mathrm{N \cdot m}$	Joules, even though the units are dimensionally identical
$I$ , rotational inertia	$\mathrm{kg \cdot m^2}$	Anything else

The bridge equations are where a degree-for-radian slip bites. $a_T = r\alpha$ and $v = r\omega$ hold only with the angular quantity in radians, because the radian is a ratio of arc length to radius and is therefore dimensionless. Substitute degrees and the linear result is off by a factor of $180/\pi$ , roughly 57. Revolutions per minute is the other trap: convert to rad/s first, since one revolution is $2\pi$ rad. The radian never appears in the unit of torque or of rotational inertia, and it is because it appears and disappears silently that it has to be tracked by hand.

Linear and rotational analyses, performed independently

EK 5.6.A.3 is the shortest statement in the topic and the one that determines how free-response questions are graded: to fully describe a rotating rigid system, linear and rotational analyses may need to be performed independently.

Read it as an instruction. Some systems need two separate applications of Newton's second law, one translational and one rotational, joined by a constraint:

Draw a free-body diagram for each object that translates, and a force diagram for each rigid system that rotates, per skill 1.A. The force diagram must show where each force acts, since that is what determines the torques.
Write $\sum F = m a$ for the translating parts. Different objects get different equations.
Write $\sum \tau = I \alpha$ for the rotating part, about its actual axis.
Write the constraint that links them. For a string that does not slip over a pulley of radius $R$ , the constraint is $a = R\alpha$ . For a wheel rolling without slipping, the center of mass acceleration and the angular acceleration are linked the same way, $a_{\text{cm}} = R\alpha$ .
Solve the system of equations together.

The most common error in this pattern is assuming the tension in a string over a real pulley equals the weight hanging from it. If the pulley has rotational inertia, spinning it up takes a net torque, and that torque comes from a tension that must therefore be less than the hanging weight. Worked example 1 puts numbers on it: 8.4 N of tension against a 19.6 N weight, and assuming otherwise inflates the angular acceleration by more than a factor of two.

There is a second reason to do both. A rotational analysis gives you $\alpha$ but says nothing about the force at the axle or pivot, which comes only from applying $\sum F = m a_{\text{cm}}$ to the whole system using the center of mass acceleration that $\alpha$ implies. Worked example 3 does exactly that for a falling rod.

The connections themselves come from Topic 5.2, which owns $v = r\omega$ and $a_T = r\alpha$ , and the translational half is Topic 2.5.

Common traps and what the graphs show

Six errors account for most of the lost marks in this topic.

**Using $I$ about the wrong axis.** The most expensive error, because the answer is wrong by a clean factor that looks plausible. Name the axis in writing before you compute $I$ .
Using the force instead of the torque. A force applied at the axle produces zero angular acceleration however large it is, because its lever arm is zero. Run every force through $\tau = r F \sin\theta$ first.
Forgetting to net the torques. Two torques of equal magnitude on opposite sides of the axis may cancel or may add, depending on the geometry. Sign each, then sum.
Assuming string tension equals hanging weight. True only for a massless pulley, where the rotational inertia is zero.
**Degrees inside $v = r\omega$ or $a_T = r\alpha$ .** Wrong by a factor near 57.
Reporting torque in joules. Newton meters. The units are dimensionally the same and the quantities are not.

Skill 1.A asks for the representations, and the graph reading follows from them:

Graph	Meaning
Slope of $\omega$ versus $t$	Angular acceleration $\alpha$
Area under $\alpha$ versus $t$	Change in angular velocity $\Delta\omega$
Slope of $\theta$ versus $t$	Angular velocity $\omega$
Slope of $\alpha$ versus $\tau_{\text{net}}$ , for one system	$1 / I_{\text{sys}}$

The last row is the lab. Hang known masses over a pulley to apply known torques to a rotating platform, measure the angular acceleration, and plot $\alpha$ against $\tau_{\text{net}}$ : the reciprocal of the slope is the rotational inertia, and a straight line through the origin is itself evidence for the second law. That is the kind of claim skill 3.C asks you to justify.

Unit 5 is weighted at 10 to 15 percent of the AP Physics 1 multiple-choice section and about 15 to 20 class periods, and this topic closes it; Unit 6 then applies energy and momentum to the same rotating systems. Next: Topic 5.4 for the rotational inertia this law consumes, Topic 5.5 for the case where the net torque vanishes, and the rotational kinematics guide for the equations that take over once $\alpha$ is known.

A hanging block and a pulley with real rotational inertia

A light string is wrapped around a pulley of radius 0.200 m and rotational inertia $0.0600\ \mathrm{kg \cdot m^2}$ , mounted on a fixed frictionless axle. A 2.00 kg block hangs from the free end and is released from rest. The string does not slip. Find the block's acceleration, the tension in the string, and the pulley's angular acceleration. Use $g = 9.8\ \mathrm{m/s^2}$ .

This is EK 5.6.A.3 in its purest form: one object translates, one rotates, and the analyses are done independently and then joined. Take downward as positive for the block, and the pulley's rotation in the direction the falling block turns it as positive.
Linear analysis, on the block. Two forces act on it: gravity $mg = (2.00)(9.8) = 19.6\ \mathrm{N}$ down, and the string tension $T$ up. So $\sum F = mg - T = ma$ , which is $19.6 - T = 2.00\,a$ .
Rotational analysis, on the pulley. The axle force acts at the axis, so its lever arm is zero and it exerts no torque. Only the tension does, applied tangentially at the rim: $\tau_{\text{net}} = T R$ , so $\alpha = \frac{TR}{I}$ . The constraint: the string does not slip, so the block's acceleration equals the tangential acceleration at the rim, $a = R\alpha$ , therefore $\alpha = a/R$ . That is one of the equations that demands radians.
Combine the rotational equation with the constraint to get the tension in terms of $a$ : $T R = I \frac{a}{R}$ , so $T = \frac{I a}{R^2} = \frac{(0.0600)a}{(0.200)^2} = \frac{0.0600}{0.0400}a = 1.50\,a$ . The quantity $I/R^2 = 1.50\ \mathrm{kg}$ acts like an extra mass the block has to drag into motion.
Substitute into the block's equation: $19.6 - 1.50\,a = 2.00\,a$ , so $19.6 = 3.50\,a$ and $a = \frac{19.6}{3.50} = 5.60\ \mathrm{m/s^2}$ .
Back-substitute: $T = (1.50)(5.60) = 8.40\ \mathrm{N}$ , and $\alpha = \frac{a}{R} = \frac{5.60}{0.200} = 28.0\ \mathrm{rad/s^2}$ .
Check both laws independently. Block: $mg - T = 19.6 - 8.40 = 11.2\ \mathrm{N}$ , and $ma = (2.00)(5.60) = 11.2\ \mathrm{N}$ . Pulley: $\tau = TR = (8.40)(0.200) = 1.68\ \mathrm{N \cdot m}$ , and $I\alpha = (0.0600)(28.0) = 1.68\ \mathrm{N \cdot m}$ . Both balance.
Now price the classic error. Assuming $T = mg = 19.6\ \mathrm{N}$ gives $\tau = (19.6)(0.200) = 3.92\ \mathrm{N \cdot m}$ and $\alpha = 3.92 / 0.0600 = 65.3\ \mathrm{rad/s^2}$ , which is 2.33 times the correct value. The tension is below the weight precisely because the block is accelerating.

The block accelerates at 5.60 m/s squared, the tension is 8.40 N, and the pulley's angular acceleration is 28.0 rad/s squared. The tension is well below the block's 19.6 N weight, which is the whole point of giving the pulley a rotational inertia. Setting $I = 0$ in the same algebra returns $a = 9.8\ \mathrm{m/s^2}$ and $T = 0$ , which is the massless-pulley limit.

The same torque on the same rod, about two different axes

A uniform rod has mass 1.50 kg and length 1.20 m. The exam provides the rotational inertia of a uniform rod about a perpendicular axis through its center as $I_{\text{cm}} = \frac{1}{12}ML^2$ . A net torque of $0.360\ \mathrm{N \cdot m}$ is applied to the rod, first about a perpendicular axis through its center, then about a perpendicular axis through one end. Find the angular acceleration in each case and compare.

Axis through the center. $I_{\text{cm}} = \frac{1}{12}(1.50)(1.20)^2 = \frac{1}{12}(1.50)(1.44) = \frac{2.16}{12} = 0.180\ \mathrm{kg \cdot m^2}$ .
Apply the printed equation: $\alpha_{\text{cm}} = \frac{\tau_{\text{net}}}{I_{\text{cm}}} = \frac{0.360}{0.180} = 2.00\ \mathrm{rad/s^2}$ .
Axis through one end. The new axis is parallel to the first and offset by $d = L/2 = 0.600\ \mathrm{m}$ , so use the parallel axis theorem from EK 5.4.B.2: $I' = I_{\text{cm}} + Md^2 = 0.180 + (1.50)(0.600)^2 = 0.180 + (1.50)(0.360) = 0.180 + 0.540 = 0.720\ \mathrm{kg \cdot m^2}$ .
Sanity-check against the closed form for a rod about its end, $\frac{1}{3}ML^2 = \frac{1}{3}(1.50)(1.44) = 0.720\ \mathrm{kg \cdot m^2}$ . They agree. Apply the second law again with the same torque: $\alpha_{\text{end}} = \frac{0.360}{0.720} = 0.500\ \mathrm{rad/s^2}$ .
Compare, which is skill 2.C. The rotational inertia went up by a factor of $0.720 / 0.180 = 4.00$ , and the angular acceleration went down by a factor of $2.00 / 0.500 = 4.00$ . That is EK 5.6.A.2's inverse proportionality holding exactly, since the torque was unchanged.
Notice what did not change: the mass. It is 1.50 kg in both calculations, but on its own it predicted nothing. The mass distribution relative to the chosen axis did all the work, which is the gap between $I$ and $m$ that makes the analogy with $F = ma$ imperfect.
Also notice that the center-of-mass axis gave the smaller $I$ and therefore the larger $\alpha$ . EK 5.4.B.1 guarantees this: rotational inertia in a given plane is at a minimum about the axis through the center of mass, so no other parallel axis can spin faster under the same torque.

About the center the rod angularly accelerates at 2.00 rad/s squared; about one end, at 0.500 rad/s squared. Same rod, same mass, same net torque, and a factor of four between the answers. The number quoted for a rotational inertia is meaningless until the axis is named, which is exactly where the analogy with mass and $F = ma$ stops holding.

A falling rod, and why its tip beats gravity

The same uniform rod, mass 1.50 kg and length 1.20 m, is now hinged at one end and held horizontal, then released from rest. At the instant of release, find the rod's angular acceleration, the linear acceleration of its far tip, and the force the hinge exerts on the rod. Its rotational inertia about the hinged end is $0.720\ \mathrm{kg \cdot m^2}$ . Use $g = 9.8\ \mathrm{m/s^2}$ .

Rotational analysis first. The hinge force acts at the axis and so exerts no torque about it. The only torque comes from gravity, which for a uniform rod acts effectively at the center of mass, 0.600 m from the hinge.
The rod is horizontal at the instant of release, so the weight is perpendicular to the rod and $\sin\theta = \sin 90^\circ = 1$ : $\tau_{\text{net}} = Mg\frac{L}{2} = (1.50)(9.8)(0.600) = 8.82\ \mathrm{N \cdot m}$ .
Apply the second law about the hinge, with the same axis for both the torque and the rotational inertia: $\alpha = \frac{\tau_{\text{net}}}{I} = \frac{8.82}{0.720} = 12.25\ \mathrm{rad/s^2}$ .
Now the tip. It sits the full length from the axis, so $a_T = L\alpha = (1.20)(12.25) = 14.7\ \mathrm{m/s^2}$ , larger than $g = 9.8\ \mathrm{m/s^2}$ . The end of the rod starts downward faster than a ball dropped beside it: balance a coin there and the rod pulls away beneath it.
Linear analysis, independently. The center of mass sits at 0.600 m, so its acceleration is $a_{\text{cm}} = \frac{L}{2}\alpha = (0.600)(12.25) = 7.35\ \mathrm{m/s^2}$ downward, which is less than $g$ .
Apply the translational second law to the whole rod. The net downward force must be $M a_{\text{cm}} = (1.50)(7.35) = 11.025\ \mathrm{N}$ , while gravity alone supplies $Mg = (1.50)(9.8) = 14.70\ \mathrm{N}$ downward.
The difference has to come from the hinge: $F_{\text{hinge}} = 14.70 - 11.025 = 3.675\ \mathrm{N}$ upward, or $3.68\ \mathrm{N}$ to three significant figures. The rotational analysis could never have produced this number, because the hinge force was deliberately outside the torque equation. That is what EK 5.6.A.3 means by performing the two analyses independently.
Both accelerations are correct at once because the rod is rigid: every point shares the same $\alpha$ , and $a_T = r\alpha$ makes the linear acceleration grow with distance from the axis. The rod is not in free fall, and the hinge is the reason.

The angular acceleration is 12.25 rad/s squared, the far tip accelerates downward at 14.7 m/s squared, and the hinge pushes up on the rod with 3.68 N. The tip out-accelerates gravity while the center of mass under-accelerates it, which is only contradictory if you forget that the hinge is exerting a force. The rotational analysis gave $\alpha$ , and only the linear analysis gave the hinge force.

Frequently asked questions

What is Newton's second law in rotational form?

It states that the angular acceleration of a system equals the net torque exerted on it divided by its rotational inertia. The AP Physics 1 equation sheet prints it as alpha equals the sum of the torques over the rotational inertia of the system. Rearranged it is the familiar net torque equals I times alpha. The CED adds that the angular acceleration is in the same direction as the net torque, is directly proportional to it, and is inversely proportional to the rotational inertia.

Is net torque equals I alpha on the AP Physics 1 equation sheet?

Not in that arrangement. The Table of Information prints the relationship solved for angular acceleration, as alpha for the system equals the sum of the torques divided by the rotational inertia of the system, which mirrors how the translational second law is printed as acceleration equals net force over mass. Writing it as net torque equals I alpha is perfectly acceptable in a solution, but do not waste exam time searching the sheet for that form.

How is torque equals I alpha different from F equals ma?

Algebraically they are the same shape, and each translational symbol has a rotational partner. The real difference is that mass is a property of an object alone, while rotational inertia is a property of an object and a chosen axis together. A rod has one mass but many rotational inertias, smallest about its center of mass and larger about any parallel axis by M times d squared. So a rotational inertia quoted without naming its axis is an incomplete statement, and the axis used for the torques must be the same one.

Why does a hoop accelerate more slowly than a disk under the same torque?

Because rotational inertia depends on how far the mass sits from the axis, not just on how much mass there is. The AP Physics 1 CED gives this exact example: a hoop has more rotational inertia than a solid disk of the same mass and radius, since all of the hoop's mass is at the rim while the disk's is spread inward. Angular acceleration is inversely proportional to rotational inertia, so the same net torque produces a smaller angular acceleration for the hoop.

Why is the string tension less than the hanging weight in a pulley problem?

Because a pulley with rotational inertia needs a net torque to spin up, and the only torque available comes from the string tension. If the tension equaled the weight, the block would have no net force and would not accelerate, yet it clearly does. Working the two second laws together for a 2.00 kg block on a pulley of rotational inertia 0.0600 kilogram meter squared and radius 0.200 m gives a tension of 8.40 N against a weight of 19.6 N. Tension equals weight only for an ideal massless pulley.

Do I use radians or degrees for angular acceleration?

Radians per second squared, always. Angular displacement, angular velocity, and angular acceleration are all radian quantities in AP Physics 1, and the equations linking them to linear motion, such as v equals r omega and tangential acceleration equals r alpha, are only valid in radians. Degrees appear separately, inside the torque equation, where the angle between the force and the position vector is a geometric angle and the exam's trigonometry table lists it in degrees.

Do I need to memorize rotational inertia formulas for AP Physics 1?

No. The Topic 5.4 boundary statement says students do not need to know the rotational inertia of extended rigid systems, as these will be provided within the exam. What you do need is to calculate the rotational inertia for systems of five or fewer objects arranged in a two-dimensional configuration, using the sum of m r squared, plus a qualitative sense of how moving mass away from the axis raises the rotational inertia.