AP Physics 1 Unit 5: Rotational Motion — Free Review Games.

Torque is the tendency of a force to cause rotation about an axis: torque = rF*sin(theta).

Torque is measured in Newton-meters (N*m), though it is not the same as a Joule.

Angular velocity (omega) is the rate at which an object rotates, measured in radians per second.

Moment of inertia (I) is the rotational analog of mass, depending on both mass and how it is distributed relative to the axis of rotation.

Angular momentum is L = I*omega, the rotational analog of linear momentum.

Torque = rF*sin(90) = 0.5(40)(1) = 20 N*m.

A ring (I = MR^2) has all mass at radius R, giving it a larger moment of inertia than a solid disk (I = 1/2 MR^2).

The rotational form of Newton's Second Law is net torque = I*alpha, where alpha is angular acceleration.

With no external torque, angular momentum (L = I*omega) is conserved. Decreasing I (pulling arms in) increases omega.

Linear velocity of a point at distance r from the axis is v = r*omega.

Total KE = 1/2 mv^2 + 1/2 I*omega^2 = 1/2(4)(25) + 1/2(2/5)(4)(0.01)(50^2) = 50 + 1/2(0.016)(2500) = 50 + 20 = 70 J.

I = mr^2 = 2(1)^2 = 2 kg*m^2. L = I*omega = 2(3) = 6 kg*m^2/s.

For a uniform rod pivoted at one end, I = 1/3 ML^2 (derived using the parallel axis theorem from the center value of 1/12 ML^2).

alpha = torque/I = 10/0.5 = 20 rad/s^2.

Conservation of angular momentum: I1*omega1 = (I1+I2)*omega_f. 2(6) = (2+4)*omega_f. 12 = 6*omega_f. omega_f = 2 rad/s.

Curling the right-hand fingers in the direction of the rotation (counterclockwise) makes the thumb point out of the page, which is the direction assigned to the torque vector by convention. The choice 'Into the page' is wrong because that direction corresponds to a clockwise rotation, not counterclockwise. Students should remember that torque, like angular velocity, is a vector defined by the right-hand rule and points along the rotation axis, not in the plane of motion.

Angular velocity measures how fast an angular position changes with time, so its unit is radians divided by seconds. 'Meters per second' is wrong because that unit describes linear speed, not the rate of change of an angle. Recognizing that angular quantities use radians instead of meters helps distinguish rotational variables from their linear counterparts.

Angular momentum is defined as \(L = I\omega\), and since \(I\) has units of \(\text{kg·m}^2\) and \(\omega\) has units of \(1/\text{s}\), the product carries units of \(\text{kg·m}^2/\text{s}\). The unit '\(\text{kg·m/s}\)' is wrong because that is the unit of linear momentum, not angular momentum. Keeping the units of rotational and translational quantities separate prevents common mix-ups on the exam.

Rotational equilibrium is defined by a zero net torque, which means the object's angular momentum is not changing, though it may still be spinning at a constant rate. The option 'The net force is zero' describes translational equilibrium, a related but distinct condition that does not by itself guarantee zero net torque. Students should keep torque balance and force balance as two separate conditions when analyzing equilibrium problems.

Torque depends on the perpendicular component of the force relative to the position vector, which is captured by the sine of the angle between them, giving \(\tau = rF\sin\theta\). The expression '\(\tau = rF\cos\theta\)' is wrong because cosine gives the component of force along the position vector, which produces no rotational effect. Remembering that only the perpendicular force component contributes to torque is essential for correctly setting up torque problems.

Torque depends on the lever arm distance from the pivot, and since that distance is zero at the pivot itself, the torque produced is zero regardless of the force's strength. The option 'Maximum' is wrong because torque actually increases with distance from the pivot, so the maximum lever arm—not zero distance—produces maximum torque. This illustrates why moving a force farther from the axis of rotation, not increasing the force alone, is often the most effective way to increase torque.

Since one full rotation covers an angle of \(2\pi\) radians in a time \(T\), dividing the angle by the time gives \(\omega = \frac{2\pi}{T}\). The choice '\(\omega = \frac{T}{2\pi}\)' is wrong because it inverts the relationship, which would give units of seconds squared per radian rather than radians per second. This relationship mirrors the linear analog \(v = \frac{2\pi r}{T}\) and is essential for converting between period and angular velocity.

Angular acceleration is defined as \(\alpha = \frac{d\omega}{dt}\), the rate at which angular velocity itself changes over time. The option 'The rate of change of angular position' is wrong because that description defines angular velocity, not angular acceleration. Keeping the hierarchy of position, velocity, and acceleration consistent between linear and rotational motion helps avoid confusing these related but distinct quantities.

Moment of inertia depends on how far each bit of mass is located from the axis of rotation, so objects with the same total mass but different shapes or mass distributions will have different moments of inertia. The option 'different angular velocities' is wrong because moment of inertia is a property of an object's mass distribution and does not depend on how fast it happens to be spinning. This is why a hollow cylinder resists changes in rotation more than a solid cylinder of the same mass and radius.

Rolling without slipping requires that the linear speed of the object's center equal the product of the radius and the angular velocity, expressed as \(v = r\omega\), so that the contact point momentarily has zero velocity relative to the surface. The choice '\(v = r/\omega\)' is wrong because dividing rather than multiplying gives incorrect units and does not represent the physical constraint of rolling motion. This relationship is the key link connecting translational and rotational kinematics for rolling objects.

Rotational kinetic energy is the rotational analog of translational kinetic energy, replacing mass with moment of inertia and linear velocity with angular velocity, giving \(\frac{1}{2}I\omega^2\). The expression '\(\frac{1}{2}mv^2\)' is wrong because it describes translational kinetic energy, not rotational kinetic energy. Recognizing the parallel structure between linear and rotational energy formulas makes it easier to remember both.

Angular momentum is conserved whenever the net external torque acting on a system is zero, just as linear momentum is conserved when net external force is zero. The option 'The net external force on the system is zero' is wrong because that condition governs conservation of linear momentum, not angular momentum, and the two are not automatically linked. This distinction is critical for correctly applying conservation laws to rotating systems like spinning skaters or colliding disks.

The lever arm is specifically the perpendicular distance from the pivot to the line along which the force acts, which is why torque depends on \(\sin\theta\) when the force is not applied perpendicular to the position vector. The option 'the distance from the pivot to the point where the force is applied' is wrong because that is simply \(r\), which only equals the lever arm when the force is applied perpendicular to the position vector. Correctly identifying the lever arm, rather than just the distance to the application point, is essential for accurate torque calculations.

Using \(\tau = rF\sin\theta = (0.8)(30)\sin(60^\circ) \approx 20.8\ \text{N·m}\) correctly accounts for only the perpendicular component of the force contributing to rotation. The choice '\(24\ \text{N·m}\)' is wrong because it comes from simply multiplying \(r\) and \(F\) without including the sine of the angle, ignoring that the force is not applied perpendicular to the lever. Always multiplying by \(\sin\theta\) when the angle is not \(90^\circ\) is essential for accurate torque calculations.

The hoop has \(I = MR^2\) because all its mass sits at radius \(R\), while the disk has \(I = \frac{1}{2}MR^2\) because its mass is spread from the center outward, giving it a smaller effective average radius. The option 'They are equal' is wrong because mass distribution matters greatly for rotational inertia even when total mass and outer radius are identical. This comparison illustrates the general principle that concentrating mass farther from the axis increases moment of inertia.

Using the rotational form of Newton's second law, \(\tau = I\alpha = (4)(3) = 12\ \text{N·m}\), gives the required net torque directly. The choice '\(1.33\ \text{N·m}\)' is wrong because it results from dividing rather than multiplying \(I\) and \(\alpha\), reversing the correct relationship. This formula is the rotational parallel to \(F = ma\) and should be applied the same way, multiplying inertia by acceleration.

For a point mass moving in a circle, angular momentum is $L = mvr = (2)(4)(0.5) = 4\ \text{kg·m}^2/\text{s}$, combining its linear momentum with its distance from the center. The choice '\(8\ \text{kg·m}^2/\text{s}\)' is wrong because it omits the correct multiplication by the 0.5 m radius, effectively treating the radius as 1 m. This formula, $L=mvr$, is the go-to expression for angular momentum whenever a mass moves in a circular path.

Since moment of inertia for a point mass is \(I = mr^2\), doubling \(r\) increases \(I\) by a factor of \(2^2 = 4\) because the radius enters the formula squared. The choice '2' is wrong because it treats the relationship as linear in \(r\) rather than quadratic. This squared dependence on distance is why moving mass even a little farther from an axis has a large effect on rotational inertia.

The parallel axis theorem states \(I = I_{cm} + Md^2\), and with \(d = L/2\) this gives \(I = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{3}ML^2\). The choice '\(\frac{1}{12}ML^2\)' is wrong because it is only the moment of inertia about the center of mass, without adding the extra term accounting for the shifted axis. Whenever an axis is not through the center of mass, the parallel axis theorem must be applied to get the correct moment of inertia.

Using $KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(2)(5)^2 = 25\ \text{J}$ correctly squares the angular velocity before multiplying by the moment of inertia. The choice '\(10\ \text{J}\)' is wrong because it comes from computing \(I\omega\) instead of \(\frac{1}{2}I\omega^2\), missing both the squaring and the factor of one-half. Always square the angular velocity term, just as linear kinetic energy squares linear velocity.

Rearranging \(v = r\omega\) gives \(\omega = v/r = 6/0.3 = 20\ \text{rad/s}\), correctly dividing the linear speed by the radius. The choice '\(1.8\ \text{rad/s}\)' is wrong because it results from multiplying \(v\) and \(r\) instead of dividing them. This conversion between linear and angular velocity is one of the most frequently tested relationships in rotational motion problems.

The right-side force produces a clockwise torque of \(10 \times 0.4 = 4\ \text{N·m}\), while the left-side force produces a counterclockwise torque of \(15 \times 0.2 = 3\ \text{N·m}\), so the net torque is \(4 - 3 = 1\ \text{N·m}\) clockwise. The choice '\(7\ \text{N·m}\) clockwise' is wrong because it incorrectly adds the two torque magnitudes instead of recognizing they act in opposite rotational directions and should be subtracted. When combining multiple torques, always assign a positive or negative sign based on rotational direction before summing.

Balancing requires equal torques on each side, so \(450 \times d = 300 \times 1.5\), giving \(d = 450/450 = 1.0\ \text{m}\). The choice '2.25 m' is wrong because it comes from multiplying the two weights and distances incorrectly rather than solving the torque balance equation for the unknown distance. This kind of two-sided torque balance is the classic seesaw setup used to test static rotational equilibrium.

Since no external torque acts on the system, angular momentum is conserved: \(I_1\omega_1 = I_2\omega_2\), so \(\omega_2 = (8)(2)/4 = 4\ \text{rad/s}\). The choice '\(1\ \text{rad/s}\)' is wrong because it incorrectly assumes angular velocity decreases when moment of inertia decreases, when conservation of \(L\) actually requires it to increase. This inverse relationship between \(I\) and \(\omega\) under conserved angular momentum is the same principle behind a spinning skater speeding up when pulling in her arms.

Rotational work is given by \(W = \tau\theta = (5)(3) = 15\ \text{J}\), the rotational analog of \(W = F d\) in linear motion. The choice '\(1.67\ \text{J}\)' is wrong because it divides torque by angular displacement instead of multiplying them. This formula lets you connect torque and angular displacement directly to energy transferred, useful for rotational work-energy problems.

Since \(L = I\omega\) and both wheels share the same \(\omega\), the angular momentum scales directly with moment of inertia, so \(L_1 = 2L_2\). The choice 'Both wheels have equal angular momentum' is wrong because it ignores that angular momentum depends on \(I\), not just \(\omega\), and the wheels have different moments of inertia. This proportionality is a direct consequence of the definition \(L = I\omega\) and should be applied whenever comparing spinning objects.

Each mass is 1 m from the center, so each contributes \(I = mr^2 = 3(1)^2 = 3\ \text{kg·m}^2\), and summing both gives a total of \(6\ \text{kg·m}^2\). The choice '\(12\ \text{kg·m}^2\)' is wrong because it likely uses the full rod length of 2 m as the radius for each mass instead of the correct 1 m distance from the center. When masses are placed symmetrically on a rod, remember to use the distance from the axis, not the total rod length, for each mass's contribution.

The 200 N weight produces a torque of \(200 \times 2 = 400\ \text{N·m}\) about the pivot, so an upward force \(F\) at 1 m must produce an equal opposing torque: \(F \times 1 = 400\), giving \(F = 400\ \text{N}\). The choice '200 N' is wrong because it fails to account for the shorter lever arm on the left side, which requires a larger force to produce the same torque as the weight on the longer right side. When lever arms are unequal, the applied force must scale inversely with distance to balance the torque.

At the point of contact, the linear speeds of the two gears must match, so \(r_A\omega_A = r_B\omega_B\), giving \(\omega_B = (0.1)(30)/0.3 = 10\ \text{rad/s}\). The choice '30 rad/s' is wrong because it assumes the gears share the same angular velocity, ignoring that meshed gears of different radii must have different angular velocities to keep their contact speeds equal. This inverse relationship between radius and angular velocity for meshed gears is a common application of rolling-contact constraints.

With \(\vec{r}\) along \(+x\) and \(\vec{F}\) along \(-y\), the cross product \(\hat{x} \times (-\hat{y}) = -\hat{z}\) points into the page, consistent with the force rotating the point clockwise as viewed. The choice 'Out of the page' is wrong because that direction corresponds to a counterclockwise rotation, which is the opposite sense produced by a downward force to the right of the pivot. Practicing the cross product with the right-hand rule ensures correct torque directions in more complex multi-force problems.

For a solid cylinder, \(I = \frac{1}{2}mr^2\), so rotational KE equals \(\frac{1}{4}mv^2\) while translational KE equals \(\frac{1}{2}mv^2\), making the rotational fraction \(\frac{1/4}{1/4+1/2} = \frac{1}{3}\) of the total. The choice '\(1/2\)' is wrong because it would only apply if rotational and translational kinetic energy were exactly equal, which is not the case for a solid cylinder's specific moment of inertia. This kind of energy-fraction calculation depends entirely on the object's shape through its moment of inertia coefficient.

Just as net force equals the rate of change of linear momentum, net torque equals the rate of change of angular momentum, expressed as $\tau_{net} = \frac{dL}{dt}$. The choice '$\tau_{net} = \frac{d\omega}{dt}$' is wrong because that expression equals angular acceleration, not torque, unless it is multiplied by a constant moment of inertia. This torque-angular momentum relationship is the rotational analog of Newton's second law in impulse-momentum form and underlies conservation of angular momentum when torque is zero.

Since both forces act at the same distance of 3 m from the pivot, balancing the torques \(F \times 3 = 40 \times 3\) requires \(F = 40\ \text{N}\), matching the hanging weight exactly. The choice '20 N' is wrong because it would only balance the torque if the applied force acted at twice the distance of the hanging weight, which is not the case here since both lever arms are equal. When two lever arms are equal, the balancing force must equal the opposing weight, a useful shortcut for symmetric torque problems.

With no external torque acting on the skater, angular momentum is conserved: \(I_1\omega_1 = I_2\omega_2\), so \(\omega_2 = (5)(3)/2 = 7.5\ \text{rad/s}\). The choice '1.2 rad/s' is wrong because it incorrectly divides \(I_2\) by \(I_1\omega_1\) rather than solving the conservation equation properly for \(\omega_2\). This numeric case is the quantitative version of the qualitative skater-spin principle: decreasing \(I\) must increase \(\omega\) to keep \(L\) constant.

The solid sphere has \(I = \frac{2}{5}MR^2\) compared to the hollow shell's \(I = \frac{2}{3}MR^2\), so for the same drop in potential energy, less of the solid sphere's energy is diverted into rotational kinetic energy, leaving more for translational speed. The option 'They arrive at the same time because mass and radius are equal' is wrong because the moment of inertia coefficient, not just mass and radius, determines how energy splits between rotation and translation. This principle explains why objects with smaller rotational inertia coefficients consistently win races down inclines regardless of their mass or radius.

Initial angular momentum is \(L = I\omega = (200)(2) = 400\ \text{kg·m}^2/\text{s}\), and after the child moves to the edge the total moment of inertia becomes \(200 + 40(2)^2 = 360\ \text{kg·m}^2\), so \(\omega_f = 400/360 \approx 1.11\ \text{rad/s}\). The choice '2 rad/s' is wrong because it ignores that the child's added moment of inertia at the edge increases the system's total \(I\), which must decrease \(\omega\) to conserve angular momentum. This problem shows that adding mass farther from the axis, not just changing shape, also conserves angular momentum by slowing rotation.

The angular acceleration is \(\alpha = \tau/I = 2\ \text{rad/s}^2\), so \(\omega = \alpha t = 8\ \text{rad/s}\), and the angular momentum gained equals the angular impulse \(\tau t = 6 \times 4 = 24\ \text{kg·m}^2/\text{s}\), which matches \(L = I\omega = (3)(8) = 24\). The choice '\(\omega = 2\ \text{rad/s}\), \(L = 6\ \text{kg·m}^2/\text{s}\)' is wrong because it uses the angular acceleration value directly as the final angular velocity, forgetting to multiply by the 4 s time interval. This problem shows that angular impulse \(\tau t\) and \(\Delta L = I\Delta\omega\) must give the same numerical result, a good way to check your work.

Since \(\tau = I\alpha\) and both objects share the same \(\alpha\), torque scales directly with moment of inertia, so \(\tau_B/\tau_A = I_B/I_A = 6/2 = 3\). The choice 'B requires the same torque as A' is wrong because it ignores that a larger moment of inertia requires proportionally more torque to achieve the identical angular acceleration. This proportional relationship between torque and moment of inertia at fixed angular acceleration mirrors \(F=ma\) in linear dynamics.

The sliding block converts all of its gravitational potential energy into translational kinetic energy, giving \(v = \sqrt{2gh}\), whereas the rolling sphere must divide its energy between translational and rotational kinetic energy, resulting in a smaller final speed of \(v = \sqrt{\frac{10}{7}gh}\). The option 'They arrive with equal speed because energy is conserved in both cases' is wrong because energy conservation alone does not guarantee equal speeds, since the two systems store the converted energy in different forms. This comparison highlights that having a rotational degree of freedom always reduces final translational speed compared to pure sliding, even though total energy is conserved in both scenarios.

For the falling block, Newton's second law gives \(mg - T = ma\) since gravity exceeds tension as the block accelerates downward, and for the pulley the string tension provides the torque that causes angular acceleration, giving \(Tr = I\alpha\). The choice '\(Tr = I\omega\)' is wrong because torque produces a change in angular velocity over time, meaning it must be set equal to \(I\alpha\), not directly to the instantaneous angular velocity \(\omega\). Coupling a linear Newton's second law equation for the hanging mass with a rotational one for the pulley, connected through \(a = r\alpha\), is the standard method for solving massive-pulley Atwood problems.

Since angular momentum $L = mvr$ is conserved with no external torque about the planet, $v_{perigee}r_{perigee} = v_{apogee}r_{apogee}$, and because $r_{perigee}$ is half of $r_{apogee}$, the speed at perigee must be twice the speed at apogee to keep the product constant. The choice 'The same speed at both points' is wrong because it would violate conservation of angular momentum unless the orbit were perfectly circular, which it is not in this elliptical case. This inverse relationship between orbital speed and distance from the central body is a direct consequence of angular momentum conservation, applicable to any object moving under a central force.

The initial total angular momentum is the platform's $L_{platform} = I\omega = 12\ \text{kg·m}^2/\text{s}$ plus the ball's $L_{ball} = mvr = (1)(8)(2) = 16\ \text{kg·m}^2/\text{s}$, giving a total of \(28\ \text{kg·m}^2/\text{s}\); dividing by the new moment of inertia \(4 + 1(2)^2 = 8\ \text{kg·m}^2\) gives \(\omega_f = 3.5\ \text{rad/s}\). The choice '3 rad/s' is wrong because it ignores the additional angular momentum contributed by the incoming ball, treating the collision as if only the platform's original momentum mattered. Inelastic rotational collisions require adding the angular momenta of all colliding objects before dividing by the combined final moment of inertia.

Since \(\tau = Fr\) and \(\alpha = \tau/I\), doubling the radius while keeping the force and the disk's moment of inertia constant doubles the torque and therefore doubles the angular acceleration, giving \(\alpha_2 = 2\alpha_1\). The choice '\(\alpha_2 = 4\alpha_1\)' is wrong because it would require the force itself, not just the radius, to scale with \(r\), which is not the case here. This shows that increasing the lever arm at constant force is a linear, not quadratic, way to increase torque and angular acceleration.

Although angular momentum \(L = I\omega\) remains constant since no external torque acts on her, rotational kinetic energy \(\frac{1}{2}I\omega^2\) increases because the skater performs internal muscular work pulling her arms inward against the outward tendency of the rotating mass. The choice 'Her rotational kinetic energy stays constant because angular momentum is conserved' is wrong because conservation of \(L\) does not imply conservation of kinetic energy, since \(I\) and \(\omega\) change in a way that increases the \(\omega^2\) term faster than \(I\) decreases. This distinction between conserved angular momentum and non-conserved kinetic energy in internal-work situations is a subtle but important exam concept.

The first force produces a clockwise torque of \(20 \times 3 = 60\ \text{N·m}\), and the second force produces a counterclockwise torque of \(30 \times 1 \times \sin(30^\circ) = 15\ \text{N·m}\), leaving a net clockwise torque of \(60 - 15 = 45\ \text{N·m}\) that a third force must cancel by supplying \(45\ \text{N·m}\) counterclockwise. The choice '\(60\ \text{N·m}\) counterclockwise' is wrong because it ignores the partial cancellation already provided by the second force's counterclockwise torque, overcounting the torque that must still be balanced. Multi-force equilibrium problems require summing all torques with correct signs before determining what a final force must contribute.