Newton's Second Law in Rotational Form - AP Physics C Mechanics

Torque is $\tau = rF\sin\theta$ and rotational Newton's Second Law is $\tau_{\text{net}} = I\alpha$ , where $I$ resists changes in rotation. A wheel has $I=2.0\,\text{kg}\cdot\text{m}^2$ and experiences a net torque $\tau_{\text{net}}=6.0\,\text{N}\cdot\text{m}$ . Calculate the angular acceleration $\alpha$ .
Answer: $3.0\,\text{rad/s}^2$ This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and calculating angular acceleration from given torque and moment of inertia. Newton's Second Law in rotational form is expressed as τ = Iα, where net torque causes angular acceleration proportional to the moment of inertia. Rearranging this equation gives α = τ/I, allowing us to find angular acceleration when torque and moment of inertia are known. Choice B is correct because it accurately applies this relationship: α = 6.0 N·m / 2.0 kg·m² = 3.0 rad/s². Choice D is incorrect because it gives the torque value with units of N·m rather than the requested angular acceleration in rad/s², showing confusion about what quantity to calculate. To help students: Practice identifying what quantity is being asked for; emphasize unit analysis to verify answers; create analogies between F = ma and τ = Iα to reinforce conceptual understanding.
A time-varying torque given by $\tau(t) = kt^2$ , where $k$ is a positive constant, is applied to a rigid body with rotational inertia $I$ . The body starts from rest at $t=0$ . What is the angular acceleration of the body as a function of time?
Answer: $\frac{kt^2}{I}$ Newton's second law for rotation is $\tau_{net}(t) = I\alpha(t)$ . The angular acceleration is therefore $\alpha(t) = \frac{\tau_{net}(t)}{I}$ . Substituting the given expression for the torque, we find $\alpha(t) = \frac{kt^2}{I}$ .
A disc's torque is $\tau=rF\sin\theta$ and $\tau=I\alpha$ . With $r=0.40\,\text{m}$ , $F=20\,\text{N}$ tangential, and $I=2.0\,\text{kg}\cdot\text{m}^2$ , calculate $\alpha$ .
Answer: $4.0\,\text{rad/s}^2$ This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and its application to torque and rotational dynamics. For a tangential force, the torque calculation simplifies because θ = 90° and sin90° = 1, making τ = rF. This torque then determines the angular acceleration through Newton's Second Law in rotational form. Choice B is correct because τ = rF = 0.40 × 20 = 8.0 N·m (since the force is tangential), then α = τ/I = 8.0/2.0 = 4.0 rad/s². Choice C is incorrect because it likely uses the torque value as the angular acceleration, confusing the two different quantities. To help students: Emphasize that tangential means perpendicular to the radius, simplifying calculations; practice identifying when sinθ = 1; use unit analysis to distinguish between torque (N·m) and angular acceleration (rad/s²).
Two identical wheels are rolling without slipping down identical inclined planes. Wheel A is a solid disk and wheel B is a ring (hoop). Both wheels have the same mass $m$ and radius $R$ . If wheel A reaches the bottom of the incline with angular velocity $\omega_A$ , what is the angular velocity $\omega_B$ of wheel B when it reaches the bottom?
Answer: $\omega_B = \sqrt{\frac{3}{4}}\omega_A$ For rolling motion down an incline, energy conservation gives $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ . With $v = R\omega$ , this becomes $gh = \frac{1}{2}R^2\omega^2(1 + \frac{I}{mR^2})$ . For the disk, $I_A = \frac{1}{2}mR^2$ , so $\omega_A^2 = \frac{4gh}{3R^2}$ . For the ring, $I_B = mR^2$ , so $\omega_B^2 = \frac{2gh}{2R^2} = \frac{gh}{R^2}$ . Therefore, $\frac{\omega_B^2}{\omega_A^2} = \frac{3}{4}$ , giving $\omega_B = \sqrt{\frac{3}{4}}\omega_A$ .
A turntable (uniform disk) of mass $M$ and radius $R$ rotates freely about a vertical axis with angular velocity $\omega_0$ . A person of mass $m$ walks from the center to the edge of the turntable along a radial path. If the person then walks around the circumference with speed $v$ relative to the turntable, what is the angular velocity of the turntable?
Answer: $\frac{\frac{1}{2}M\omega_0 - \frac{mv}{R}}{\frac{1}{2}M + m}$ This problem tests conservation of angular momentum in a rotating system where mass distribution changes. When you see rotating objects with people or masses moving on them, immediately think about how angular momentum is conserved for the entire system. Initially, only the turntable rotates with angular momentum $L_i = I_{disk}\omega_0 = \frac{1}{2}MR^2\omega_0$ . When the person walks to the edge, they become part of the rotating system. The key insight is that the person's motion has two components: rotation with the turntable at angular velocity $\omega$ , plus additional motion around the circumference at speed $v$ relative to the turntable. The final angular momentum consists of the turntable rotating at $\omega$ plus the person's total angular momentum. The person's speed relative to the ground is $\omega R + v$ , so their angular momentum is $mR(\omega R + v) = mR^2\omega + mRv$ . The total final angular momentum is $L_f = \frac{1}{2}MR^2\omega + mR^2\omega + mRv$ . Setting $L_i = L_f$ and solving: $\frac{1}{2}MR^2\omega_0 = R^2\omega(\frac{1}{2}M + m) + mRv$ , which gives $\omega = \frac{\frac{1}{2}M\omega_0 - \frac{mv}{R}}{\frac{1}{2}M + m}$ . This is answer D. Choice A incorrectly separates the terms. Choice B adds instead of subtracts the velocity term, missing that relative motion opposes the turntable's rotation. Choice C uses incorrect moment of inertia ( $M$ instead of $\frac{1}{2}M$ ) for the disk. Remember: in rotating systems, carefully account for all contributions to angular momentum and use the correct moments of inertia for each object's geometry.
A thin, uniform rod of mass $M$ and length $L$ is pivoted at one end. It is held horizontally and released from rest. The rotational inertia of the rod about its end is $I = \frac{1}{3}ML^2$ . What is the initial angular acceleration of the rod immediately after release?
Answer: $\frac{3g}{2L}$ The torque is produced by the gravitational force acting at the rod's center of mass, which is a distance $L/2$ from the pivot. The torque is $\tau = F_{g}r_{\perp} = Mg(\frac{L}{2})$ . Using Newton's second law for rotation, $\tau = I\alpha$ , we have $Mg\frac{L}{2} = (\frac{1}{3}ML^2)\alpha$ . Solving for $\alpha$ gives $\alpha = \frac{MgL/2}{ML^2/3} = \frac{3g}{2L}$ .
A solid cylinder of mass $M$ and radius $R$ rolls without slipping down a rough inclined plane of angle $\theta$ . The cylinder is then placed on an identical inclined plane, but this time it slides without rolling (frictionless surface). What is the ratio of the acceleration when sliding to the acceleration when rolling?
Answer: $\frac{3}{2}$ When you encounter rolling motion problems, you need to analyze two scenarios: rolling (with rotational motion) versus sliding (pure translation). The key is applying Newton's second law while considering the constraint of rolling without slipping. For the rolling case, two forces act on the cylinder: the component of gravity down the incline ( $Mg\sin\theta$ ) and friction up the incline ( $f$ ). Newton's second law gives: $Mg\sin\theta - f = Ma_{roll}$ . The rolling constraint means $a = \alpha R$ , and for rotational motion: $fR = I\alpha = \frac{1}{2}MR^2 \cdot \frac{a}{R} = \frac{1}{2}MRa$ . Therefore, $f = \frac{1}{2}Ma_{roll}$ . Substituting back: $Mg\sin\theta - \frac{1}{2}Ma_{roll} = Ma_{roll}$ , which gives $a_{roll} = \frac{2g\sin\theta}{3}$ . For the sliding case, there's no friction or rotational motion. Only gravity's component acts: $Mg\sin\theta = Ma_{slide}$ , so $a_{slide} = g\sin\theta$ . The ratio is: $\frac{a_{slide}}{a_{roll}} = \frac{g\sin\theta}{\frac{2g\sin\theta}{3}} = \frac{3}{2}$ , confirming answer D. Answer A ( $\frac{2}{3}$ ) inverts this ratio. Answer B ( $1$ ) incorrectly assumes both accelerations are equal, ignoring rotational effects. Answer C ( $2$ ) might result from incorrectly doubling one acceleration or using wrong rotational inertia. Remember: rolling objects always accelerate slower than sliding ones because some gravitational energy goes into rotation. For solid cylinders, sliding acceleration is always $\frac{3}{2}$ times the rolling acceleration.
A rigid body with rotational inertia $I$ experiences a net torque $\tau$ , resulting in an angular acceleration $\alpha$ . If the mass of the body is doubled, with its shape and dimensions remaining identical, what is the new angular acceleration for the same net torque $\tau$ ?
Answer: $\frac{1}{2}\alpha$ Rotational inertia $I$ is directly proportional to mass $M$ (e.g., $I=kMR^2$ ). If the mass is doubled while shape and dimensions are unchanged, the new rotational inertia $I'$ will be $2I$ . From Newton's second law for rotation, $\alpha = \tau/I$ . The new angular acceleration $\alpha'$ will be $\alpha' = \tau/I' = \tau/(2I) = \frac{1}{2}(\tau/I) = \frac{1}{2}\alpha$ .
An Atwood machine consists of two blocks of mass $m_1$ and $m_2$ ( $m_1 > m_2$ ) connected by a light string over a solid disk pulley of mass $M$ and radius $R$ . The string does not slip, and the pulley has rotational inertia $I = \frac{1}{2}MR^2$ . What is the magnitude of the pulley's angular acceleration?
Answer: $\frac{(m_1-m_2)g}{(m_1+m_2+\frac{1}{2}M)R}$ The net torque on the pulley is $\tau_{net} = (T_1-T_2)R = I\alpha$ . The linear equations for the masses are $m_1g - T_1 = m_1a$ and $T_2 - m_2g = m_2a$ . Using $a=R\alpha$ , we solve for $T_1$ and $T_2$ and substitute into the torque equation. This yields $(m_1-m_2)g = (m_1+m_2+\frac{I}{R^2})a$ . Substituting $I=\frac{1}{2}MR^2$ and $a=R\alpha$ gives $\alpha = \frac{(m_1-m_2)g}{(m_1+m_2+\frac{1}{2}M)R}$ .
A uniform solid cylinder of mass $M$ and radius $R$ rolls without slipping down an incline that makes an angle $\theta$ with the horizontal. The rotational inertia of the cylinder is $I = \frac{1}{2}MR^2$ . What is the angular acceleration of the cylinder?
Answer: $\frac{2g \sin\theta}{3R}$ The linear equation of motion is $Mg\sin\theta - f = Ma$ , where $f$ is the static friction force. The torque equation about the center of mass is $\tau = fR = I\alpha$ . For rolling without slipping, $a = R\alpha$ . Substituting $f = I\alpha/R = Ia/R^2$ into the linear equation gives $Mg\sin\theta - \frac{Ia}{R^2} = Ma$ . Solving for $a$ yields $a = \frac{Mg\sin\theta}{M + I/R^2}$ . With $I = \frac{1}{2}MR^2$ , this becomes $a = \frac{g\sin\theta}{1 + 1/2} = \frac{2}{3}g\sin\theta$ . Finally, $\alpha = a/R = \frac{2g \sin\theta}{3R}$ .
A thin, uniform rod of mass $M$ and length $L$ is pivoted at one end and hangs vertically. It is displaced by a small angle $\theta$ from the vertical and released. What is the magnitude of its initial angular acceleration? The rotational inertia of the rod about its end is $I = \frac{1}{3}ML^2$ .
Answer: $\frac{3g \sin\theta}{2L}$ The torque is provided by the component of the gravitational force perpendicular to the rod, acting at the center of mass ( $L/2$ from the pivot). The torque is $\tau = (Mg\sin\theta)(L/2)$ . Using $\tau = I\alpha$ , we have $\frac{1}{2}MgL\sin\theta = (\frac{1}{3}ML^2)\alpha$ . Solving for $\alpha$ gives $\alpha = \frac{MgL\sin\theta/2}{ML^2/3} = \frac{3g\sin\theta}{2L}$ .
A uniform rod of length $L$ and mass $M$ is pivoted at its center. A force $F$ is applied perpendicularly to the rod at one end, and a force $2F$ is applied perpendicularly at the other end, in the opposite direction. The rotational inertia of the rod about its center is $I = \frac{1}{12}ML^2$ . What is the magnitude of the initial angular acceleration of the rod?
Answer: $\frac{18F}{ML}$ Both forces produce torques in the same rotational direction. The lever arm for each force is $L/2$ . The net torque is $\tau_{net} = F(L/2) + 2F(L/2) = \frac{3}{2}FL$ . Using $\tau_{net} = I\alpha$ , we have $\frac{3}{2}FL = (\frac{1}{12}ML^2)\alpha$ . Solving for $\alpha$ gives $\alpha = \frac{3FL/2}{ML^2/12} = \frac{18F}{ML}$ .
A solid disk and a hoop have the same mass $M$ and radius $R$ . They are both initially at rest and are free to rotate about a fixed, frictionless axle through their centers. The same constant net torque is applied to both. After one second, which of the following correctly compares their angular accelerations, $\alpha_{disk}$ and $\alpha_{hoop}$ ?
Answer: $\alpha_{disk} > \alpha_{hoop}$ because the disk has a smaller rotational inertia. According to Newton's second law for rotation, $\alpha = \tau_{net}/I$ . Since both objects experience the same net torque, the object with the smaller rotational inertia will have the greater angular acceleration. The rotational inertia of a solid disk is $I_{disk} = \frac{1}{2}MR^2$ , while that of a hoop is $I_{hoop} = MR^2$ . Thus, $I_{disk} < I_{hoop}$ , which means $\alpha_{disk} > \alpha_{hoop}$ .
A wheel's torque is $\tau=rF\sin\theta$ and also $\tau=I\alpha$ . If $\alpha=4.0\,\text{rad/s}^2$ and $I=0.75\,\text{kg}\cdot\text{m}^2$ , what is $\tau_{\text{net}}$ ?
Answer: $3.0\,\text{N}\cdot\text{m}$ This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and its application to torque and rotational dynamics. Newton's Second Law in rotational form states τ = Iα, where torque equals the product of moment of inertia and angular acceleration. This relationship allows us to calculate any of the three quantities when the other two are known. Choice B is correct because τ_net = Iα = 0.75 × 4.0 = 3.0 N·m, demonstrating proper application of the rotational form of Newton's Second Law. Choice C is incorrect because it has units of angular acceleration (rad/s²) rather than torque (N·m), showing confusion between the quantities in the equation. To help students: Practice identifying which quantity to solve for in τ = Iα; emphasize unit analysis to catch errors; create a comparison table showing linear (F = ma) and rotational (τ = Iα) forms of Newton's Second Law.
Torque is $\tau = rF\sin\theta$ and $\tau_{\text{net}} = I\alpha$ connects net torque to angular acceleration. A disc has $I=0.50\,\text{kg}\cdot\text{m}^2$ and two tangential forces act at the rim: $8\,\text{N}$ clockwise and $3\,\text{N}$ counterclockwise at $r=0.25\,\text{m}$ . What is $\alpha$ (clockwise positive)?
Answer: $2.5\,\text{rad/s}^2$ This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form with multiple torques and sign conventions. When multiple torques act, the net torque is the algebraic sum considering direction: τ_net = τ_clockwise - τ_counterclockwise = (0.25 m)(8 N) - (0.25 m)(3 N) = 2.0 - 0.75 = 1.25 N·m clockwise. Applying Newton's Second Law τ = Iα gives the angular acceleration. Choice C is correct because α = τ_net/I = 1.25 N·m / 0.50 kg·m² = 2.5 rad/s² in the clockwise (positive) direction. Choice A is incorrect because it has the wrong sign, suggesting counterclockwise rotation when the net torque is clockwise. To help students: Practice problems with multiple torques; establish clear sign conventions before solving; use free-body diagrams adapted for rotation showing all torques and their directions.
On a disc, torque satisfies $\tau=rF\sin\theta$ and $\tau=I\alpha$ . If $F$ stays constant and tangential, what happens to $\tau$ when $r$ doubles?
Answer: $\tau$ doubles This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and its application to torque and rotational dynamics. Torque (τ) is the rotational equivalent of force, calculated as τ = rFsinθ, where r is the lever arm and θ is the angle between force and lever arm. For a tangential force (θ = 90°), the torque equation simplifies to τ = rF since sin90° = 1. Choice A is correct because when r doubles while F remains constant and tangential, the torque τ = rF also doubles, demonstrating the direct proportionality between torque and lever arm. Choice B is incorrect because it suggests an inverse relationship, which would only occur if we were considering angular acceleration with changing moment of inertia. To help students: Use visual demonstrations showing how increasing the lever arm increases the rotational effect; emphasize that torque is directly proportional to both force and lever arm distance; practice problems where students identify which variables change and predict the effect on torque.
A disc has $\tau=rF\sin\theta$ and $\tau=I\alpha$ . A $10\,\text{N}$ force acts at $r=0.30\,\text{m}$ at $\theta=30^\circ$ ; $I=0.50\,\text{kg}\cdot\text{m}^2$ . Find $\alpha$ .
Answer: $3.0\,\text{rad/s}^2$ This question tests AP Physics C: Mechanics skills, specifically understanding Newton's Second Law in rotational form and its application to torque and rotational dynamics. This problem requires calculating torque using τ = rFsinθ with a non-perpendicular force, then applying τ = Iα to find angular acceleration. The angle θ is crucial because it determines the perpendicular component of the force that creates torque. Choice A is correct because τ = rFsinθ = 0.30 × 10 × sin30° = 0.30 × 10 × 0.5 = 1.5 N·m, then α = τ/I = 1.5/0.50 = 3.0 rad/s². Choice B is incorrect because it likely uses sin30° = 1 instead of 0.5, doubling the calculated torque and angular acceleration. To help students: Emphasize that only the perpendicular component of force creates torque; practice problems with various angles; use vector diagrams to visualize force components and identify the angle θ in the torque equation.
A flywheel with rotational inertia $I$ is rotating with angular velocity $\omega_0$ . A constant frictional torque $\tau_f$ is exerted on the axle, causing the flywheel to slow down and stop. The angular acceleration $\alpha$ of the flywheel is:
Answer: $\frac{\tau_f}{I}$ Newton's second law for rotation states that the net torque equals the rotational inertia times the angular acceleration: $\tau_{net} = I\alpha$ . In this case, the only torque is the frictional torque, so $\tau_{net} = \tau_f$ . Therefore, the magnitude of the angular acceleration is $\alpha = \frac{\tau_f}{I}$ . The acceleration is negative if $\omega_0$ is positive, but the question asks for the magnitude.
A uniform thin hoop of mass $M$ and radius $R$ is pivoted at a point on its rim. It is released from rest with its center of mass at the same height as the pivot. The rotational inertia of the hoop about its center is $MR^2$ . What is the initial angular acceleration of the hoop?
Answer: $\frac{g}{2R}$ First, find the rotational inertia about the pivot on the rim using the parallel axis theorem: $I_{pivot} = I_{cm} + Md^2 = MR^2 + MR^2 = 2MR^2$ . The torque is provided by gravity acting at the center of mass, which is a distance $d=R$ horizontally from the pivot. So, $\tau = MgR$ . Using Newton's second law for rotation, $\tau = I_{pivot}\alpha$ , we get $MgR = (2MR^2)\alpha$ . Solving for $\alpha$ gives $\alpha = \frac{MgR}{2MR^2} = \frac{g}{2R}$ .
A wheel with rotational inertia $I=3.0 \text{ kg} \cdot \text{m}^2$ is mounted on a fixed axle. A tangential force of $6.0 \text{ N}$ is applied to the wheel's rim, which has a radius of $0.50 \text{ m}$ . A constant frictional torque of $1.5 \text{ N} \cdot \text{m}$ opposes the motion. What is the magnitude of the wheel's angular acceleration?
Answer: $0.50 \text{ rad/s}^2$ First, calculate the torque from the applied force: $\tau_{app} = F \cdot R = (6.0 \text{ N})(0.50 \text{ m}) = 3.0 \text{ N} \cdot \text{m}$ . The net torque is the applied torque minus the opposing frictional torque: $\tau_{net} = \tau_{app} - \tau_f = 3.0 \text{ N} \cdot \text{m} - 1.5 \text{ N} \cdot \text{m} = 1.5 \text{ N} \cdot \text{m}$ . Now, use Newton's second law for rotation, $\tau_{net} = I\alpha$ , to find the angular acceleration: $\alpha = \frac{\tau_{net}}{I} = \frac{1.5 \text{ N} \cdot \text{m}}{3.0 \text{ kg} \cdot \text{m}^2} = 0.50 \text{ rad/s}^2$ .

Start this quiz