Newton's Second Law in Rotational Form - College Physics

A flywheel with moment of inertia $I = 0.5$ kg⋅m $^2$ is initially rotating at $10$ rad/s. A constant braking torque is applied for $4$ seconds, after which the flywheel rotates at $2$ rad/s in the same direction. During the next $2$ seconds, the braking torque is removed and a driving torque of $3$ N⋅m is applied. What is the angular acceleration during this second phase?
Answer: $\alpha = 6$ rad/s $^2$ During the second phase, only the driving torque of $3$ N⋅m acts on the flywheel. Using Newton's second law in rotational form: $\tau = I\alpha$ , so $3 = 0.5 \times \alpha$ , giving $\alpha = 6$ rad/s $^2$ . The information about the first phase (braking) is irrelevant for finding the acceleration during the second phase - it only establishes the initial conditions for the second phase. Choice A uses the wrong torque value. Choice B confuses the time interval with the calculation. Choice D incorrectly includes effects from the first phase.
A hollow cylinder (thin cylindrical shell) of mass $m$ and radius $R$ is rolling without slipping on a horizontal surface. A constant horizontal force $F$ is applied to the center of mass. What is the angular acceleration of the cylinder?
Answer: $\frac{F}{2mR}$ When you encounter rolling motion problems with applied forces, you need to analyze both translational and rotational motion simultaneously, using the no-slip condition to connect them. For a rolling cylinder, apply Newton's second law for translation and rotation. The applied force $F$ causes linear acceleration $a = F/m$ . However, the key insight is that friction at the contact point also affects the motion. When force $F$ is applied horizontally, static friction $f$ acts backward at the contact point to maintain the rolling condition. For translation: $F - f = ma$ For rotation about the center: $fR = I\alpha$ The moment of inertia for a thin cylindrical shell is $I = mR^2$ , so $fR = mR^2\alpha$ . The no-slip condition requires $a = R\alpha$ , which gives us $\alpha = a/R$ . Substituting: $fR = mR^2(a/R) = mRa$ , so $f = ma$ . From the translational equation: $F - ma = ma$ , therefore $F = 2ma$ and $a = F/(2m)$ . Finally: $\alpha = a/R = \frac{F}{2mR}$ Choice A ( $\frac{F}{mR}$ ) ignores friction entirely, treating this as pure rotation about a fixed axis. Choice C ( $\frac{2F}{mR}$ ) incorrectly doubles the effect of the applied force. Choice D ( $\frac{F}{4mR}$ ) likely comes from using the wrong moment of inertia or misapplying the rolling condition. Remember: in rolling motion problems, always account for friction forces and use the constraint $a = R\alpha$ to connect linear and angular motion.
A uniform rod of mass $m$ and length $L$ is pivoted at one end and hangs vertically in equilibrium. A horizontal force $F$ is applied perpendicular to the rod at a distance $d$ from the pivot point. What is the initial angular acceleration of the rod?
Answer: $\frac{3Fd}{mL^2}$ When you encounter rotational motion problems involving forces and pivots, you need to apply Newton's second law for rotation: $\tau = I\alpha$ , where torque equals moment of inertia times angular acceleration. To find the initial angular acceleration, you must calculate the net torque about the pivot. The applied horizontal force $F$ creates a torque $\tau_F = Fd$ (force times perpendicular distance). Since the rod starts in equilibrium, gravity's torque is initially zero because the gravitational force acts through the pivot point when the rod hangs vertically. Therefore, the net torque is simply $Fd$ . For a uniform rod pivoted at one end, the moment of inertia is $I = \frac{1}{3}mL^2$ . Using $\tau = I\alpha$ : $Fd = \frac{1}{3}mL^2 \alpha$ $\alpha = \frac{3Fd}{mL^2}$ This confirms answer B is correct. Answer A $\frac{Fd}{mL^2}$ incorrectly uses $I = mL^2$ instead of the proper $I = \frac{1}{3}mL^2$ for a rod pivoted at its end. Answer C $\frac{Fd}{\frac{1}{3}mL^2}$ appears to use the correct moment of inertia but presents it incorrectly—this actually equals $\frac{3Fd}{mL^2}$ , making it equivalent to B but written confusingly. Answer D incorrectly includes gravitational torque, but gravity produces no initial torque when the rod hangs vertically. Remember: always identify the correct moment of inertia formula for the specific geometry and pivot location. For rods, $I = \frac{1}{12}mL^2$ (center) versus $I = \frac{1}{3}mL^2$ (end) makes a crucial difference.
A uniform solid sphere of mass $m$ and radius $r$ rolls without slipping down an inclined plane of angle $\theta$ . What is the linear acceleration of the sphere's center of mass?
Answer: $\frac{5}{7}g\sin\theta$ When you encounter a rolling object on an inclined plane, you're dealing with both translational and rotational motion simultaneously. The key insight is that rolling without slipping creates a constraint that links these two types of motion, and you must apply both Newton's second law and rotational dynamics. For a sphere rolling down an incline, two forces act on it: gravitational force ( $mg$ ) and friction. The component $mg\sin\theta$ pulls the sphere down the incline, while friction $f$ acts up the incline, causing rotation. Using Newton's second law: $mg\sin\theta - f = ma$ , where $a$ is the linear acceleration. For rotation about the center of mass: $fr = I\alpha$ , where $I = \frac{2}{5}mr^2$ for a solid sphere. The rolling constraint gives us $a = r\alpha$ , so $\alpha = \frac{a}{r}$ . Substituting into the rotational equation: $f = \frac{I\alpha}{r} = \frac{2ma}{5}$ Plugging this back into the force equation: $mg\sin\theta - \frac{2ma}{5} = ma$ Solving for $a$ : $mg\sin\theta = ma + \frac{2ma}{5} = \frac{7ma}{5}$ Therefore: $a = \frac{5g\sin\theta}{7}$ Choice A ( $g\sin\theta$ ) ignores rotational inertia entirely. Choice B ( $\frac{3}{5}g\sin\theta$ ) uses an incorrect moment of inertia. Choice D ( $\frac{2}{3}g\sin\theta$ ) applies to rolling cylinders, not spheres. Remember: rolling problems always require considering both translational and rotational motion. The moment of inertia determines how much the acceleration differs from the frictionless case.
A wheel of radius $R$ and moment of inertia $I$ about its center is subjected to three simultaneous torques: $\tau_1 = 15$ N⋅m (clockwise), $\tau_2 = 8$ N⋅m (counterclockwise), and a friction torque $\tau_f = 2$ N⋅m opposing motion. If the wheel is initially rotating counterclockwise with angular velocity $\omega_0 = 5$ rad/s, what is the correct expression for the wheel's angular acceleration?
Answer: $\alpha = \frac{-15 + 8 - 2}{I} = \frac{-9}{I}$ rad/s $^2$ Using $\tau_{net} = I\alpha$ with counterclockwise as positive: $\tau_1 = -15$ N⋅m (clockwise), $\tau_2 = +8$ N⋅m (counterclockwise), and $\tau_f = -2$ N⋅m (opposes current counterclockwise motion). Therefore: $\tau_{net} = -15 + 8 - 2 = -9$ N⋅m, giving $\alpha = \frac{-9}{I}$ rad/s $^2$ . The negative sign indicates clockwise angular acceleration. Choice A incorrectly treats friction as always negative. Choice C adds all torques positively. Choice D incorrectly treats friction as aiding motion.
A rotating platform (moment of inertia $I_0$ ) spins at angular velocity $\omega_0$ . A person of mass $m$ drops vertically onto the platform at distance $r$ from the center and immediately begins rotating with the platform. If we want to find the angular acceleration of the platform during the brief time interval when the person is landing, which form of Newton's second law should be applied?
Answer: Apply $\tau_{external} = I_{total}\alpha$ where $I_{total} = I_0 + mr^2$ and $\tau_{external}$ includes the impulse torque from the person During the landing process, we must consider the system as it transitions from just the platform to platform plus person. The person exerts torques on the platform through normal and friction forces at the contact point. Newton's second law in rotational form requires $\tau_{external} = I\alpha$ where $I$ is the moment of inertia of the system being analyzed. Since the person becomes part of the system, $I_{total} = I_0 + mr^2$ . Choice B incorrectly excludes the person from the system. Choice C uses conservation of angular momentum, which applies after the interaction, not during. Choice D oversimplifies the interaction forces.
A pulley system consists of a uniform solid disk of mass $M$ and radius $R$ with two masses $m_1$ and $m_2$ (where $m_1 > m_2$ ) hanging on opposite sides. Assuming the string doesn't slip on the pulley, what is the magnitude of angular acceleration of the pulley?
Answer: $\frac{(m_1 - m_2)g}{(m_1 + m_2 + M/2)R}$ When you encounter a pulley system with masses on both sides, you're dealing with a rotational dynamics problem that requires analyzing both the linear motion of the masses and the rotational motion of the pulley. Start by applying Newton's second law to each component. For the masses, the net force drives their acceleration: $m_1g - T_1 = m_1a$ and $T_2 - m_2g = m_2a$ , where $T_1$ and $T_2$ are the tensions and $a$ is the linear acceleration. Since the string doesn't slip, the constraint relationship gives us $a = \alpha R$ , where $\alpha$ is the angular acceleration. For the pulley, apply the rotational version of Newton's second law: $\tau = I\alpha$ . The net torque comes from the tension difference: $(T_1 - T_2)R = I\alpha$ . For a uniform solid disk, $I = \frac{1}{2}MR^2$ . The key insight is that the string not slipping creates equal tensions on each side, so $T_1 = T_2 = T$ . Combining the equations: $m_1g - T = m_1\alpha R$ and $T - m_2g = m_2\alpha R$ . Adding these eliminates $T$ : $(m_1 - m_2)g = (m_1 + m_2)\alpha R$ . Including the pulley's rotational inertia: $(m_1 - m_2)g = (m_1 + m_2 + \frac{M}{2})\alpha R$ . Therefore, $\alpha = \frac{(m_1 - m_2)g}{(m_1 + m_2 + M/2)R}$ , which is answer A. Answer B incorrectly uses $M$ instead of $M/2$ , forgetting that $I = \frac{1}{2}MR^2$ for a solid disk. Answer C completely ignores the pulley's inertia. Answer D uses the wrong force term $(m_1 + m_2)g$ instead of the net force $(m_1 - m_2)g$ . Remember: always include the rotational inertia of the pulley and use the correct moment of inertia formula for the given geometry.
A uniform disk of mass $M$ and radius $R$ is mounted horizontally and can rotate freely about a vertical axis through its center. A small object of mass $m$ slides across the disk along a diameter. As the object moves from the center to the edge, what happens to the angular velocity of the disk-object system?
Answer: It decreases because the moment of inertia of the system increases When you encounter problems involving rotating systems where mass redistributes, conservation of angular momentum is your key tool. Since the disk can rotate freely with no external torques acting on the system, angular momentum $L = I\omega$ must remain constant throughout the motion. Initially, when the object is at the center, the system's moment of inertia is just that of the disk: $I_i = \frac{1}{2}MR^2$ . As the object slides outward to radius $r$ , it contributes $mr^2$ to the total moment of inertia, making $I_f = \frac{1}{2}MR^2 + mr^2$ . Since the object moves from center to edge, the final moment of inertia is definitely larger than the initial value. With angular momentum conserved ( $L_i = L_f$ ), we have $I_i\omega_i = I_f\omega_f$ . Since $I_f > I_i$ , the angular velocity must decrease: $\omega_f < \omega_i$ . This confirms answer B is correct. Looking at the wrong answers: A incorrectly assumes angular momentum increases, but no external torques act on the system. C misses that while angular momentum stays constant, angular velocity changes because moment of inertia changes. D mentions friction, but even if friction exists between the object and disk, these are internal forces that don't change the system's total angular momentum. Remember this pattern: in isolated rotating systems, when mass moves farther from the rotation axis, angular velocity decreases to conserve angular momentum. Think of a figure skater extending their arms during a spin.
A compound pendulum consists of a uniform rod of length $L$ and mass $M$ pivoted at distance $d$ from its center. A small mass $m$ is attached at one end of the rod. For small oscillations, the equation of motion is $\tau = -I\alpha$ where $\alpha = \ddot{\theta}$ . What is the correct expression for the restoring torque $\tau$ about the pivot?
Answer: $\tau = -[Mgd + mg(L/2 \pm d)]\sin\theta$ The restoring torque comes from the gravitational forces on both the rod and the attached mass. For the rod: its center of mass is at distance $d$ from the pivot, so $\tau_{rod} = -Mgd\sin\theta$ . For the attached mass: it's at distance $(L/2 \pm d)$ from the pivot (depending on which end and which side of center the pivot is), so $\tau_{mass} = -mg(L/2 \pm d)\sin\theta$ . Total: $\tau = -[Mgd + mg(L/2 \pm d)]\sin\theta$ . Choice A treats both masses as if located at the pivot distance $d$ . Choice C ignores the pivot location. Choice D has redundant $\sin\theta$ terms and incorrect structure.
A uniform solid disk of mass $M$ and radius $R$ is initially at rest. A constant tangential force $F$ is applied at the rim for time $t$ . What is the angular velocity of the disk at the end of this time interval?
Answer: $\frac{2Ft}{MR}$ This problem tests rotational dynamics, specifically the relationship between torque, angular momentum, and angular velocity. When you see a rotating object with forces applied at the rim, think about torque and rotational motion equations. To find the angular velocity, start with the rotational analog of Newton's second law. The tangential force $F$ applied at radius $R$ creates a torque $\tau = FR$ . This torque causes angular acceleration $\alpha = \frac{\tau}{I}$ , where $I$ is the moment of inertia. For a uniform solid disk, the moment of inertia is $I = \frac{1}{2}MR^2$ . Therefore, the angular acceleration is: $\alpha = \frac{FR}{\frac{1}{2}MR^2} = \frac{2F}{MR}$ Since the disk starts from rest and experiences constant angular acceleration for time $t$ , the final angular velocity is: $\omega = \alpha t = \frac{2F}{MR} \cdot t = \frac{2Ft}{MR}$ Choice A ( $\frac{Ft}{MR}$ ) incorrectly uses $I = MR^2$ instead of the correct $I = \frac{1}{2}MR^2$ for a solid disk. Choice C ( $\frac{Ft}{2MR}$ ) makes the opposite error, using $I = 2MR^2$ . Choice D ( $\frac{FtR}{M}$ ) treats this like linear motion, incorrectly applying $F = ma$ without considering rotational inertia. Study tip: Always identify the correct moment of inertia formula for the specific geometry (solid disk, hollow cylinder, rod, etc.) before solving rotational dynamics problems. The factor of $\frac{1}{2}$ for solid disks is frequently tested.
A solid sphere of radius $R$ and mass $M$ rolls without slipping down an inclined plane of angle $\theta$ . At the point of contact, the friction force is $f$ . Which equation correctly applies Newton's second law in rotational form about the sphere's center of mass?
Answer: $f \cdot R = I\alpha$ When applying Newton's second law in rotational form about the center of mass, only forces that create torques about that point contribute. The weight $Mg$ acts at the center of mass, so it creates zero torque about the center of mass. The normal force acts through the center of mass (perpendicular to the line from contact point to center), so it also creates zero torque. Only the friction force $f$ , acting at distance $R$ from the center, creates a torque: $\tau = fR = I\alpha$ . Choice A incorrectly includes weight torque. Choice C and D incorrectly combine forces before calculating torque.
A thin ring of mass $m$ and radius $R$ is initially at rest. A tangential force $F$ is applied at a point on the rim for a time $t$ . What is the final angular velocity of the ring?
Answer: $\frac{Ft}{mR}$ When you encounter rotational motion problems involving forces and time, you're dealing with angular momentum and torque concepts. The key is connecting linear impulse-momentum to its rotational analog. Start with the rotational impulse-momentum theorem: the change in angular momentum equals the torque multiplied by time. Since the ring starts from rest, $\Delta L = L_f - 0 = I\omega_f$ , where $I$ is the moment of inertia and $\omega_f$ is the final angular velocity. The torque from the tangential force is $\tau = FR$ (force times perpendicular distance from the axis). For a thin ring, all mass is concentrated at radius $R$ , so $I = mR^2$ . Applying the impulse-momentum theorem: $I\omega_f = \tau t$ , which gives us $mR^2\omega_f = FRt$ . Solving for $\omega_f$ : $\omega_f = \frac{FRt}{mR^2} = \frac{Ft}{mR}$ . Choice A is correct. Choice B ( $\frac{FtR}{m}$ ) incorrectly omits the $R$ in the denominator from the moment of inertia—this might result from confusing rotational with linear momentum. Choice C ( $\frac{2Ft}{mR}$ ) suggests using the wrong moment of inertia, perhaps confusing a thin ring with a solid disk (which has $I = \frac{1}{2}mR^2$ ). Choice D ( $\frac{Ft}{2mR}$ ) makes the opposite error, introducing an incorrect factor of 2 in the denominator. Remember: always identify the correct moment of inertia for the specific geometry, and use $\tau t = I\omega_f$ for rotational impulse problems starting from rest.
A compound object consists of a thin rod of mass $m$ and length $L$ with a point mass $M$ attached at one end. The object rotates about an axis perpendicular to the rod at the other end (the end without the point mass). What is the moment of inertia of this system about the rotation axis?
Answer: $\frac{1}{3}mL^2 + ML^2$ When you encounter compound objects rotating about an axis, you need to find the moment of inertia for each component separately, then add them together using the principle of superposition. For this system, you have two components: a thin rod of mass $m$ and length $L$ , and a point mass $M$ . The rotation axis is at one end of the rod (the end without the point mass). Start with the rod. When a uniform thin rod rotates about an axis through one end (perpendicular to the rod), its moment of inertia is $\frac{1}{3}mL^2$ . This is a standard formula you should memorize - it's different from rotation about the center, which would give $\frac{1}{12}mL^2$ . Next, consider the point mass $M$ at the other end of the rod. Since it's located a distance $L$ from the rotation axis, its moment of inertia is simply $I = Mr^2 = ML^2$ . The total moment of inertia is the sum: $\frac{1}{3}mL^2 + ML^2$ , which is answer A. Looking at the wrong answers: B uses $\frac{1}{12}mL^2$ , which would be correct if the rod rotated about its center, not its end. C uses $mL^2$ for the rod, treating it incorrectly as a point mass at distance $L$ . D omits $L^2$ from the point mass term, forgetting that moment of inertia depends on the square of the distance. Key strategy: Memorize the standard moments of inertia for common shapes, especially noting whether the axis is at the center or end. Always remember that moments of inertia add linearly for compound objects.
A compound pulley system consists of a solid cylinder of radius $R$ and mass $M$ with two different radii sections: an inner radius $r$ and outer radius $R$ . A rope wound around the inner section supports a hanging mass $m_1$ , while a rope around the outer section supports mass $m_2$ . If both masses descend with the same angular acceleration $\alpha$ of the pulley, which equation correctly applies Newton's second law in rotational form?
Answer: $T_1 r - T_2 R = I\alpha$ For the pulley, we apply $\tau_{net} = I\alpha$ . The tensions create torques about the center: $T_1$ creates clockwise torque $T_1 r$ and $T_2$ creates counterclockwise torque $T_2 R$ . Taking counterclockwise as positive: $\tau_{net} = T_1 r - T_2 R = I\alpha$ . The weights $m_1 g$ and $m_2 g$ don't directly create torques on the pulley - they affect the tensions through the constraint equations for the hanging masses. Choice A incorrectly includes weight terms. Choice B has wrong signs. Choice C mixes hanging mass equations with pulley torque equation.
A bicycle wheel (treated as a thin ring) of mass $m$ and radius $R$ is spinning with angular velocity $\omega_0$ . A constant frictional torque $\tau_f$ acts to slow it down. What is the angular velocity after time $t$ ?
Answer: $\omega_0 - \frac{\tau_f t}{mR^2}$ When you encounter rotational motion problems involving changing angular velocity, you're dealing with rotational dynamics and the rotational analog of Newton's second law. To find how angular velocity changes under constant torque, start with the fundamental relationship: $\tau = I\alpha$ , where $\tau$ is torque, $I$ is moment of inertia, and $\alpha$ is angular acceleration. For a thin ring, $I = mR^2$ . Since the frictional torque opposes motion, $-\tau_f = mR^2 \alpha$ . Solving for angular acceleration: $\alpha = -\frac{\tau_f}{mR^2}$ Since angular acceleration is the rate of change of angular velocity, and this acceleration is constant, you can use: $\omega = \omega_0 + \alpha t$ Substituting: $\omega = \omega_0 - \frac{\tau_f t}{mR^2}$ This confirms answer B is correct. Answer A ( $\omega_0 - \frac{\tau_f t}{mR}$ ) incorrectly uses $mR$ instead of the proper moment of inertia $mR^2$ . This suggests confusing linear and rotational quantities. Answer C ( $\omega_0 - \frac{\tau_f t R}{m}$ ) has the radius in the numerator, which would give incorrect units and doesn't follow from the rotational dynamics equation. Answer D ( $\omega_0 + \frac{\tau_f t}{mR^2}$ ) has the correct form but wrong sign—friction always opposes motion, so it must decrease angular velocity. Study tip: Always remember that for rotational problems, you need the moment of inertia ( $I = mR^2$ for rings), and friction/resistance forces always oppose the direction of motion.
A uniform solid cylinder of mass $M$ and radius $R$ is mounted on a frictionless axle through its center. A string is wound around the cylinder and a mass $m$ hangs from the string. When the system is released, what is the angular acceleration of the cylinder?
Answer: $\frac{2mg}{(M+2m)R}$ This problem tests your understanding of rotational dynamics when linear and rotational motion are coupled. When you see a mass hanging from a string wound around a rotating object, you need to analyze both the linear motion of the hanging mass and the rotational motion of the cylinder simultaneously. Start by applying Newton's second law to the hanging mass: $mg - T = ma$ , where $T$ is the tension in the string. For the cylinder, use the rotational version: $\tau = I\alpha$ . The torque is $\tau = TR$ , and for a solid cylinder, $I = \frac{1}{2}MR^2$ , so $TR = \frac{1}{2}MR^2\alpha$ . The key insight is that the string doesn't slip, creating a constraint: $a = \alpha R$ . This links the linear acceleration of the mass to the angular acceleration of the cylinder. Solving the system: From the cylinder equation, $T = \frac{1}{2}MR\alpha$ . Substituting into the mass equation and using $a = \alpha R$ : $mg - \frac{1}{2}MR\alpha = m\alpha R$ . Solving for $\alpha$ : $mg = \alpha R(m + \frac{M}{2}) = \alpha R(\frac{2m + M}{2})$ , which gives $\alpha = \frac{2mg}{(M + 2m)R}$ . Answer A neglects the hanging mass's inertia. Answer B incorrectly uses $M + m$ instead of $M + 2m$ in the denominator, missing the factor from the cylinder's moment of inertia. Answer C has the wrong numerator, missing the factor of 2. Remember: In coupled rotation problems, always write separate equations for each moving part, identify the constraint relationship, then solve the system simultaneously.
A door of mass $m$ , height $h$ , and width $w$ can rotate about vertical hinges at one edge. The door can be modeled as a uniform rectangular plate. A horizontal force $F$ is applied perpendicular to the door at the edge opposite the hinges. What is the angular acceleration of the door?
Answer: $\frac{3F}{mw}$ When you encounter rotational motion problems, you need to apply Newton's second law for rotation: $\tau = I\alpha$ , where torque equals moment of inertia times angular acceleration. First, calculate the torque. Since force $F$ is applied perpendicular to the door at the edge opposite the hinges, the lever arm is the full width $w$ . Therefore, $\tau = Fw$ . Next, find the moment of inertia. For a uniform rectangular plate rotating about an edge (not the center), you must use the parallel axis theorem. The moment of inertia about the center is $I_{center} = \frac{m(h^2 + w^2)}{12}$ . Using the parallel axis theorem: $I = I_{center} + md^2$ , where $d = \frac{w}{2}$ is the distance from center to edge. This gives $I = \frac{m(h^2 + w^2)}{12} + m(\frac{w}{2})^2 = \frac{m(h^2 + 3w^2)}{12}$ . For most doors, height greatly exceeds width ( $h >> w$ ), so $h^2 + 3w^2 \approx h^2$ . Therefore, $I \approx \frac{mh^2}{12}$ . Solving for angular acceleration: $\alpha = \frac{\tau}{I} = \frac{Fw}{\frac{mh^2}{12}} = \frac{12Fw}{mh^2}$ . Wait - this doesn't match any option exactly because we made an approximation. Actually, let me reconsider the geometry. For a door rotating about its vertical edge, the relevant dimension for rotational inertia is primarily the width, not the height. The correct moment of inertia is $I = \frac{mw^2}{3}$ for rotation about one edge. Therefore: $\alpha = \frac{Fw}{\frac{mw^2}{3}} = \frac{3F}{mw}$ , which is answer C. Options A and B incorrectly use $h^2$ in the denominator, while D is missing the factor of 3 from the moment of inertia calculation. Remember: always identify the correct axis of rotation and use the appropriate moment of inertia formula - don't assume rotation about the center of mass.
A massless rod of length $L$ has two point masses: mass $m$ at one end and mass $2m$ at the other end. The rod rotates about an axis perpendicular to the rod and passing through the center of mass. If a torque $\tau$ is applied about this axis, what is the angular acceleration?
Answer: $\frac{3\tau}{2mL^2}$ When you encounter rotational dynamics problems involving torque and angular acceleration, you need to apply Newton's second law for rotation: $\tau = I\alpha$ , where $I$ is the moment of inertia and $\alpha$ is the angular acceleration. First, find the center of mass location. With mass $m$ at one end and $2m$ at the other, the center of mass is closer to the heavier mass. Setting one end as the origin, the center of mass is at $x_{cm} = \frac{m \cdot 0 + 2m \cdot L}{m + 2m} = \frac{2L}{3}$ from the lighter mass. Next, calculate the moment of inertia about the center of mass. The lighter mass $m$ is at distance $\frac{2L}{3}$ from the axis, and the heavier mass $2m$ is at distance $\frac{L}{3}$ from the axis. Therefore: $I = m\left(\frac{2L}{3}\right)^2 + 2m\left(\frac{L}{3}\right)^2 = m \cdot \frac{4L^2}{9} + 2m \cdot \frac{L^2}{9} = \frac{6mL^2}{9} = \frac{2mL^2}{3}$ Using $\tau = I\alpha$ , the angular acceleration is $\alpha = \frac{\tau}{I} = \frac{3\tau}{2mL^2}$ , confirming answer B. Answer A assumes $I = mL^2$ , which ignores the mass distribution. Answer C incorrectly inverts the fraction from the correct calculation. Answer D would result from assuming both masses are at distance $\frac{L}{2}$ from the center. Remember: always locate the center of mass first when the rotation axis passes through it, then calculate distances from that point to find the moment of inertia.
A playground merry-go-round (modeled as a uniform disk) of mass $M$ and radius $R$ is initially at rest. A child of mass $m$ runs tangentially at speed $v$ and jumps onto the edge of the merry-go-round. Immediately after the child lands, what is the angular velocity of the system?
Answer: $\frac{2mv}{(M + 2m)R}$ When you encounter a collision problem involving rotational motion, you're dealing with conservation of angular momentum. The key insight is that angular momentum before the collision equals angular momentum after the collision. Before the collision, only the running child has angular momentum. Since the child moves tangentially at the edge of the merry-go-round, their angular momentum is $L_i = mvR$ (where $mvR$ comes from $L = r \times p$ for tangential motion). After the collision, both the child and merry-go-round rotate together with the same angular velocity $\omega$ . The total moment of inertia includes the merry-go-round (uniform disk: $I = \frac{1}{2}MR^2$ ) plus the child treated as a point mass at distance $R$ ( $I = mR^2$ ). So $I_{total} = \frac{1}{2}MR^2 + mR^2$ . Applying conservation of angular momentum: $mvR = I_{total}\omega = (\frac{1}{2}MR^2 + mR^2)\omega$ Solving for $\omega$ : $\omega = \frac{mvR}{\frac{1}{2}MR^2 + mR^2} = \frac{mvR}{R^2(\frac{M}{2} + m)} = \frac{mv}{(\frac{M}{2} + m)R}$ Wait—this matches option B, but the correct answer is C. Let me recalculate: $\frac{mv}{(\frac{M}{2} + m)R} = \frac{2mv}{(M + 2m)R}$ after multiplying numerator and denominator by 2. Option A incorrectly uses $M + m$ instead of $\frac{M}{2} + m$ for the moment of inertia. Option B is algebraically equivalent to C but not simplified. Option D ignores the child's contribution to the final moment of inertia. Remember: always include the moment of inertia formula for uniform disks ( $\frac{1}{2}MR^2$ , not $MR^2$ ) and account for all rotating masses in the final system.
A wheel of radius $R$ and moment of inertia $I$ is initially rotating with angular velocity $\omega_0$ . A constant braking torque $\tau$ (opposing the motion) is applied. How long does it take for the wheel to come to rest?
Answer: $\frac{I\omega_0}{\tau}$ When you encounter rotational motion problems involving torque and angular velocity, think about the rotational analog of Newton's second law: $\tau = I\alpha$ , where torque equals moment of inertia times angular acceleration. Since the braking torque $\tau$ opposes motion, we have $-\tau = I\alpha$ , so the angular acceleration is $\alpha = -\frac{\tau}{I}$ (negative because it's decelerating). Using the kinematic equation $\omega = \omega_0 + \alpha t$ , when the wheel stops, $\omega = 0$ . Substituting: $0 = \omega_0 + \left(-\frac{\tau}{I}\right)t$ . Solving for time: $t = \frac{I\omega_0}{\tau}$ . This confirms answer A is correct. The stopping time is proportional to both the initial angular velocity and moment of inertia, but inversely proportional to the braking torque. Answer B ( $\frac{\tau}{I\omega_0}$ ) incorrectly places torque in the numerator and $I\omega_0$ in the denominator, suggesting stronger braking takes longer—the opposite of reality. Answer C ( $\frac{I\omega_0^2}{\tau}$ ) incorrectly squares the initial angular velocity, which would give units of time × angular velocity rather than just time. Answer D ( $\frac{\tau R}{I\omega_0}$ ) unnecessarily includes the radius $R$ , which doesn't affect how long deceleration takes—only the torque and rotational inertia matter. Remember: in rotational kinematics, always check your units and use the rotational analogs of linear motion equations. The moment of inertia $I$ plays the same role as mass in linear motion.

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