Torque and Work - College Physics

1. A hollow cylinder and a solid disk, both with mass $M$ and radius $R$, roll without slipping down identical inclines. At the bottom, which statement about their rotational kinetic energies is correct?
   Answer: The hollow cylinder has greater rotational kinetic energy because it has larger moment of inertia When objects roll down inclines, you need to analyze both translational and rotational motion using energy conservation. The key insight is that objects with different mass distributions will have different moments of inertia, leading to different energy distributions between linear and rotational motion. Using conservation of energy, the initial gravitational potential energy converts to both translational kinetic energy ($\frac{1}{2}Mv^2$) and rotational kinetic energy ($\frac{1}{2}I\omega^2$). For rolling without slipping, $v = \omega R$, so we can write: $Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2 + \frac{1}{2}I\frac{v^2}{R^2}$ The hollow cylinder has $I = MR^2$, while the solid disk has $I = \frac{1}{2}MR^2$. Solving for velocity, you find that objects with larger moments of inertia have lower linear velocities but higher rotational kinetic energies. The hollow cylinder's larger moment of inertia means more energy goes into rotation, making (A) correct. (B) is wrong because while the disk does reach the bottom first (due to lower rotational inertia), this doesn't give it more rotational kinetic energy—it actually has less. (C) is incorrect because equal mass and radius don't guarantee equal rotational kinetic energy when the mass distributions differ. (D) correctly identifies that the hollow cylinder has lower linear velocity, but this supports rather than contradicts why it has greater rotational kinetic energy. Study tip: Remember that for rolling objects, higher moment of inertia means more energy stored in rotation and less in translation—the energy "prefers" to be rotational when $I$ is large.
2. A uniform rod of length $L$ and mass $M$ is pivoted at one end and released from rest in a horizontal position. What is the angular velocity when the rod reaches the vertical position?
   Answer: $\sqrt{\frac{3g}{L}}$ When you see a rod pivoting from rest, this is a classic rotational energy problem. You'll want to use conservation of energy, comparing the initial gravitational potential energy to the final rotational kinetic energy. Initially, the rod's center of mass is at height $\frac{L}{2}$ above its lowest point. When vertical, the center of mass drops by $\frac{L}{2}$, so the change in potential energy is $\Delta PE = Mg\frac{L}{2}$. This potential energy converts to rotational kinetic energy: $KE_{rot} = \frac{1}{2}I\omega^2$. For a uniform rod pivoted at one end, the moment of inertia is $I = \frac{1}{3}ML^2$. Setting up conservation of energy: $Mg\frac{L}{2} = \frac{1}{2} \cdot \frac{1}{3}ML^2\omega^2$ Solving for $\omega$: $\frac{MgL}{2} = \frac{ML^2\omega^2}{6}$ $3g = L\omega^2$ $\omega = \sqrt{\frac{3g}{L}}$ This confirms answer C is correct. Option A, $\sqrt{\frac{g}{L}}$, would result from incorrectly using $I = ML^2$ (point mass formula) instead of the proper rod formula. Option B, $\sqrt{\frac{2g}{L}}$, comes from mistakenly using the center-of-mass moment of inertia $I = \frac{1}{12}ML^2$, which applies when the pivot is at the center. Option D, $\sqrt{\frac{6g}{L}}$, results from algebraic errors in the energy equation setup. Remember: always identify the correct moment of inertia formula based on the pivot location and mass distribution. For uniform rods, $I = \frac{1}{3}ML^2$ at the end, $I = \frac{1}{12}ML^2$ at the center.
3. A uniform wheel of radius $R = 0.40$ m and mass $M = 8.0$ kg rolls without slipping down an incline. When the wheel has rotated through $5.0$ complete revolutions, it has moved a distance $s = 12.6$ m along the incline. What is the work done by the friction force during this motion?
   Answer: $W_f = 0$ J because rolling without slipping means no sliding occurs When analyzing rolling motion problems, the key insight is understanding what work means: it's the dot product of force and displacement, $W = \vec{F} \cdot \vec{d}$. For work to be done, the force must have a component in the direction of motion. In rolling without slipping, the friction force acts at the contact point between the wheel and incline. Here's the crucial point: at any instant, the contact point has zero velocity relative to the incline surface. This is what "without slipping" means - there's no sliding motion at the contact point. Since the point where friction acts has zero displacement relative to the surface, the work done by friction is zero. Think of it this way: friction prevents sliding, but if there's no sliding, friction does no work. Option A incorrectly assumes that because friction provides the torque for rolling, it must do positive work. While friction does create torque, work and torque are different concepts - torque doesn't require displacement. Option B falls into the trap of thinking friction always opposes motion. Although friction points up the incline (opposite to the wheel's center-of-mass motion), work depends on motion at the point of force application, not the center of mass. Option D makes the error of calculating work using the wheel's center-of-mass displacement rather than the displacement at the contact point where friction acts. Remember this pattern: in pure rolling motion, friction does zero work because there's no relative motion at the contact point. This is a fundamental principle that distinguishes rolling from sliding motion.
4. A uniform solid disk of radius $R = 0.50$ m and mass $M = 4.0$ kg is initially at rest. A constant tangential force $F = 12$ N is applied at the rim for a time interval $t = 3.0$ s. What is the total work done by this force during this interval?
   Answer: $W = 324$ J The work done by a torque is $W = \tau \theta$, where $\theta$ is the angular displacement. First, find the torque: $\tau = FR = 12 \times 0.50 = 6.0$ N⋅m. For a solid disk, $I = \frac{1}{2}MR^2 = \frac{1}{2}(4.0)(0.50)^2 = 0.50$ kg⋅m². The angular acceleration is $\alpha = \tau/I = 6.0/0.50 = 12$ rad/s². Using kinematics, $\theta = \frac{1}{2}\alpha t^2 = \frac{1}{2}(12)(3.0)^2 = 54$ rad. Therefore, $W = \tau \theta = 6.0 \times 54 = 324$ J. Choice B uses incorrect moment of inertia (treating as point mass). Choice C uses linear work formula $Fd$ incorrectly. Choice D uses wrong kinematic equation.
5. A flywheel in the form of a solid disk is spinning at $\omega_0 = 60$ rad/s when a brake applies a constant torque $\tau = -8.0$ N⋅m (opposing the motion). The flywheel has moment of inertia $I = 2.0$ kg⋅m². How much work is done by the brake torque during the first $2.0$ seconds of braking?
   Answer: $W = -112$ J This problem tests your understanding of rotational dynamics and the work-energy theorem for rotating objects. When you see a flywheel with changing angular velocity, think about how torque relates to angular acceleration and how work connects to energy changes. Start by finding the angular acceleration using $\tau = I\alpha$: $\alpha = \frac{\tau}{I} = \frac{-8.0}{2.0} = -4.0$ rad/s². Next, determine the angular displacement during the first 2.0 seconds using $\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 60(2.0) + \frac{1}{2}(-4.0)(2.0)^2 = 120 - 8.0 = 112$ rad. The work done by the brake torque is $W = \tau \theta = (-8.0)(112) = -112$ J. The negative sign confirms that the brake does negative work, removing energy from the system. Answer A (-80 J) likely comes from incorrectly calculating the displacement as just $\omega_0 t = 60 \times 2 = 120$ rad, then multiplying by torque incorrectly. Answer B (-96 J) might result from using the average angular velocity incorrectly or making calculation errors. Answer C (-32 J) appears to use only the acceleration term: $\frac{1}{2}\alpha t^2 = 8$ rad, then multiplying by torque. The correct answer is D: $W = -112$ J. Remember: for rotational work problems, always use the complete kinematic equation for angular displacement when acceleration is constant. Don't forget that work equals torque times angular displacement, just as linear work equals force times linear displacement.
6. A thin rod of mass $M = 2.0$ kg and length $L = 1.2$ m is pivoted at its center and initially at rest. Two forces are applied simultaneously: $F_1 = 8.0$ N perpendicular to the rod at one end, and $F_2 = 6.0$ N perpendicular to the rod at the other end in the opposite rotational direction. After the rod has rotated $45°$, what is the net work done by both forces?
   Answer: $W_{net} = 0.94$ J When you encounter rotational motion problems involving work, remember that work done by a torque equals the torque multiplied by the angular displacement: $W = \tau \theta$. First, let's find the net torque. Since the rod is pivoted at its center, each force creates a torque with moment arm $r = L/2 = 0.6$ m. The torques are: - $\tau_1 = F_1 \times r = 8.0 \times 0.6 = 4.8$ N⋅m - $\tau_2 = F_2 \times r = 6.0 \times 0.6 = 3.6$ N⋅m Since the forces cause rotation in opposite directions, the net torque is $\tau_{net} = 4.8 - 3.6 = 1.2$ N⋅m. Converting the angular displacement to radians: $\theta = 45° \times \frac{\pi}{180°} = \frac{\pi}{4}$ rad. Therefore, the net work done is: $W_{net} = \tau_{net} \times \theta = 1.2 \times \frac{\pi}{4} = 0.94$ J. This confirms answer B is correct. Answer A (1.88 J) likely comes from doubling the correct answer, perhaps by incorrectly adding torques instead of subtracting. Answer C (5.5 J) appears to result from using the full length L instead of L/2 for the moment arm. Answer D (3.7 J) might come from using degrees instead of radians in the calculation. Study tip: Always convert angles to radians when calculating rotational work, and remember that opposing torques subtract, not add. When forces act at opposite ends causing opposite rotations, find the net torque first, then calculate work.
7. A flywheel is spinning at $\omega_0 = 100 \text{ rad/s}$. A constant braking torque reduces its angular velocity to $50 \text{ rad/s}$ after it rotates through $150 \text{ rad}$. What fraction of the original kinetic energy was dissipated?
   Answer: $\frac{3}{4}$ When you encounter rotational motion problems involving energy changes, focus on the relationship between rotational kinetic energy and angular velocity. Rotational kinetic energy is $KE = \frac{1}{2}I\omega^2$, so energy depends on the square of angular velocity. First, find the angular acceleration using kinematics. With $\omega_0 = 100 \text{ rad/s}$, $\omega_f = 50 \text{ rad/s}$, and $\theta = 150 \text{ rad}$, use $\omega_f^2 = \omega_0^2 + 2\alpha\theta$: $(50)^2 = (100)^2 + 2\alpha(150)$ $2500 = 10000 + 300\alpha$ $\alpha = -25 \text{ rad/s}^2$ Now compare the kinetic energies. The initial energy is $KE_0 = \frac{1}{2}I(100)^2 = 5000I$. The final energy is $KE_f = \frac{1}{2}I(50)^2 = 1250I$. The fraction of energy dissipated is: $\frac{KE_0 - KE_f}{KE_0} = \frac{5000I - 1250I}{5000I} = \frac{3750I}{5000I} = \frac{3}{4}$ This confirms answer C. Answer A ($\frac{1}{4}$) represents the fraction of energy remaining, not dissipated. Answer B ($\frac{1}{2}$) might tempt you since the angular velocity was halved, but energy depends on $\omega^2$, not $\omega$. Answer D ($\frac{2}{3}$) has no clear physical basis for this problem. Remember: when angular velocity changes by a factor, kinetic energy changes by the square of that factor. Here, $\omega$ was halved, so final energy is $(\frac{1}{2})^2 = \frac{1}{4}$ of the original, meaning $\frac{3}{4}$ was lost.
8. A wheel of radius $R$ and moment of inertia $I$ is initially spinning with angular velocity $\omega_0$. A tangential force $F$ is applied at the rim to slow it down. After the wheel has rotated through angle $\theta$, what is its angular velocity?
   Answer: $\sqrt{\omega_0^2 - \frac{2FR\theta}{I}}$ When you encounter rotational motion problems involving forces and energy, think about whether to use kinematic equations or energy methods. Since this problem asks for final angular velocity after a given angular displacement with a known force, rotational kinematics with energy considerations is the most direct approach. The tangential force $F$ creates a torque $\tau = FR$ that opposes the wheel's rotation. This torque produces an angular acceleration $\alpha = \tau/I = FR/I$ in the direction opposite to the initial motion. Using the rotational kinematic equation $\omega^2 = \omega_0^2 + 2\alpha\theta$, and substituting the negative acceleration (since the force slows the wheel): $\omega^2 = \omega_0^2 - 2\frac{FR}{I}\theta = \omega_0^2 - \frac{2FR\theta}{I}$. Taking the square root gives $\omega = \sqrt{\omega_0^2 - \frac{2FR\theta}{I}}$. Option A is correct as shown above. Option B has the wrong sign—it shows $+\frac{2FR\theta}{I}$, which would mean the force is speeding up the wheel rather than slowing it down. Option C uses a linear relationship $\omega = \omega_0 - \frac{FR\theta}{I}$, which incorrectly applies the kinematic equation $v = v_0 + at$ instead of $v^2 = v_0^2 + 2as$. Option D is missing the factor of 2 in the denominator, suggesting confusion about which kinematic equation to use. Remember: rotational problems often require the $\omega^2$ kinematic equations, especially when dealing with angular displacement. The factor of 2 comes from the $2\alpha\theta$ term, just like in linear motion.
9. A solid cylinder and a hollow cylinder, both with the same mass $M$ and radius $R$, are released simultaneously from rest at the top of an inclined plane of height $h$. Both roll without slipping down the incline. What is the ratio of the work done by gravity on the solid cylinder to the work done by gravity on the hollow cylinder when they reach the bottom?
   Answer: The ratio is $1:1$ since both have the same mass and height The work done by gravity depends only on the vertical displacement and the weight of the object: $W = mgh$. Since both cylinders have the same mass $M$ and fall through the same height $h$, gravity does the same work $Mgh$ on each cylinder. The different moments of inertia affect how this energy is distributed between translational and rotational kinetic energy, and they affect the acceleration down the incline, but they do not change the total work done by gravity. Choice B incorrectly relates work to moment of inertia. Choice C confuses work with time. Choice D incorrectly assumes work depends on energy distribution.
10. A rod of length $L = 2.0$ m and mass $M = 3.0$ kg can rotate about a fixed axis through one end. A force $F = 15$ N is applied perpendicular to the rod at a distance $d = 1.5$ m from the axis. If the rod rotates through an angle $\theta = 90°$ from rest, what is the work done by the applied force?
    Answer: $W = 35$ J When you encounter rotational motion problems involving work, remember that work can be calculated as $W = \tau \theta$ when a constant torque acts through an angular displacement, where $\tau$ is torque and $\theta$ is the angle in radians. First, calculate the torque produced by the applied force. Since the force is applied perpendicular to the rod, $\tau = F \times d = 15 \text{ N} \times 1.5 \text{ m} = 22.5 \text{ N}\cdot\text{m}$. Next, convert the angular displacement to radians: $\theta = 90° = \frac{\pi}{2} \text{ radians} \approx 1.57 \text{ rad}$. Now calculate the work: $W = \tau \theta = 22.5 \times 1.57 = 35.3 \text{ J} \approx 35 \text{ J}$. This confirms answer C is correct. Let's examine why the other answers are wrong. Answer A (24 J) likely comes from incorrectly using the angle in degrees rather than radians: $22.5 \times \frac{90}{90} = 22.5 \text{ J}$, then rounding poorly. Answer B (22 J) represents just the torque value without multiplying by the angular displacement—a common error when students confuse torque with work. Answer D (18 J) might result from using the wrong distance or force value in calculations. Remember this key strategy: In rotational work problems, always convert angles to radians before calculating $W = \tau \theta$. Also, torque and work have different units (N·m vs J), so don't confuse the two—work requires multiplying torque by angular displacement.
11. A bicycle wheel of radius $R$ rolls without slipping with its center moving at constant speed $v$. A point on the rim of the wheel traces out a curve called a cycloid. What is the maximum speed of any point on the rim relative to the ground?
    Answer: $2v$ When analyzing rolling motion, you need to understand that each point on a rolling wheel has two velocity components: one from the wheel's rotation and one from its translation through space. These velocities add vectorially, creating different speeds for different points on the rim. For a wheel rolling without slipping, the center moves at speed $v$, and the wheel rotates with angular velocity $\omega = v/R$. Any point on the rim has a translational velocity $v$ (same as the center) plus a rotational velocity of magnitude $v$ relative to the center. The key insight is determining when these velocities add constructively. At the top of the wheel, both the translational and rotational velocities point in the same direction (forward), so they add directly: $v + v = 2v$. This gives the maximum speed of any point on the rim relative to the ground. Looking at the wrong answers: Choice (A) $v$ represents the speed of the wheel's center or points on the rim when rotational and translational components are perpendicular. Choice (B) $\sqrt{2}v$ would result from incorrectly applying the Pythagorean theorem when the velocity components are actually parallel, not perpendicular. Choice (D) $v/2$ has no physical basis in this rolling motion scenario. Remember that for rolling motion problems, the maximum rim speed always occurs at the top of the wheel where translational and rotational velocities are parallel and additive. This "$2v$ rule" for maximum rim speed in rolling motion appears frequently in physics problems.
12. A uniform disk of radius $R$ and mass $M$ is initially at rest. A constant tangential force $F$ is applied at the rim for a time $t$. What is the work done by this force?
    Answer: $F^2t^2R^2/M$ When you encounter rotational motion problems involving work and energy, you need to connect linear force concepts with rotational kinematics. The key insight is that work equals the change in kinetic energy, and for rotation, you must use rotational kinetic energy. Since the tangential force $F$ acts at radius $R$, it creates torque $\tau = FR$. For a uniform disk, the moment of inertia is $I = \frac{1}{2}MR^2$. The angular acceleration is $\alpha = \frac{\tau}{I} = \frac{FR}{\frac{1}{2}MR^2} = \frac{2F}{MR}$. After time $t$, the angular velocity becomes $\omega = \alpha t = \frac{2Ft}{MR}$. The rotational kinetic energy is $KE = \frac{1}{2}I\omega^2 = \frac{1}{2} \cdot \frac{1}{2}MR^2 \cdot \left(\frac{2Ft}{MR}\right)^2 = \frac{F^2t^2R^2}{M}$. Since the disk started from rest, this equals the work done. Choice A, $\frac{1}{2}F^2t^2R^2/M$, incorrectly applies the $\frac{1}{2}$ factor twice—once in the kinetic energy formula and again in the moment of inertia. Choice C, $FRt$, incorrectly treats this as linear work ($F \cdot d$) where distance is $Rt$, ignoring rotational effects. Choice D, $\frac{1}{2}FRt$, makes the same linear assumption but adds an erroneous $\frac{1}{2}$ factor. Remember: for rotational work problems, always use the work-energy theorem with rotational kinetic energy $\frac{1}{2}I\omega^2$. Don't confuse this with linear motion formulas—the geometry and energy distribution are fundamentally different.
13. A disk of mass $M$ and radius $R$ is rotating about a fixed axis through its center. A small mass $m$ is attached to the rim of the disk. If the system's angular velocity is $\omega$, what is the total rotational kinetic energy?
    Answer: $\frac{1}{2}(\frac{1}{2}MR^2 + mR^2)\omega^2$ When analyzing rotational motion problems involving multiple objects, you need to find the total rotational kinetic energy by considering each rotating component separately, then adding their contributions. The rotational kinetic energy formula is $KE_{rot} = \frac{1}{2}I\omega^2$, where $I$ is the moment of inertia. For this system, you have two rotating objects: the disk itself and the small mass on the rim. For the disk, the moment of inertia is $I_{disk} = \frac{1}{2}MR^2$ (standard formula for a solid disk about its center). For the small mass $m$ at distance $R$ from the axis, it acts as a point mass with $I_{mass} = mR^2$. The total moment of inertia is $I_{total} = \frac{1}{2}MR^2 + mR^2$, giving total kinetic energy: $KE_{total} = \frac{1}{2}(\frac{1}{2}MR^2 + mR^2)\omega^2$. Looking at the wrong answers: Option B incorrectly separates the $\frac{1}{2}$ factor, applying it to each term individually instead of factoring it out from the total moment of inertia. Option C uses $\frac{1}{4}$ for the disk's contribution, which would come from incorrectly applying the $\frac{1}{2}$ factor twice to the disk term. Option D incorrectly treats both the disk and point mass as having the same moment of inertia formula. Study tip: Always identify each rotating component separately, find its individual moment of inertia using the appropriate formula, then sum them before applying the $\frac{1}{2}I\omega^2$ kinetic energy formula to the total system.
14. A wheel with moment of inertia $I$ is accelerated from rest by a motor that provides constant power $P$. After time $t$, what is the wheel's angular velocity?
    Answer: $\sqrt{\frac{2Pt}{I}}$ When you encounter rotational motion problems involving constant power, you need to connect power, energy, and kinematics. Power is the rate of energy transfer, so $P = \frac{dE}{dt}$, where the energy here is rotational kinetic energy. Since the wheel starts from rest, all the work done by the motor becomes rotational kinetic energy: $E = \frac{1}{2}I\omega^2$. With constant power $P$, the total energy after time $t$ is $E = Pt$. Setting these equal: $Pt = \frac{1}{2}I\omega^2$ Solving for $\omega$: $\omega^2 = \frac{2Pt}{I}$, so $\omega = \sqrt{\frac{2Pt}{I}}$ This confirms answer B is correct. Answer A ($\sqrt{\frac{Pt}{I}}$) represents the common error of forgetting the factor of $\frac{1}{2}$ in the rotational kinetic energy formula. Students sometimes use $E = I\omega^2$ instead of $E = \frac{1}{2}I\omega^2$. Answer C ($\frac{Pt}{I}$) would result from incorrectly assuming $P = I\omega$ or confusing this with the linear relationship $\omega = \alpha t$ for constant angular acceleration problems. Answer D ($\frac{2Pt}{I}$) comes from the same kinetic energy mistake as option A, but without taking the square root—treating angular velocity as if it's linear in energy rather than proportional to the square root of energy. Remember: constant power problems always involve energy methods, not force/torque methods. The key insight is that $P \cdot t = \Delta E$, then use the appropriate kinetic energy formula.
15. A disk of radius $R$ and mass $M$ rotates about its center with angular velocity $\omega$. If the radius were doubled while keeping the mass and angular velocity constant, by what factor would the rotational kinetic energy change?
    Answer: $4$ When you encounter rotational motion problems involving energy changes, focus on how the moment of inertia depends on the mass distribution. Rotational kinetic energy is given by $KE_{rot} = \frac{1}{2}I\omega^2$, where $I$ is the moment of inertia and $\omega$ is angular velocity. For a solid disk rotating about its center, the moment of inertia is $I = \frac{1}{2}MR^2$. Initially, the rotational kinetic energy is $KE_1 = \frac{1}{2} \cdot \frac{1}{2}MR^2\omega^2 = \frac{1}{4}MR^2\omega^2$. When the radius doubles to $2R$ while keeping mass $M$ and angular velocity $\omega$ constant, the new moment of inertia becomes $I_2 = \frac{1}{2}M(2R)^2 = \frac{1}{2}M \cdot 4R^2 = 2MR^2$. The new kinetic energy is $KE_2 = \frac{1}{2} \cdot 2MR^2\omega^2 = MR^2\omega^2$. Comparing the energies: $\frac{KE_2}{KE_1} = \frac{MR^2\omega^2}{\frac{1}{4}MR^2\omega^2} = 4$. The answer is (B) 4. Choice (A) 2 incorrectly assumes kinetic energy scales linearly with radius. Choice (C) 8 might result from confusing this with rotational scenarios involving different mass distributions or incorrectly cubing the radius factor. Choice (D) 16 suggests raising the radius change to the fourth power, which has no physical basis here. Remember: rotational kinetic energy depends on $I\omega^2$, and for most geometric shapes, the moment of inertia scales with $R^2$. When radius changes by a factor, the energy typically changes by that factor squared.
16. A solid cylinder rolls without slipping down an inclined plane of height $h$. What fraction of the total kinetic energy at the bottom is rotational kinetic energy?
    Answer: $\frac{1}{3}$ When analyzing rolling objects on inclined planes, you need to consider both translational and rotational motion. The key insight is understanding how energy is distributed between these two types of motion for different geometric shapes. As the solid cylinder rolls down the incline, gravitational potential energy converts to kinetic energy. The total kinetic energy has two components: translational kinetic energy ($KE_{trans} = \frac{1}{2}mv^2$) and rotational kinetic energy ($KE_{rot} = \frac{1}{2}I\omega^2$). For a solid cylinder, the moment of inertia is $I = \frac{1}{2}mr^2$, and the rolling condition gives us $v = r\omega$. Substituting these relationships: $KE_{rot} = \frac{1}{2} \cdot \frac{1}{2}mr^2 \cdot \frac{v^2}{r^2} = \frac{1}{4}mv^2$ The total kinetic energy is: $KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$ Therefore, the fraction that is rotational is: $\frac{KE_{rot}}{KE_{total}} = \frac{\frac{1}{4}mv^2}{\frac{3}{4}mv^2} = \frac{1}{3}$ This confirms answer B is correct. Answer A ($\frac{1}{4}$) incorrectly represents the ratio of rotational to translational energy, not rotational to total. Answer C ($\frac{1}{2}$) would apply to a thin ring or hollow cylinder. Answer D ($\frac{2}{3}$) represents the translational fraction, not rotational. Remember: the moment of inertia determines energy distribution in rolling motion. Solid cylinders have $I = \frac{1}{2}mr^2$, leading to the $\frac{1}{3}$ rotational fraction pattern.
17. A uniform rod of mass $M$ and length $L$ is free to rotate about a pivot at one end. The rod is released from rest at an angle $\theta$ from the vertical. When the rod swings to the vertical position, what is its angular velocity?
    Answer: $\sqrt{\frac{3g(1-\cos\theta)}{L}}$ When you see a rotating rigid body problem involving energy changes, think conservation of energy. This rod starts with gravitational potential energy and converts it to rotational kinetic energy as it swings down. Set up your energy conservation equation. Initially, the rod has potential energy and no kinetic energy (released from rest). When vertical, it has kinetic energy and lower potential energy. The center of mass of a uniform rod is at $L/2$, so the height change of the center of mass is $\frac{L}{2}(1-\cos\theta)$. Initial energy: $E_i = Mg\frac{L}{2}(1-\cos\theta)$ Final energy: $E_f = \frac{1}{2}I\omega^2$ For a rod rotating about one end, the moment of inertia is $I = \frac{1}{3}ML^2$. Setting $E_i = E_f$: $Mg\frac{L}{2}(1-\cos\theta) = \frac{1}{2} \cdot \frac{1}{3}ML^2\omega^2$ Solving for $\omega$: $\omega^2 = \frac{3g(1-\cos\theta)}{L}$ $\omega = \sqrt{\frac{3g(1-\cos\theta)}{L}}$ This is answer C. Answer A uses the wrong moment of inertia ($\frac{1}{12}ML^2$ for rotation about the center). Answer B has the correct moment of inertia but incorrect algebra in the energy equation. Answer D contains an extra factor of 2, likely from incorrectly handling the height calculation or the rotational kinetic energy formula. Remember: for rotating rigid bodies, always identify the correct moment of inertia for the specific axis of rotation. A rod about one end is $\frac{1}{3}ML^2$, not $\frac{1}{12}ML^2$ (which is for rotation about the center).
18. A uniform rod of mass $M$ and length $L$ rotates about an axis through its center. If a force $F$ is applied perpendicular to the rod at a distance $L/4$ from the center, what is the power delivered when the rod's angular velocity is $\omega$?
    Answer: $\frac{FL\omega}{4}$ When you encounter rotational motion problems involving power, you need to connect linear and rotational concepts. Power in rotational systems is given by $P = \tau\omega$, where $\tau$ is torque and $\omega$ is angular velocity. First, find the torque. Torque equals force times the perpendicular distance from the axis of rotation: $\tau = F \times r$. Since the force is applied at distance $L/4$ from the center, the torque is $\tau = F \times \frac{L}{4} = \frac{FL}{4}$. Now calculate the power: $P = \tau\omega = \frac{FL}{4} \times \omega = \frac{FL\omega}{4}$. This confirms answer A is correct. Looking at the wrong answers: Answer B ($\frac{FL\omega}{2}$) incorrectly uses $L/2$ as the distance, which would be the case if the force were applied at the end of the rod. Answer C ($FL\omega$) mistakenly uses the full length $L$ as the moment arm, ignoring that the force is only applied at $L/4$ from center. Answer D ($\frac{FL\omega}{8}$) appears to confuse the geometry, possibly mixing up the distance relationships or incorrectly applying a factor of 2. The key insight is that torque depends on the actual perpendicular distance from the rotation axis to where the force is applied, not the total length of the object. Always identify this distance carefully in rotational problems, and remember that rotational power follows the same $P = \tau\omega$ relationship that linear power follows $P = Fv$.
19. A flywheel with moment of inertia $I$ is spinning with angular velocity $\omega_0$. A constant braking torque $\tau$ is applied. How much work is done by the braking torque when the flywheel comes to rest?
    Answer: $-\frac{1}{2}I\omega_0^2$ When you encounter rotational motion problems involving work and energy, think about how the work-energy theorem applies to rotating objects. Just as work equals the change in kinetic energy for linear motion, the same principle holds for rotational motion. The flywheel initially has rotational kinetic energy $KE_i = \frac{1}{2}I\omega_0^2$ and finally has zero kinetic energy when it stops. By the work-energy theorem, the work done by the braking torque equals the change in kinetic energy: $W = KE_f - KE_i = 0 - \frac{1}{2}I\omega_0^2 = -\frac{1}{2}I\omega_0^2$. The negative sign is crucial—it indicates that the braking torque does negative work, removing energy from the system as expected for a braking force. Choice A ($\frac{1}{2}I\omega_0^2$) gives the magnitude of the initial kinetic energy but misses the essential negative sign. This would imply the torque adds energy to the system, which contradicts the braking action. Choice C ($\tau \omega_0$) incorrectly uses the formula for instantaneous power ($P = \tau \omega$) rather than work. Power and work have different units and meanings. Choice D ($-\tau \omega_0$) also confuses work with power, applying the negative sign to the wrong expression. Remember: when analyzing rotational work problems, use the work-energy theorem first. The sign of your answer should always make physical sense—braking forces do negative work, while driving forces do positive work.
20. A solid sphere rolls without slipping up an inclined plane. At the bottom of the incline, the sphere has both translational kinetic energy $KE_{trans} = 60$ J and rotational kinetic energy $KE_{rot} = 24$ J. As the sphere rolls up the incline, what is the work done by gravity when the sphere has risen to a height where its total kinetic energy is exactly half of its initial total kinetic energy?
    Answer: $W_g = -42$ J When a sphere rolls without slipping on an incline, you're dealing with energy conservation between kinetic and gravitational potential energy. The key insight is that work done by gravity equals the negative change in gravitational potential energy. First, find the initial and final total kinetic energies. Initially: $KE_{total} = 60 + 24 = 84$ J. When the total kinetic energy becomes half: $KE_{final} = 42$ J. Using conservation of energy, the decrease in kinetic energy equals the increase in gravitational potential energy: $\Delta KE = KE_{final} - KE_{initial} = 42 - 84 = -42$ J. Since $\Delta KE + \Delta PE = 0$, we have $\Delta PE = +42$ J. The work done by gravity is $W_g = -\Delta PE = -42$ J, confirming answer B. Answer A ($-30$ J) incorrectly assumes the work equals half the initial translational kinetic energy, missing the rotational component. Answer C ($-84$ J) represents the negative of the total initial kinetic energy, which would only be correct if the sphere came to a complete stop. Answer D ($-60$ J) uses only the initial translational kinetic energy, ignoring the rotational energy entirely. Study tip: In rolling motion problems, always account for both translational and rotational kinetic energy. Remember that work done by gravity equals the negative change in gravitational potential energy, and use energy conservation to connect kinetic and potential energy changes.

Start this quiz