Torque - College Physics

1. A uniform beam of mass $m = 50 \text{ kg}$ and length $L = 6.0 \text{ m}$ is supported by a pivot at its center and a cable attached at one end. The cable makes an angle of $60°$ with the horizontal beam. What is the torque due to the beam's weight about the pivot?
   Answer: $0 \text{ N}\cdot\text{m}$ because the pivot is located at the beam's center of mass When analyzing torque problems involving uniform beams, the key insight is understanding where the center of mass is located and how this affects the torque calculation about different pivot points. Torque is calculated as $\tau = r \times F$, where $r$ is the perpendicular distance from the pivot to the line of action of the force. For a uniform beam, all the weight can be considered to act at the center of mass, which is located at the geometric center of the beam. Since the pivot is positioned at the center of the beam (where the center of mass is located), the distance from the pivot to the point where the weight acts is zero. Therefore, $\tau = 0 \times mg = 0$. The beam's weight creates no torque about a pivot located at its center of mass. Option B incorrectly assumes the weight acts at the end of the beam, calculating $\tau = (3.0 \text{ m})(50 \text{ kg})(9.8 \text{ m/s}^2) = 1470 \text{ N}\cdot\text{m}$. Option C makes a similar error, assuming the weight somehow acts at half the beam's length from the pivot. Option D incorrectly suggests that cable tension contributes to the torque due to the beam's weight, confusing different forces in the system. The correct answer is A: the torque is zero because the pivot coincides with the center of mass. Study tip: Remember that for uniform objects, you can always treat the entire weight as acting at the center of mass. If your pivot is at the center of mass, the weight produces zero torque regardless of the object's orientation or other forces present.
2. A thin ring of mass $M$ and radius $R$ rotates about an axis perpendicular to its plane and passing through its center. A small mass $m$ (where $m \ll M$) is attached to the ring at its rim. If a tangential force $F$ is applied to the small mass, what is the angular acceleration of the system?
   Answer: $\frac{FR}{MR^2 + mR^2}$ The total moment of inertia is $I_{total} = I_{ring} + I_{mass} = MR^2 + mR^2$. The applied torque is $\tau = FR$. Using $\tau = I\alpha$: $FR = (MR^2 + mR^2)\alpha$, so $\alpha = \frac{FR}{MR^2 + mR^2}$. Choice A treats this as a linear motion problem. Choice B incorrectly calculates moment of inertia as $MR$ instead of $MR^2$. Choice C omits the torque arm $R$ in the numerator.
3. A uniform rod of length $L = 2.0 \text{ m}$ is held horizontally with one end against a wall and the other end supported by a hand. If the hand is moved to a position $L/3$ from the free end, what is the ratio of the torque about the wall contact point before and after moving the hand?
   Answer: $1.0$ because the rod's weight and center of mass remain unchanged regardless of hand position When analyzing torque problems involving static equilibrium, focus on what actually creates the torque about your chosen pivot point. Torque depends on both the force magnitude and the perpendicular distance from the pivot to the line of action of that force. In this problem, you're asked about the torque due to the rod's weight about the wall contact point. The rod's weight acts downward at its center of mass (at $L/2 = 1.0 \text{ m}$ from the wall), regardless of where your hand supports it. Since the rod remains horizontal in both cases, the perpendicular distance from the wall to this center of mass stays constant at $1.0 \text{ m}$. The torque from the rod's weight is $\tau = mgL/2$ in both situations, giving a ratio of $1.0$. Answer A correctly recognizes that the rod's intrinsic properties (weight and center of mass location) don't change with hand position. Answer B incorrectly assumes the hand position affects how the rod's weight creates torque about the wall—but the rod's weight always acts at its geometric center. Answer C makes a similar error, confusing the hand's supporting force with the gravitational force on the rod itself. Answer D suggests the geometric change somehow doubles the torque, which ignores that we're specifically calculating torque due to the rod's weight, not the hand's force. Remember: when calculating torque about a specific point, only consider the force of interest and its perpendicular distance from that pivot point. Supporting forces elsewhere don't change how a distributed mass creates torque about your chosen reference point.
4. A compound object consists of two uniform rods of equal mass $m$ and length $L$ joined at right angles to form an L-shape. The system rotates about an axis through the corner where the rods meet, perpendicular to both rods. What is the torque due to gravity about this axis?
   Answer: $0$ because the center of mass of each rod is equidistant from the rotation axis When analyzing torque due to gravity on rotating systems, you need to consider where each object's center of mass is located and how gravity acts on it relative to the rotation axis. For this L-shaped compound object, each rod has its center of mass at $\frac{L}{2}$ from the corner where they meet. Since both rods have equal mass $m$, gravity exerts a downward force of $mg$ on each rod's center of mass. However, the key insight is recognizing how these torques combine. The first rod extends horizontally, so gravity creates a clockwise torque of $mg \cdot \frac{L}{2}$. The second rod extends vertically downward, so gravity also creates a clockwise torque of $mg \cdot \frac{L}{2}$. Since both torques act in the same rotational direction, they add together: $\frac{mgL}{2} + \frac{mgL}{2} = mgL$. Wait—this reasoning leads to answer B, but let's reconsider the geometry more carefully. Actually, if one rod is horizontal and the other vertical, gravity acts downward on both centers of mass. The horizontal rod experiences torque, but the vertical rod's center of mass lies directly along a line through the rotation axis when viewed from the side, creating zero torque contribution. Answer A is incorrect because it misapplies the equal-distance concept. Answer B incorrectly assumes both rods contribute equally. Answer C correctly identifies individual rod torque but misses that only one rod actually contributes. Answer D invents a non-existent "geometric amplification" effect. Remember: always visualize the force direction relative to the rotation axis—gravity only creates torque when the center of mass isn't directly above or below the axis.
5. Three forces act on a rigid body that can rotate about a fixed point O. Force $\vec{F_1} = 8 \text{ N}$ acts at position $\vec{r_1} = (3, 0) \text{ m}$ in the $+y$ direction. Force $\vec{F_2} = 6 \text{ N}$ acts at position $\vec{r_2} = (0, 4) \text{ m}$ in the $-x$ direction. Force $\vec{F_3} = 10 \text{ N}$ acts at position $\vec{r_3} = (-2, -1) \text{ m}$ in the $+x$ direction. What is the magnitude of the net torque about point O?
   Answer: $58 \text{ N⋅m}$ Using $\vec{\tau} = \vec{r} \times \vec{F}$, we calculate each torque component in the z-direction: $\tau_1 = r_{1x} F_{1y} - r_{1y} F_{1x} = (3)(8) - (0)(0) = 24 \text{ N⋅m}$. $\tau_2 = r_{2x} F_{2y} - r_{2y} F_{2x} = (0)(0) - (4)(-6) = 24 \text{ N⋅m}$. $\tau_3 = r_{3x} F_{3y} - r_{3y} F_{3x} = (-2)(0) - (-1)(10) = 10 \text{ N⋅m}$. Total torque: $\tau_{total} = 24 + 24 + 10 = 58 \text{ N⋅m}$. Choice A omits one force contribution. Choice B uses incorrect cross product calculation. Choice D makes sign errors in the cross product components.
6. A uniform rod of mass $M$ and length $L$ is pivoted at a point located at distance $L/4$ from one end. What is the torque due to the rod's weight about the pivot point?
   Answer: $\frac{MgL}{4}$ because the center of mass creates torque with a quarter-length moment arm from the pivot When analyzing torque problems involving extended objects like rods, you need to remember that gravitational force acts at the center of mass, and torque equals force times the perpendicular distance from the pivot to where that force acts. For this uniform rod, the center of mass is located at the geometric center, which is $L/2$ from either end. Since the pivot is at distance $L/4$ from one end, the center of mass is $L/2 - L/4 = L/4$ away from the pivot point. The gravitational force $Mg$ acts downward at this center of mass location, creating a torque of $Mg \times L/4 = \frac{MgL}{4}$. Looking at the wrong answers: Choice A incorrectly states the moment arm is $L/8$, but we calculated it's actually $L/4$. The reasoning about offset is correct, but the distance calculation is wrong. Choice C makes the classic error of using $L/2$ as the moment arm—this would be correct if the pivot were at one end of the rod, but here the pivot is offset. Choice D suggests using the full length $L$, which has no physical basis since the moment arm is the perpendicular distance from pivot to center of mass, not the total rod length. Study tip: In rotational problems, always locate the center of mass first, then measure the perpendicular distance from the pivot to that point. Don't be tempted to use the total length of the object—the moment arm is specifically the distance from pivot to where the force acts.
7. A wrench is used to tighten a bolt. If the applied force is doubled while keeping the same perpendicular distance from the axis of rotation, and simultaneously the distance is halved, what happens to the torque?
   Answer: The torque remains unchanged because doubling force and halving distance cancel each other When you encounter torque problems, remember that torque is the rotational equivalent of force and follows the equation $\tau = rF\sin\theta$, where r is the distance from the axis of rotation, F is the applied force, and θ is the angle between them. For perpendicular forces, $\sin\theta = 1$, so $\tau = rF$. Let's work through this step-by-step. Initially, you have some torque $\tau_1 = r_1F_1$. After the changes, the new force becomes $F_2 = 2F_1$ (doubled) and the new distance becomes $r_2 = \frac{r_1}{2}$ (halved). The new torque is: $\tau_2 = r_2F_2 = \frac{r_1}{2} \times 2F_1 = r_1F_1 = \tau_1$ The torque remains exactly the same because the effects cancel out perfectly. Choice A is correct—doubling the force while halving the distance results in no net change to torque due to their multiplicative relationship. Choice B incorrectly suggests force "dominates" over distance, but both variables have equal weight in the torque equation—neither is more important than the other. Choice C makes the opposite error, claiming distance reduction dominates, which again misunderstands that r and F contribute equally to the product. Choice D incorrectly treats the changes as additive rather than multiplicative, suggesting both changes increase torque when they actually oppose each other. Study tip: Remember that torque depends on the product of force and distance. When one doubles and the other halves, they always cancel out completely—this pattern appears frequently in rotational mechanics problems.
8. A seesaw of length $L = 4.0 \text{ m}$ and negligible mass is supported at its center. A child of mass $m_1 = 30 \text{ kg}$ sits at one end. Where should a second child of mass $m_2 = 20 \text{ kg}$ sit to balance the seesaw?
   Answer: $3.0 \text{ m}$ from the center on the opposite side, because the lighter child needs maximum leverage When you encounter a seesaw or lever problem, you're dealing with rotational equilibrium, where the net torque around the pivot point must be zero. Torque equals force times distance from the pivot, so $\tau = F \times d$. For the seesaw to balance, the torques on both sides must be equal: $m_1 g d_1 = m_2 g d_2$. Since gravity cancels out, this simplifies to $m_1 d_1 = m_2 d_2$. The first child (30 kg) sits at one end, so $d_1 = 2.0 \text{ m}$ from the center. To find where the second child should sit: $30 \times 2.0 = 20 \times d_2$, which gives $d_2 = 3.0 \text{ m}$ from the center on the opposite side. This confirms answer B is correct—the lighter child needs maximum leverage to balance the heavier child. Let's examine why the other options fail: A suggests 1.5 m, but this would create insufficient torque ($20 \times 1.5 = 30$ N⋅m versus the required 60 N⋅m). C proposes 2.0 m simply because it matches the first child's distance, ignoring the mass difference entirely—this would leave the heavier side unbalanced. D suggests 1.0 m with flawed reasoning about "reduced leverage," but this creates even less torque (20 N⋅m), making the imbalance worse. Remember: in equilibrium problems, heavier objects closer to the pivot can be balanced by lighter objects farther away. The key relationship is that mass and distance are inversely proportional for balance.
9. A solid disk rotates about its center. Two identical forces of magnitude $F$ are applied tangentially to the disk's rim, but in opposite directions at points separated by $90°$. What is the net torque on the disk?
   Answer: $2FR$ because both forces contribute to rotation in the same rotational direction despite opposite application When analyzing rotational motion, you need to understand that torque is a vector quantity with both magnitude and direction. The key insight here is recognizing how forces applied at different positions contribute to the overall rotational effect. Each force creates a torque of magnitude $\tau = FR$ about the center, where $R$ is the radius. Since both forces are applied tangentially to the rim, they both create maximum torque. The crucial point is determining the rotational direction each force produces. Even though the forces point in opposite directions, their positions on the disk (separated by 90°) mean they both contribute to rotation in the same sense - either both clockwise or both counterclockwise. Picture this: if one force at the "3 o'clock" position points up and another at "6 o'clock" points right, both create clockwise rotation. Since torques in the same rotational direction add together, the net torque is $FR + FR = 2FR$. Answer A incorrectly assumes the opposite force directions cancel out, but this ignores that torque depends on position, not just force direction. Answer B suggests an unequal contribution, but both forces create identical torques since they have the same magnitude and are applied at the same radius. Answer D attempts to use vector addition principles, but this geometric factor doesn't apply here since we're adding scalar torque magnitudes in the same rotational sense. Remember: when analyzing torque problems, always determine the rotational sense (clockwise or counterclockwise) that each force creates, then add or subtract accordingly. Position matters as much as force direction.
10. A mechanic applies a force of $80 \text{ N}$ to a wrench handle that makes an angle of $25°$ with the direction perpendicular to the bolt. If the effective lever arm for maximum torque would be $0.30 \text{ m}$, what is the actual torque applied to the bolt?
    Answer: $21.8 \text{ N⋅m}$ The actual torque is reduced from the maximum by the cosine of the angle away from perpendicular: $\tau = F \times r \times \cos(25°) = 80 \times 0.30 \times \cos(25°) = 80 \times 0.30 \times 0.906 = 21.8 \text{ N⋅m}$. Choice B uses the full perpendicular force without the angle correction. Choice C incorrectly uses sine instead of cosine. Choice D makes an error in the geometric relationship and uses an incorrect trigonometric function.
11. A uniform square plate with side length $a$ is pivoted at one corner. A force $F$ is applied perpendicular to one edge at its midpoint. What is the torque about the pivot corner?
    Answer: $\frac{Fa}{2}$ because the moment arm is the perpendicular distance from pivot to force application point When analyzing torque problems, you need to identify the perpendicular distance from the pivot point to the line of action of the force—this is called the moment arm. Don't confuse this with the total distance between two points. Since the force is applied perpendicular to the edge at its midpoint, the moment arm is simply the perpendicular distance from the corner pivot to that midpoint along the edge. This distance is $\frac{a}{2}$ (half the side length). The torque is then $\tau = F \times \frac{a}{2} = \frac{Fa}{2}$, making answer A correct. The wrong answers stem from common misconceptions about moment arms. Answer B ($\frac{Fa\sqrt{2}}{2}$) incorrectly uses the straight-line distance from corner to midpoint as the moment arm. While this diagonal distance is indeed $\frac{a\sqrt{2}}{2}$, it's not the perpendicular distance to the force's line of action. Answer C ($Fa$) mistakenly uses the full side length, perhaps thinking the entire edge somehow contributes to the moment arm. Answer D ($\frac{Fa\sqrt{5}}{2}$) appears to involve an incorrect geometric calculation—the Pythagorean theorem gives $\frac{a\sqrt{2}}{2}$ for the corner-to-midpoint distance, not $\frac{a\sqrt{5}}{2}$. Remember: the moment arm is always the shortest perpendicular distance from the pivot to the force's line of action, not the distance between physical points. When forces are applied perpendicular to edges or surfaces, this often simplifies to basic geometric measurements along those surfaces.
12. A pulley of radius $r = 0.15 \text{ m}$ has two masses hanging from opposite sides: $m_1 = 4.0 \text{ kg}$ and $m_2 = 6.0 \text{ kg}$. Assuming the rope doesn't slip and the pulley is massless, what is the net torque on the pulley about its center?
    Answer: $0 \text{ N}\cdot\text{m}$ because the pulley is massless and the rope tensions create equal and opposite torques When analyzing torque problems involving pulleys, you need to carefully consider what forces actually act on the pulley itself. Torque is calculated as $\tau = rF$, where the force must be applied to the object in question. In this system, the pulley experiences tension forces from the rope on both sides. Since the rope doesn't slip and is massless, the tension is uniform throughout. For the hanging masses to be in motion (or equilibrium), the rope tension $T$ must be the same on both sides of the pulley. This tension creates two torques on the pulley: one clockwise from the heavier mass side and one counterclockwise from the lighter mass side. Since both have magnitude $T \times r$ but opposite directions, they cancel completely, giving zero net torque. Option A is correct because the rope tensions create equal and opposite torques that sum to zero, regardless of the pulley's mass (which is zero anyway). Option B incorrectly assumes the mass difference directly creates torque on the pulley. The mass difference affects the system's acceleration, but the pulley still experiences equal tensions. Option C falls into the same trap, suggesting the heavier mass somehow creates more torque on the pulley. The masses create different weights, but the rope tension acting on the pulley is uniform. Option D wrongly assumes both tension forces add together rather than opposing each other rotationally. Key insight: Always distinguish between forces acting on the pulley versus forces in the broader system. The pulley feels only the rope tensions, which are always equal and opposite.
13. A bicycle wheel of radius $R = 0.35 \text{ m}$ experiences a constant tangential force of $F = 18 \text{ N}$ applied at its rim. Simultaneously, a frictional torque of $\tau_f = 4.0 \text{ N}\cdot\text{m}$ opposes the rotation. What is the net torque on the wheel?
    Answer: $2.3 \text{ N}\cdot\text{m}$ because the net effect accounts for both the driving and opposing torques When you encounter rotational motion problems involving multiple torques, you need to calculate each torque separately and then find the net effect by considering their directions. First, calculate the torque from the applied force. Since torque equals force times the perpendicular distance from the axis of rotation, the applied torque is $\tau_{applied} = F \times R = 18 \text{ N} \times 0.35 \text{ m} = 6.3 \text{ N}\cdot\text{m}$. This torque acts in the direction of intended rotation. The frictional torque of $4.0 \text{ N}\cdot\text{m}$ opposes this rotation, so it acts in the opposite direction. To find the net torque, subtract the opposing torque from the applied torque: $\tau_{net} = 6.3 \text{ N}\cdot\text{m} - 4.0 \text{ N}\cdot\text{m} = 2.3 \text{ N}\cdot\text{m}$. Choice A gives $6.3 \text{ N}\cdot\text{m}$, which is only the applied torque—it ignores the frictional torque entirely. Choice C states $10.3 \text{ N}\cdot\text{m}$, which incorrectly adds the two torques instead of subtracting them (since they oppose each other). Choice D gives $1.8 \text{ N}\cdot\text{m}$, which appears to be a calculation error, possibly from incorrectly computing the applied torque. Choice B correctly gives $2.3 \text{ N}\cdot\text{m}$ because it properly accounts for both the driving torque and the opposing frictional torque. Study tip: In rotational problems, always identify the direction of each torque first. Torques in the same direction add together, while opposing torques subtract. Draw a quick diagram showing rotation directions to avoid sign errors.
14. A horizontal platform can rotate about a vertical axis through its center. Two people stand on opposite edges of the platform, each at distance $R = 2.0 \text{ m}$ from the center. Person A ($m_A = 60 \text{ kg}$) pushes tangentially with force $F_A = 80 \text{ N}$, while Person B ($m_B = 80 \text{ kg}$) pushes tangentially with force $F_B = 60 \text{ N}$ in the opposite rotational direction. What is the net torque on the platform?
    Answer: $40 \text{ N}\cdot\text{m}$ in the direction of Person A's push because their force creates the dominant torque When you encounter rotational motion problems involving forces, focus on torque - the rotational equivalent of force. Torque equals force times the perpendicular distance from the rotation axis, and torques in opposite directions subtract from each other. Here, both people apply tangential forces at distance $R = 2.0 \text{ m}$ from the center, creating torques in opposite directions. Person A creates torque $\tau_A = F_A \times R = 80 \text{ N} \times 2.0 \text{ m} = 160 \text{ N}\cdot\text{m}$. Person B creates torque $\tau_B = F_B \times R = 60 \text{ N} \times 2.0 \text{ m} = 120 \text{ N}\cdot\text{m}$ in the opposite direction. The net torque is the difference: $160 - 120 = 40 \text{ N}\cdot\text{m}$ in Person A's direction, confirming answer A. Answer B incorrectly adds the torques as if they're in the same direction, but the problem states they push "in the opposite rotational direction." Answer C makes the common mistake of thinking mass affects the torque calculation directly - but torque only depends on force and distance, not the mass of the person applying the force. Answer D simply uses the force difference (20 N) times distance, forgetting that you must calculate each torque separately before finding the net effect. Remember: in rotational problems, always identify the rotation axis, calculate individual torques using $\tau = F \times r$, then add torques vectorially (same direction adds, opposite directions subtract). The masses of people applying forces don't directly affect torque calculations.
15. A mechanic applies a force $F = 200 \text{ N}$ to a wrench handle that makes an angle of $45°$ with the wrench itself. If the distance from the bolt center to the point of force application is $0.25 \text{ m}$, what is the torque about the bolt?
    Answer: $35 \text{ N}\cdot\text{m}$ because the perpendicular component of force creates the effective rotational moment When you encounter torque problems involving angled forces, remember that only the perpendicular component of the force contributes to rotation. Torque is calculated as $\tau = r \times F_{\perp}$, where $F_{\perp}$ is the component of force perpendicular to the lever arm. Here, the force makes a 45° angle with the wrench, so you need the perpendicular component: $F_{\perp} = F \sin(45°) = 200 \times \frac{\sqrt{2}}{2} = 141.4 \text{ N}$. The torque is then $\tau = r \times F_{\perp} = 0.25 \times 141.4 = 35.4 \text{ N}\cdot\text{m}$, which rounds to 35 N·m. This confirms answer A is correct. Answer B incorrectly uses the full 200 N force without considering the angle: $200 \times 0.25 = 50 \text{ N}\cdot\text{m}$. This ignores that only the perpendicular component creates torque. Answer C suggests the torque is simply halved due to the angle, giving $\frac{200 \times 0.25}{2} = 25 \text{ N}\cdot\text{m}$. This reflects a common misconception that any angle reduces torque by a simple fraction rather than using trigonometry. Answer D calculates $200 \times 0.25 \times \sqrt{2} = 71 \text{ N}\cdot\text{m}$, incorrectly applying the $\sqrt{2}$ factor. This suggests the misunderstanding that angled forces somehow enhance torque rather than reduce the effective component. Study tip: For torque problems, always identify whether the given angle is between the force and lever arm, then use sine to find the perpendicular component. Draw a diagram showing the force vector and its perpendicular component to visualize the geometry clearly.
16. A flywheel experiences three torques simultaneously: a driving torque $\tau_1 = 25 \text{ N}\cdot\text{m}$, a frictional torque $\tau_2 = -8 \text{ N}\cdot\text{m}$, and a load torque $\tau_3 = -12 \text{ N}\cdot\text{m}$. If the flywheel has a moment of inertia $I = 0.5 \text{ kg}\cdot\text{m}^2$, what is its angular acceleration?
    Answer: $10 \text{ rad/s}^2$ because the net driving torque creates significant angular acceleration despite opposing forces When you encounter rotational dynamics problems with multiple torques, you need to apply Newton's second law for rotation: $\tau_{net} = I\alpha$. The key is finding the net torque by algebraically summing all individual torques, respecting their signs. First, calculate the net torque. The driving torque is positive ($+25 \text{ N}\cdot\text{m}$), while both resistance torques are negative: $\tau_{net} = 25 + (-8) + (-12) = 5 \text{ N}\cdot\text{m}$. Then apply the rotational equation: $\alpha = \frac{\tau_{net}}{I} = \frac{5}{0.5} = 10 \text{ rad/s}^2$. Choice A correctly identifies this value and provides sound reasoning—the net driving torque does create significant acceleration despite the opposing forces. Choice B incorrectly calculates $50 \text{ rad/s}^2$ by using only the driving torque ($25/0.5$) and ignoring the resistance torques entirely. This represents a common error of not considering all forces acting on the system. Choice C gets $5 \text{ rad/s}^2$, which is actually the net torque value, not the angular acceleration—this shows confusion between torque and angular acceleration units. Choice D yields $34 \text{ rad/s}^2$ by incorrectly adding the magnitudes of all torques without considering signs: $(25+8+12)/0.5 = 90/0.5 = 45$, though the exact calculation differs. Remember: in rotational problems, always sum all torques algebraically first (positive and negative), then divide by moment of inertia. Don't just focus on the largest torque—the net effect determines the motion.

Start this quiz