10 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 10
A block of candle wax keeps its shape when it is set on a table. When the wax is melted, the same wax molecules form a liquid that flows and spreads out across the table. Which statement best accounts for the solid block keeping its shape?
Answer and reasoning
AEach molecule of the solid wax is itself solid, while each molecule of melted wax is liquid. A student who thinks each particle has the properties of the whole substance picks this. Solid and melted wax contain the same molecules; being solid is a property of the system, set by how the molecules interact, not a property of any one molecule.
BThe solid block has more mass than the liquid it melts into, so it holds together much better. A student who thinks mass changes with state picks this. Melting neither adds nor removes molecules, so the block and the liquid it becomes have the same mass. The difference in behavior comes from the interactions between the molecules.
CThe block is one solid object, so the molecules inside it play no part in its shape. A student who thinks a solid object is not a system of parts picks this. The block can be modeled as a single object for some questions, such as how it slides, but its ability to keep its shape is explained only by the interactions between its molecules.
DNeighboring molecules in the solid interact strongly enough to hold one another in place.Correct The wax block is a system of molecules, and its rigidity is a property of that system that comes from how the molecules interact. In the solid, each molecule is held near a fixed position by its neighbors; in the liquid, the same molecules can slide past one another, so the liquid flows.
A student wants to find the time Earth takes to orbit the Sun once. The student models Earth, with its oceans, its atmosphere and its slowly churning interior, as a single object. Which evaluation of this modeling choice is correct?
Answer and reasoning
AAppropriate: the orbital period depends only on how Earth moves as a whole.Correct A system can be treated as a single object when the properties and interactions of its parts do not matter for the behavior being analyzed. Earth's orbital period depends only on the motion of Earth as a whole, so the motions of its oceans, air and interior can be ignored, however large Earth is.
BNot appropriate: Earth is much too large to be treated as one single object. A student who thinks only physically small systems can be single objects picks this. Size is not the test: the question concerns only Earth's motion as a whole, which does not depend on its parts, so the object model works for a planet.
CNot appropriate: Earth's oceans, air and interior move relative to one another. A student who thinks a system with many or moving parts must be analyzed part by part picks this. Those internal motions do not affect how long Earth as a whole takes to go once around the Sun, so they can be ignored for this question.
DAppropriate: any system can be modeled as a single object, whatever the question. A student who thinks the object model always works picks this. The conclusion is right for this question, but the reason is wrong: to explain the tides or Earth's changing shape, the parts matter and Earth cannot be a single object. The model is justified only because this question does not involve the parts.
Logs burn in an open fireplace. Smoke and hot gases rise up the chimney, and a little ash is left behind. The system is the logs and whatever remains of them in the fireplace. Which statement about the mass of this system is correct?
Answer and reasoning
AIt stays constant, since mass is conserved in every physical and chemical change. A student who reads 'mass is conserved' as 'every system keeps its mass' picks this. Mass is conserved overall, but this system is open: the gases carry mass out of it, so its mass decreases. Only a system that includes all the gases keeps a constant mass.
BIt stays constant, since the gases that leave the fireplace have no mass of their own. A student who thinks gases have no mass picks this. A gas is matter made of particles, each with mass, and the gases leaving the fireplace carry away most of the mass of the wood.
CIt decreases, as the gases produced carry mass across the system's boundary.Correct Burning turns most of the wood, together with oxygen from the air, into gases. Those gases are matter: as they rise up the chimney they carry mass out of the system, so the mass of what remains in the fireplace decreases. The total mass of the remains plus all the gases and smoke is unchanged.
DIt decreases, as the burning destroys the wood's matter and its mass disappears. A student who thinks burning destroys matter picks this. The mass does leave the system, but it is not destroyed: the matter of the wood is rearranged into gases and smoke, which carry that mass into the environment.
Two small objects, P and Q, each of mass m, move along the x-axis. P's position is xP = bt², where b is a positive constant, and Q moves with constant velocity. Q is then replaced by an object of mass 3m that moves in exactly the same way as Q did. By what factor does the acceleration of the center of mass of the system change?
Answer and reasoning
A×1.00 A student who takes the system's acceleration to be that of its accelerating part picks this: P still accelerates at 2b. The system's acceleration is the mass-weighted average, and Q's larger mass, with zero acceleration, pulls that average down.
B×3.00 A student who thinks a center-of-mass quantity is proportional to any one mass picks this. Q's mass appears in the total mass in the denominator, and Q's acceleration is zero, so tripling Q's mass reduces acm.
C×1.50 A student who weights each acceleration by the other object's mass picks this: after the change, (3m·2b + m·0)/(4m) = 1.5b, compared with b before. Each acceleration must be weighted by its own object's mass: P's 2b by m.
D×0.50Correct aP = d²xP/dt² = 2b and aQ = 0, and acm is the mass-weighted average of the accelerations. Before: (m·2b)/(2m) = b. After: (m·2b)/(4m) = b/2. The extra mass of the non-accelerating object halves the acceleration of the center of mass, although P accelerates exactly as before.
Working aP = d²(bt²)/dt² = 2b; aQ = 0. Differentiating xcm twice: acm = (mP aP + mQ aQ)/(mP + mQ). Before: (m·2b + m·0)/(2m) = b. After: (m·2b + 3m·0)/(4m) = b/2. Factor ×0.50. (P's acceleration taken as the system's: 2b both times, ×1.00. Tripling one mass taken to triple acm: ×3.00. Masses swapped in the weighting: (3m·2b + m·0)/(4m) = 1.5b after and b before, ×1.50.)
Two carts of equal mass roll at the same speed toward a wall along a level track. Cart 1 has a stiff steel spring bumper and bounces back from the wall; cart 2 has a soft clay bumper and stops at the wall. A student analyzes each cart's approach to the wall and then explains why the carts behave differently at the wall. Which statement about modeling the carts is correct?
Answer and reasoning
AEach can be one object at every stage, since any system can be modeled as one object. A student who thinks the object model always works picks this. A single object has no internal structure, so the model cannot explain why one cart bounces and the other stops; that difference comes from how each bumper's parts interact.
BNeither can be one object at any stage, since each cart has many parts that move. A student who thinks a system with many or moving parts can never be one object picks this. While a cart rolls toward the wall, its spinning wheels and other parts do not affect its motion as a whole, so it can be modeled as a single object.
CEach can be one object on the approach, but not to explain their behavior at the wall.Correct On the approach, each cart's internal structure does not affect its motion, so each can be treated as a single object. At the wall the carts behave differently because their bumpers are built differently: the spring is compressed and pushes back, while the clay deforms and stays deformed. Explaining that requires each cart's internal structure.
DOnly cart 2 needs its internal structure, since cart 1's steel bumper is a solid piece. A student who thinks a solid object's internal interactions play no part in its behavior picks this. The steel spring bounces the cart back because of how its parts interact as it is compressed and springs back; cart 1's internal structure matters at the wall as much as cart 2's.
A sealed plastic bottle of liquid water is placed in a freezer, and the water freezes. The system is the water in the bottle. Which statement about the effect of lowering the temperature of the system's surroundings is correct?
Answer and reasoning
AIts substructure stays the same, since the system still contains the same set of molecules. A student who thinks the parts of a system fix its arrangement picks this. The molecules are the same, but the way they are arranged and bound together changes when the water freezes: that is a change of substructure.
BThe same molecules become bound into a rigid arrangement, so its substructure changes.Correct The freezer's temperature is a variable external to the system. Lowering it changes how the water molecules are arranged and bound together: they become locked into the rigid structure of ice. The system still contains the same molecules, but its substructure has changed.
CIts mass increases as it freezes, since ice is a more solid form of matter than liquid water. A student who thinks mass changes with state picks this. The bottle is sealed, so no molecules enter or leave, and the mass of the system is unchanged; only the arrangement of the molecules changes.
DEach molecule becomes solid, while the arrangement of the molecules stays as it was. A student who thinks each particle takes on the properties of the substance picks this. A molecule is not solid or liquid; freezing changes how the molecules are arranged and bound to their neighbors, which is exactly what this option denies.
The figure shows a thin square plate of uniform thickness. Its left half is steel and its right half is aluminum, which is less dense than steel; each half is uniform. Which labeled point could be the center of mass of the plate?
Answer and reasoning
APoint QCorrect The mass is symmetric about the horizontal line through the middle, so the center of mass is on that line. The steel half is more massive, so the center of mass is to the left of the center P, toward the steel. The aluminum half also has mass, so the center of mass is between the center of the steel half, R, and the center of the plate: point Q.
BPoint P A student who places the center of mass at the geometric center of the outline picks this. The square's shape is symmetric about its vertical midline, but its mass is not: the steel half is more massive, so the center of mass is to the left of P.
CPoint R A student who places the center of mass at the center of the most massive part picks this. R is the center of mass of the steel half alone; the aluminum half also has mass and pulls the center of mass of the whole plate to the right of R.
DPoint S A student who puts the center of mass nearer the lighter part picks this. Each half's position is weighted by its own mass, so the more massive steel half pulls the center of mass toward itself, to the left of P.
Working Mass is distributed symmetrically about the horizontal line through the middle, so the center of mass lies on that line. It is not symmetric about the vertical line: the steel half has more mass, so the center of mass is left of the center P. Treat each half as a point mass at its own center: xcm is a weighted average of the two half-centers, so it lies between P and R (at R only if aluminum had no mass). With steel about 2.9 times as dense as aluminum, xcm ≈ 0.5 × (distance from P to R) to the left of P: Q.
The figure shows three small objects, A, B and C, in the xy-plane, each labeled with its mass. What is the x-coordinate of the center of mass of the three-object system?
Answer and reasoning
A1.0 m A student who averages the x-coordinates without weighting them by mass picks this: (−2 + 1 + 4)/3 m. C, the most massive object, pulls the center of mass toward itself, so the weighted average is larger.
B2.8 m A student who enters A's coordinate as a positive distance picks this: [(2.0)(2) + (1.0)(1) + (3.0)(4)]/6.0 m. A is on the negative side of the origin, so its term is (2.0 kg)(−2 m).
C1.5 mCorrect Each x-coordinate is weighted by its object's mass, with A's coordinate negative: xcm = [(2.0)(−2) + (1.0)(1) + (3.0)(4)] kg·m/(6.0 kg) = 9.0 kg·m/6.0 kg = 1.5 m. The y-coordinates do not enter the x-coordinate of the center of mass.
D3.0 m A student who divides Σ mi xi by the number of objects picks this: 9.0 kg·m/3. The sum must be divided by the total mass, 6.0 kg, which also makes the units come out in meters.
Working Read xA = −2 m, xB = 1 m, xC = 4 m. xcm = (Σ mi xi)/(Σ mi) = [(2.0)(−2) + (1.0)(1) + (3.0)(4)] kg·m/(6.0 kg) = (−4 + 1 + 12)/6.0 m = 1.5 m. (Plain average: (−2 + 1 + 4)/3 = 1.0 m. Sign of xA dropped: (4 + 1 + 12)/6.0 = 2.8 m. Divided by the number of objects: 9/3 = 3.0 m.)
A thin rod of length L lies along the x-axis from x = 0 to x = L. Its linear mass density is λ(x) = λ₀x/L, where λ₀ is a positive constant. Where is the rod's center of mass?
Answer and reasoning
A0.50L A student who puts the center of mass at the geometric middle picks this. That is right only for a uniform rod; here the density grows from 0 to λ₀ along the rod, so more of the mass is in the right-hand half.
B0.67LCorrect Each element dm = λ(x) dx is at position x. The total mass is ∫₀ᴸ (λ₀x/L) dx = λ₀L/2, and ∫ x dm = ∫₀ᴸ (λ₀x²/L) dx = λ₀L²/3. Dividing, xcm = 2L/3 ≈ 0.67L: toward the denser end, as it must be.
C0.71L A student who puts the center of mass where there is equal mass on each side picks this: ∫₀ˣ λ dx = λ₀L/4 gives x = L/√2. The center of mass is where the mass-weighted positions balance, ∫ x dm/∫ dm = 2L/3.
D0.33L A student who weights each position by the density at the mirror-image point picks this, putting the center of mass toward the lighter end. Each element dm = λ(x) dx sits at its own x, so the center of mass is toward the denser end.
Working dm = λ dx. M = ∫₀ᴸ λ₀x/L dx = λ₀L/2. ∫ x dm = ∫₀ᴸ λ₀x²/L dx = λ₀L²/3. xcm = (λ₀L²/3)/(λ₀L/2) = 2L/3 ≈ 0.67L. (Uniform rod assumed: 0.50L. Equal mass on each side: λ₀x²/(2L) = λ₀L/4 gives L/√2 ≈ 0.71L. Weighted by the density at the mirror point, ∫ x λ₀(L − x)/L dx divided by M: L/3 ≈ 0.33L.) Checked with sympy.
A gymnast leaves a trampoline twice with the same velocity of her center of mass. In jump 1 she stays straight; in jump 2 she pulls into a tight tuck and somersaults. In each jump her center of mass returns to the height at which she left the trampoline, and air resistance is negligible. How do her times in the air, t₁ and t₂, compare?
Answer and reasoning
At₁ = t₂, since her center of mass moves as the same projectile in both jumpsCorrect Modeled as a single object at her center of mass, the gymnast is a projectile in both jumps. The center of mass leaves with the same velocity and returns to the same height, so it follows the same path and takes the same time. Tucking and somersaulting move her parts around the center of mass, not the center of mass itself.
Bt₂ > t₁, since pulling into a tuck lifts her center of mass and her flight path A student who thinks changing body shape changes the path of the center of mass picks this. Tucking pulls her legs up toward her center of mass, but the center of mass itself keeps moving as the same projectile.
Ct₁ > t₂, since some of her motion is used up in turning the somersaults in the air A student who thinks spinning uses up some of the motion picks this. With the same launch velocity of the center of mass, the center of mass rises to the same height and lands at the same time, whether she spins or not.
DThey cannot be compared, since a tumbling body is not a single projectile A student who thinks a tumbling body cannot be modeled as a single object picks this. However her body rotates and changes shape, her center of mass moves as a single projectile would, so the times can be compared and are equal.
In preparation: 0 of 10 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
2.1.A.1 System Fix
System
A chosen object or collection of objects that is analyzed together; everything outside it is the environment. A system has properties, such as its total mass and the location of its center of mass, that belong to the collection as a whole.
Internal and external interactions
Interactions between objects inside a system are internal; interactions between a part of the system and something outside it are external. The properties of a system, such as rigidity, come from its internal interactions.
Students often think A property of a whole substance, such as being solid, hard or liquid, is also a property of each of its particles. In fact No. Solid wax and melted wax are made of the same molecules. Holding a shape is a property of the system of molecules, set by how they interact and are arranged, not a property of each molecule.
Students often think The mass of a substance changes when it changes state; for example, a solid has more mass than the same substance as a liquid. In fact No. Melting and freezing change how the molecules are arranged and how they interact, not how many there are, so a closed sample keeps the same mass.
2.1.A.2 Single-object model Fix
Single-object model
Treating a whole system as one object with no internal structure. It may be used when the properties and interactions of the parts do not matter for the behavior being analyzed, whatever the system's size.
Students often think A system can be treated as a single object only if it is physically small, such as a ball or a cart; a very large system, such as a planet or a ship, cannot. In fact No. What matters is whether the properties or interactions of the parts affect the behavior being analyzed, not the system's size. Earth can be treated as a single object when finding how long it takes to orbit the Sun.
Students often think A system made of many parts, or whose parts move about, cannot be treated as a single object. In fact No. If the parts' properties and interactions do not matter for the behavior being analyzed, the whole system can be treated as a single object, however many parts it has and however they move about.
2.1.A.3 Open system Fix
Open system
A system whose boundary matter crosses, so that its mass changes; energy can also be transferred across a system's boundary by interactions with the environment.
Students often think Because mass is conserved, the mass of any chosen system stays constant, even when matter leaves or enters it. In fact No. Mass is conserved overall, but a system's mass changes when matter crosses its boundary. The mass of a burning log decreases as gases leave it; the mass of the log plus all the gases and ash it produces is constant.
Students often think Gases have no mass, or negligible mass, so a gas leaving a system does not change the system's mass. In fact Yes. A gas is matter made of particles, each with mass. Gases leaving a burning log carry mass out of the log.
2.1.A.4 Velocity and acceleration of the center of mass Fix
Velocity and acceleration of the center of mass
Differentiating xcm = (Σ mi xi)/(Σ mi) with respect to time for a system of constant mass gives vcm = (Σ mi vi)/(Σ mi) and acm = (Σ mi ai)/(Σ mi): mass-weighted averages of the parts' velocities and accelerations.
Students often think The motion of a system is the motion of one particular part of it, such as the part that is pushed, the part that accelerates or the part that is easiest to see. In fact No. The motion of the system as a whole is the motion of its center of mass, which in general differs from the motion of any one part, such as the part that is pushed or the part that speeds up.
2.1.A.5 Idea 5 Fix
2.1.A.6 Substructure of a system Fix
Substructure of a system
How the parts of a system are arranged and bound together. The same parts can form a different substructure when variables outside the system, such as temperature or how hard the system is pushed, are changed.
Students often think Once a system is chosen, its substructure is fixed: if it is made of the same parts, their arrangement and the way they are bound together cannot change. In fact No. Changing a variable external to the system, such as the temperature of its surroundings, can change how the parts are arranged and bound together, even though the same parts make up the system.
2.1.B.1 Center of mass Fix
Center of mass
The mass-weighted average position of a system's mass. For a uniform object with a line of symmetry it lies on that line; it need not lie within the material of the object. SI unit of its coordinates: m.
Students often think The center of mass is at the geometric center of an object's shape or outline, whatever the distribution of its mass. In fact No. The center of mass is at the geometric center when the mass is distributed symmetrically about that point, but not in general otherwise. For example, a square plate that is steel on one half and aluminum on the other has its center of mass on the steel side of its geometric center.
Students often think The center of mass of an object must lie within its material. In fact No. The center of mass is an average position and can lie in empty space, as at the center of a ring, of a horseshoe or of a body bent into an arch.
2.1.B.2 Center-of-mass coordinate of discrete objects Fix
Center-of-mass coordinate of discrete objects
xcm = (Σ mi xi)/(Σ mi), with each xi a signed coordinate measured from one chosen origin and each extended object's mass placed at that object's own center of mass. The same form holds for y and z.
Students often think The center of mass of a system is at its most massive part (or at that part's center), so the system moves as that part does. In fact No. The center of mass is the mass-weighted average position of all the parts, so every part with mass pulls it toward itself. For two parts it lies nearer the more massive one, but in general it is not at the most massive part, and it does not in general move as that part does.
Students often think The center of mass (and its velocity and acceleration) is the plain average of the parts' positions (or velocities and accelerations); the masses do not matter. In fact No. It is the mass-weighted average: xcm = (Σ mi xi)/(Σ mi). The plain average gives the right answer only when all the masses are equal. The same holds for the velocity and acceleration of the center of mass.
2.1.B.3 Differential mass element, dm Fix
Differential mass element, dm
A piece of an object small enough that all of its mass can be treated as being at one position r⃗. The center of mass of a continuous object is r⃗cm = (∫ r⃗ dm)/(∫ dm), where ∫ dm is the total mass.
Linear mass density, λ
Mass per unit length at a point of a rod or other linear object: λ = dm/dℓ, the derivative of the mass m(ℓ) of the part between one end and position ℓ. SI unit: kg/m. For a uniform rod λ = M/L everywhere; for a nonuniform rod it varies along the rod.
Area and volume mass densities
Surface mass density σ (kg/m²) and volume mass density ρ (kg/m³). If a density function is given, the total mass is found by integrating it over the object: M = ∫ λ dℓ, ∫ σ dA or ∫ ρ dV. For a sphere with ρ depending on r, dV = 4πr² dr is the volume of a thin shell.
Students often think The mass of a solid can be found by integrating its density over r with a slice that is not its volume element, such as ∫ ρ dr or the disk's ring area ∫ ρ 2πr dr for a sphere. In fact No. Each slice must have the right volume for the object: for a sphere, a thin shell of radius r and thickness dr has volume dV = 4πr² dr, so M = ∫ ρ(r) 4πr² dr. Integrating ρ dr alone, or ρ 2πr dr (the ring of a flat disk), does not give a mass.
Students often think The mass of a nonuniform object is its density at one point, such as the center, times its total volume. In fact No. Multiplying the density at one point (for example the center) by the whole volume treats the object as uniform. When the density varies, its integral over the volume is needed.
2.1.B.4 Modeling a system as an object at its center of mass Fix
Modeling a system as an object at its center of mass
A system can be represented as a single object, with the system's total mass, located at the system's center of mass. In free fall, a spinning or flexing body's center of mass moves just as a single thrown object would.
Students often think A spinning, tumbling or flexing body cannot be modeled as a single object: it has no single, definite center of mass, or its center of mass need not move as a projectile. In fact Yes, for the motion of its center of mass. However the parts rotate or rearrange, the center of mass of a body in free flight moves as a single thrown object would.
Students often think Rearranging the parts of a system in flight, such as a gymnast pulling into a tuck, changes the path of its center of mass and how long it stays in the air. In fact No. Tucking or stretching out rearranges the parts around the center of mass but does not change the motion of the center of mass of a body in free flight.
18 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 18
Two gliders, A and B, move toward each other along a level air track. The graph shows the position x of each glider as a function of time t; each line is labeled with that glider's mass. What is the velocity of the center of mass of the system of the two gliders during the interval shown? Velocities in the +x direction are positive.
Answer and reasoning
A−0.10 m/s A student who averages the two velocities without weighting them by mass picks this: (0.40 − 0.60)/2 m/s. Glider A has three times B's mass, so its velocity counts three times as much in the average.
B+0.15 m/sCorrect Read each velocity from its line's slope: vA = +0.40 m/s and vB = −0.60 m/s. The velocity of the center of mass is the derivative of xcm, the mass-weighted average of the velocities: [(3.0 kg)(+0.40 m/s) + (1.0 kg)(−0.60 m/s)]/(4.0 kg) = +0.15 m/s. Neither glider moves at this velocity: the parts behave differently from the system as a whole.
C+0.45 m/s A student who enters both velocities as positive speeds picks this: [(3.0)(0.40) + (1.0)(0.60)]/4.0 m/s. Glider B moves in the −x direction, so its velocity is −0.60 m/s and it reduces the center of mass's velocity.
D+0.40 m/s A student who thinks the system moves as its most massive part picks glider A's velocity, +0.40 m/s. Glider B's mass also counts: its motion in the −x direction makes the center of mass move more slowly than A.
Working From the slopes: vA = (2.0 m − 0)/(5.0 s) = +0.40 m/s and vB = (3.0 m − 6.0 m)/(5.0 s) = −0.60 m/s. Differentiating xcm = (mA xA + mB xB)/(mA + mB) with respect to time: vcm = (mA vA + mB vB)/(mA + mB) = [(3.0 kg)(+0.40 m/s) + (1.0 kg)(−0.60 m/s)]/(4.0 kg) = +0.15 m/s. Each glider moves at its own constant velocity, and the system's center of mass moves at a third velocity. (Plain average: (0.40 − 0.60)/2 = −0.10 m/s. Signs dropped: (1.2 + 0.6)/4.0 = +0.45 m/s. Most massive glider: +0.40 m/s.)
A high jumper clears the bar by arching her back, so that her body is draped over the bar in an upside-down U shape, with her head and arms hanging on one side and her legs on the other. Which statement about her center of mass, at the instant her hips are directly above the bar, can be correct?
Answer and reasoning
AIt is above the bar, since a center of mass lies inside the material of a body. A student who thinks the center of mass must lie within the material picks this. Like the center of a ring, the average position of a bent body can be in empty space: with her head, arms and legs hanging down on both sides, it is below the curve of her back.
BIt is just above the bar, since a person's center of mass stays fixed at the hips. A student who thinks a person's center of mass is a fixed point of the body picks this. The center of mass depends on posture: when her head, arms and legs hang down on both sides of the bar, the average position of her mass moves down, away from her hips.
CIt has no definite location, since her body is changing shape over the bar. A student who thinks a flexing body cannot be modeled with a single center of mass picks this. At every instant her body has a definite mass distribution and so a definite center of mass, the mass-weighted average position of her parts.
DIt is below the bar, since much of her mass hangs below the bar on both sides.Correct The center of mass is the mass-weighted average position of her body, and it need not lie in the body. With her head, arms and legs hanging below the bar on both sides, the mass-weighted average height can be below the bar, outside her body, even though each part of her body passes over the bar in turn.
A uniform rod of mass 2m and length L lies along the x-axis from x = 0 to x = L. A small block of mass m is fixed to the rod at x = L/4. Where is the center of mass of the rod–block system?
Answer and reasoning
Ax = 0.42LCorrect Treat the uniform rod as a point of mass 2m at its midpoint, x = L/2, and the block as a point of mass m at x = L/4. Then xcm = (2m·L/2 + m·L/4)/(3m) = 5L/12 ≈ 0.42L: between the two points, nearer the more massive rod's midpoint.
Bx = 0.75L A student who places the rod's mass at its end, x = L, picks this: (2m·L + m·L/4)/(3m) = 3L/4. A uniform rod's mass is centered at its midpoint, L/2, so that is where its mass 2m goes in the sum.
Cx = 0.38L A student who takes the plain average of the two positions, L/2 and L/4, picks this: 3L/8. The rod has twice the block's mass, so the center of mass is nearer the rod's midpoint than the plain average is.
Dx = 0.33L A student who weights each position by the other object's mass picks this: (m·L/2 + 2m·L/4)/(3m) = L/3, nearer the lighter block. Each position is weighted by its own object's mass.
Working Replace the uniform rod by a point of mass 2m at its midpoint, x = L/2. xcm = (2m·L/2 + m·L/4)/(2m + m) = (5L/4)/3 = 5L/12 ≈ 0.42L. (Rod's mass at its end x = L: (2mL + mL/4)/(3m) = 3L/4 = 0.75L. Plain average of L/2 and L/4: 3L/8 ≈ 0.38L. Each position weighted by the other object's mass: (m·L/2 + 2m·L/4)/(3m) = L/3 ≈ 0.33L.) Checked with sympy.
Two small blocks, P and Q, of equal mass are on a straight track, with P at x = 0 and Q at x = d. Q is replaced by a block of twice its mass at the same position. By what factor does the distance from P to the center of mass of the two-block system change?
Answer and reasoning
A×2.00 A student who thinks xcm is proportional to one mass picks this, doubling d/2 to d. Q's mass is also part of the total mass in the denominator, so xcm rises from d/2 only to 2d/3.
B×1.33Correct With P at the origin, xcm = mQ d/(mP + mQ). Doubling mQ changes it from md/(2m) = d/2 to 2md/(3m) = 2d/3, a factor of 4/3. The center of mass moves toward the more massive block, but by less than a factor of 2 because the total mass in the denominator also grows.
C×1.00 A student who thinks the center of mass depends only on the positions picks this. The center of mass is a mass-weighted average, so making Q more massive moves it toward Q.
D×0.67 A student who weights each position by the other block's mass picks this: xcm = mP d/(mP + mQ) falls from d/2 to d/3. Each position is weighted by its own block's mass, so the center of mass moves toward the more massive block.
Working xcm = mQ d/(mP + mQ). Before: md/(2m) = d/2. After: 2md/(3m) = 2d/3. Factor (2d/3)/(d/2) = 4/3 ≈ ×1.33. (Proportional to mQ: ×2.00. Positions only: ×1.00. Masses swapped: mP d/(mP + mQ) goes from d/2 to d/3, ×0.67.)
A baseball bat balances on a support at the point shown in the figure. The bat is then sawed through at the balance point into a handle piece and a barrel piece. Which reasoning correctly compares the masses of the two pieces?
Answer and reasoning
AThe pieces have equal mass, since the balance point divides any object into two parts of equal mass. A student who thinks the balance point splits an object into equal masses picks this. Balance requires the mass-weighted positions on the two sides to cancel, not the masses to be equal: the handle's mass is farther from the cut, so less of it balances more mass on the barrel side.
BThe handle piece has more mass, since it is the longer piece and so contains more of the bat. A student who judges mass by length picks this. The handle piece is longer but much thinner; its mass is spread far from the cut, so it balances the barrel piece with less mass.
CThe barrel piece has more mass, since its center of mass is nearer the cut than the handle piece's.Correct With the balance point as origin, the bat's center of mass is at x = 0, so Mhandle xhandle + Mbarrel xbarrel = 0, with each piece replaced by a point at its own center of mass. The short, thick barrel piece has its center of mass close to the cut; the long, thin handle piece has its center of mass far from the cut. For the two terms to cancel, the barrel piece must have more mass.
DThe barrel piece has more mass, since the bat's center of mass is at the center of the barrel. A student who places the center of mass at the center of the most massive part picks this. The figure shows the balance point well inside the bat, toward the handle from the barrel's center; the handle's mass pulls the center of mass toward it. The comparison is right, but this reasoning is not.
Working Take the balance point (the bat's center of mass) as the origin and treat each piece as a point at its own center of mass: Mhandle·xhandle + Mbarrel·xbarrel = 0, so Mbarrel/Mhandle = |xhandle|/|xbarrel|. The handle piece is long and thin, so its center of mass is far from the cut; the barrel piece is short and thick, so its center of mass is near the cut. Hence Mbarrel > Mhandle.
Two uniform rods are joined end to end along the x-axis, as shown in the figure, with each rod's mass and the positions of its ends labeled. What is the x-coordinate of the center of mass of the two-rod system?
Answer and reasoning
A0.85 m A student who places each rod's mass at its right-hand end picks this: [(3.0)(0.60) + (1.0)(1.60)]/4.0 m. A uniform rod's mass is centered at its midpoint, 0.30 m and 1.10 m here.
B0.80 m A student who takes the geometric middle of the combined length picks this: 1.60 m/2. The left rod has three times the mass in a shorter length, so the center of mass is well to the left of the middle.
C0.40 m A student who puts the center of mass where there is equal mass on each side picks this: 2.0 kg of the 3.0 kg rod lies to the left of 0.40 m. The center of mass balances mass-weighted positions; the light rod's mass is far from 0.40 m and counts for more, so xcm is at 0.50 m.
D0.50 mCorrect Place each uniform rod's mass at its own midpoint: 3.0 kg at 0.30 m and 1.0 kg at 1.10 m. Then xcm = [(3.0 kg)(0.30 m) + (1.0 kg)(1.10 m)]/(4.0 kg) = 0.50 m, inside the more massive rod.
Working Each uniform rod's mass acts at its midpoint: rod 1 at 0.30 m, rod 2 at (0.60 + 1.60)/2 = 1.10 m. xcm = [(3.0)(0.30) + (1.0)(1.10)]/(4.0) m = 2.00/4.0 m = 0.50 m. (Masses at the right-hand ends: [(3.0)(0.60) + (1.0)(1.60)]/4.0 = 0.85 m. Middle of the whole length: 1.60/2 = 0.80 m. Point with equal mass on each side: 2.0 kg of rod 1 lies within 2.0/(3.0/0.60) = 0.40 m.)
A thin rod lies along the x-axis, starting at x = 0. The graph shows the rod's linear mass density λ as a function of position x along its whole length. What is the x-coordinate of the rod's center of mass?
Answer and reasoning
A2.4 mCorrect The graph gives λ = 1.0 kg/m + (1.0 kg/m²)x. The rod's mass is the area under the graph, M = ∫₀⁴ (1.0 + 1.0x) dx = 12 kg, and ∫ x λ dx = ∫₀⁴ (x + x²) dx = 29.3 kg·m. So xcm = (29.3 kg·m)/(12 kg) = 2.4 m.
B2.0 m A student who takes the middle of the rod picks this. The density increases along the rod, so more mass lies beyond 2.0 m than before it and the center of mass is farther along.
C2.6 m A student who puts the center of mass where half the mass is on each side picks this: ∫₀ˣ (1.0 + 1.0x) dx = 6.0 kg gives x = 2.6 m. The center of mass balances mass-weighted positions, ∫ x dm/∫ dm, which gives 2.4 m.
D4.0 m A student who puts the center of mass where the rod is densest picks this, at the end. The less dense parts of the rod also have mass and pull the center of mass back toward the origin.
Working From the graph, λ = 1.0 kg/m + (1.0 kg/m²)x from x = 0 to 4.0 m. M = ∫₀⁴ λ dx = 4.0 + 8.0 = 12 kg. ∫ x dm = ∫₀⁴ (x + x²) dx = 8.0 + 21.3 = 29.3 kg·m. xcm = 29.3/12 = 2.4 m. (Geometric middle: 2.0 m. Equal mass on each side: x + x²/2 = 6.0 gives x = −1 + √13 = 2.6 m. Densest point: 4.0 m.)
A thin, nonuniform rod of length L and total mass M lies along an axis from ℓ = 0 to ℓ = L. The mass of the part of the rod between ℓ = 0 and any position ℓ is m(ℓ) = Mℓ³/L³. What is the linear mass density of the rod at position ℓ?
Answer and reasoning
AMℓ²/L³ A student who divides the mass up to ℓ by the distance ℓ picks this: m(ℓ)/ℓ. That is the average density of the part from 0 to ℓ; the density at ℓ is the derivative dm/dℓ, three times larger here.
B3Mℓ²/L³Correct The linear mass density is the derivative of the mass function: λ = dm/dℓ = 3Mℓ²/L³. Integrating it back from 0 to L returns M, as it should. It grows with ℓ, so the rod is densest at ℓ = L.
CM/L A student who uses M/L for every rod picks this. M/L is the average density of the whole rod; this rod is nonuniform, and its density at ℓ is dm/dℓ = 3Mℓ²/L³.
DMℓ⁴/(4L³) A student who integrates the mass function instead of differentiating it picks this. The result has units kg·m, not kg/m; λ is the derivative of m(ℓ).
Working λ(ℓ) = dm/dℓ = d(Mℓ³/L³)/dℓ = 3Mℓ²/L³. Check: ∫₀ᴸ 3Mℓ²/L³ dℓ = M. (Average up to ℓ, m(ℓ)/ℓ: Mℓ²/L³. Average over the whole rod: M/L. Integrating m(ℓ): Mℓ⁴/(4L³), in kg·m.) Checked with sympy.
A thin rod lies along an axis starting at ℓ = 0. The graph shows the mass m of the part of the rod between ℓ = 0 and position ℓ, along the whole rod. What is the linear mass density of the rod at ℓ = 0.70 m?
Answer and reasoning
A3.3 kg/m A student who divides the mass up to the point by its distance from the end picks this: (2.3 kg)/(0.70 m). That is the slope of a line from the origin, the average density of the first 0.70 m, which includes the dense first section; the density at 0.70 m is the slope of the graph there.
B2.3 kg/m A student who reads the height of the m(ℓ) graph as the density picks this: m(0.70 m) = 2.3 kg. The height is a mass, in kg; the density is the slope, in kg/m.
C1.0 kg/mCorrect The linear mass density is the derivative of the mass function, λ = dm/dℓ: the slope of this graph at ℓ = 0.70 m. That point is on the second segment, whose slope is (2.6 kg − 2.0 kg)/(1.00 m − 0.40 m) = 1.0 kg/m. The first 0.40 m of the rod is much denser, 5.0 kg/m.
D2.6 kg/m A student who uses M/L everywhere picks this: 2.6 kg/1.00 m. That is the average density of the whole rod; the graph's two different slopes show that the rod is not uniform.
Working λ = dm/dℓ, the slope of the m(ℓ) graph. At ℓ = 0.70 m the graph is on its second segment: slope = (2.6 kg − 2.0 kg)/(1.00 m − 0.40 m) = 1.0 kg/m. (m(0.70 m) = 2.0 + 1.0 × 0.30 = 2.3 kg; m/ℓ = 2.3/0.70 = 3.3 kg/m; height read as λ: 2.3 kg/m; whole-rod average M/L = 2.6/1.00 = 2.6 kg/m.)
A solid sphere of radius R has a volume mass density that decreases from its center to its surface as ρ(r) = ρ₀(1 − r²/R²), where r is the distance from the center and ρ₀ is a positive constant. What is the total mass of the sphere?
Answer and reasoning
A1.33πρ₀R³ A student who multiplies the central density by the whole volume picks this, treating the sphere as uniform. The density falls to zero at the surface, so the mass is much less than ρ₀ times the volume.
B0.89πρ₀R³ A student who averages the density over the radius, getting (2/3)ρ₀, and multiplies by the volume picks this. The outer shells, where the density is lowest, hold most of the volume, so the volume-weighted average density is only 0.40ρ₀.
C0.50πρ₀R² A student who integrates with the ring element of a flat disk, 2πr dr, picks this: ∫₀ᴿ ρ₀(1 − r²/R²)2πr dr = 0.50πρ₀R². The units show the error, kg/m rather than kg: a spherical shell's volume is 4πr² dr.
D0.53πρ₀R³Correct Split the sphere into thin shells of radius r and thickness dr, each of volume 4πr² dr and density ρ₀(1 − r²/R²). Adding them: M = ∫₀ᴿ ρ₀(1 − r²/R²)4πr² dr = 4πρ₀(R³/3 − R³/5) = (8/15)πρ₀R³ ≈ 0.53πρ₀R³, 40% of the mass a uniform sphere of density ρ₀ would have.
Working Thin shell of radius r, thickness dr: dV = 4πr² dr. M = ∫₀ᴿ ρ₀(1 − r²/R²)4πr² dr = 4πρ₀(R³/3 − R³/5) = (8/15)πρ₀R³ ≈ 0.53πρ₀R³. (Central density times volume: (4/3)πρ₀R³ ≈ 1.33πρ₀R³. Average of ρ over r, (2/3)ρ₀, times volume: (8/9)πρ₀R³ ≈ 0.89πρ₀R³. Disk ring 2πr dr used: 2πρ₀(R²/2 − R²/4) = 0.50πρ₀R², in kg/m.) Checked with sympy.
A person of mass m stands at one end of a boat of mass M that floats at rest on still water. The person walks a distance L along the boat, measured relative to the boat, and stops. The horizontal force exerted by the water on the boat is negligible, so the center of mass of the person–boat system stays at rest. How far does the boat move relative to the water?
Answer and reasoning
AmL/(m + M)Correct The center of mass stays at rest, so m Δxperson + M Δxboat = 0, with both displacements measured relative to the water. The person moves L relative to the boat, so Δxperson = L + Δxboat. Solving, Δxboat = −mL/(m + M): the boat moves a distance mL/(m + M) backward while the person moves ML/(m + M) forward.
BmL/M A student who takes the person's walk, L, as the person's displacement relative to the water picks this: mL = M d. The boat moves back while the person walks, so relative to the water the person moves less than L.
CL/2 A student who keeps the plain average of the two positions fixed picks this, so that person and boat move equal distances in opposite directions. The center of mass is weighted by mass: the more massive object moves less.
D0 A student who places the center of mass at the boat, the most massive part, picks this, concluding that the boat must stay put. The person's mass also counts: as the person moves forward, the boat must move back for the center of mass to stay at rest.
Working Model the system as a single object at its center of mass, which stays at rest: m Δxp + M Δxb = 0. Relative to the water, Δxp = L + Δxb (the walk relative to the boat plus the boat's own displacement). m(L + Δxb) + M Δxb = 0, so Δxb = −mL/(m + M): the boat moves a distance mL/(m + M) opposite to the walk. (Δxp = L taken relative to the water: mL/M. Equal and opposite displacements, plain average: L/2. Center of mass at the more massive boat, so the boat stays put: 0.) Checked with sympy.
A baton is a light rod 0.40 m long with a small ball of mass 0.30 kg at one end and a small ball of mass 0.10 kg at the other. It is thrown so that it spins as it flies. At release the rod is vertical, with the 0.30 kg ball 1.20 m above the ground and the 0.10 kg ball directly above it, and the baton's center of mass is moving straight up at 6.0 m/s. Air resistance is negligible. What is the greatest height above the ground reached by the baton's center of mass? Use g = 10 m/s².
Answer and reasoning
A3.2 m A student who starts the rise from the rod's midpoint, 1.40 m, picks this. The two balls have different masses, so the center of mass is nearer the 0.30 kg ball, at 1.30 m.
B3.0 m A student who starts the rise from the 0.30 kg ball, the most massive part, picks this. The 0.10 kg ball also has mass and pulls the center of mass up to 1.30 m.
C3.1 mCorrect Locate the center of mass at release: ycm = [(0.30 kg)(1.20 m) + (0.10 kg)(1.60 m)]/(0.40 kg) = 1.30 m. Modeled as a single object at that point, the baton's center of mass rises as a projectile, by v²/(2g) = (6.0 m/s)²/(20 m/s²) = 1.80 m, whatever the spinning. Greatest height: 1.30 m + 1.80 m = 3.1 m.
D3.3 m A student who weights each ball's height by the other ball's mass picks this, starting from 1.50 m, nearer the lighter ball. Each height is weighted by its own ball's mass, which puts the center of mass at 1.30 m.
Working Center of mass at release: ycm = [(0.30)(1.20) + (0.10)(1.60)]/(0.40) m = 1.30 m. Modeled as a single object at its center of mass, the baton's center of mass moves as a projectile: rise = v²/(2g) = (6.0 m/s)²/(2 × 10 m/s²) = 1.80 m. Greatest height = 1.30 m + 1.80 m = 3.1 m. (Rod midpoint 1.40 m: 3.2 m. Heavy ball 1.20 m: 3.0 m. Masses swapped, 1.50 m: 3.3 m.)
The figure shows three systems, each made of two small objects on the x-axis, drawn to a common scale with the masses labeled. x₁, x₂ and x₃ are the x-coordinates of the centers of mass of systems 1, 2 and 3. Which ranking is correct?
Answer and reasoning
Ax₂ > x₃ > x₁ A student who takes the plain midpoint of each pair picks this: 1, 2 and 1.5 m. The masses matter: in system 1 the heavy object is at 2 m, pulling x₁ up to 1.5 m, and in system 3 it is at the origin, pulling x₃ down to 0.75 m.
Bx₁ > x₂ > x₃ A student who divides Σ mi xi by the number of objects picks this: 3.0, 2.0 and 1.5. Dividing by the total mass instead gives 1.5, 2.0 and 0.75 m.
Cx₃ > x₂ > x₁ A student who weights each position by the other object's mass picks this, putting each center of mass nearer the lighter object: 0.5, 2.0 and 2.25 m. Each center of mass lies nearer the heavier object.
Dx₂ > x₁ > x₃Correct Weight each position by its mass and divide by the total mass. System 1: (3.0 kg)(2 m)/(4.0 kg) = 1.5 m. System 2: (1.0 kg)(4 m)/(2.0 kg) = 2.0 m. System 3: (1.0 kg)(3 m)/(4.0 kg) = 0.75 m. So x₂ > x₁ > x₃.
Working xcm = (Σ mi xi)/(Σ mi). System 1: (1.0·0 + 3.0·2)/4.0 = 1.5 m. System 2: (1.0·0 + 1.0·4)/2.0 = 2.0 m. System 3: (3.0·0 + 1.0·3)/4.0 = 0.75 m. x₂ > x₁ > x₃. (Plain averages 1, 2, 1.5 m: x₂ > x₃ > x₁. Divided by the number of objects, 3.0, 2.0, 1.5: x₁ > x₂ > x₃. Masses swapped, 0.5, 2.0, 2.25 m: x₃ > x₂ > x₁.)
A thin, uniform wire of linear mass density λ is bent into a semicircle of radius R, centered on the origin and lying in the region y ≥ 0 of the xy-plane. Where is the wire's center of mass?
Answer and reasoning
Ay = 0.50R, on the y-axis A student who takes the middle of the shape's height picks this. More of the wire's length is at large y, near the top of the arc, than near y = 0, so the center of mass is above R/2.
By = 0.64R, on the y-axisCorrect By symmetry about the y-axis, xcm = 0. An element at angle θ has mass dm = λR dθ and height y = R sin θ, so ycm = (∫₀π R sin θ λR dθ)/(λπR) = 2R/π ≈ 0.64R. This point is on the line of symmetry but in the empty space inside the arc.
Cy = 1.00R, on the y-axis A student who thinks the center of mass must lie on the material picks the top of the arc. Every element of the wire except the one at the top is lower than R, so the average height is less than R; the center of mass is in empty space.
Dy = 0.71R, on the y-axis A student who puts the center of mass where half the wire is above and half below picks this: the arc above y = R/√2 has half the length. The center of mass is the mass-weighted average height, ∫ y dm/∫ dm = 2R/π.
Working By symmetry xcm = 0. Element at angle θ: dm = λR dθ, at y = R sin θ. ycm = ∫₀π R sin θ λR dθ/(λπR) = 2λR²/(λπR) = 2R/π ≈ 0.64R, on the y-axis. (Middle of the height: 0.50R. On the wire at the top: 1.00R. Equal mass above and below: arc above y = h subtends π − 2 arcsin(h/R) = π/2, so h = R/√2 ≈ 0.71R.) Checked with sympy.
A solid sphere of radius R and mass M has a volume mass density that increases in proportion to the distance r from its center: ρ = br, where b is a positive constant. How much of the sphere's mass lies within r = 0.60R?
Answer and reasoning
A0.22M A student who treats the sphere as having one density throughout picks this: the mass within 0.60R would then scale with volume, (0.60)³ = 0.22. Here the density is lower in the inner region, so the fraction is smaller still.
B0.36M A student who integrates the density over r alone picks this: ∫₀ʳ bs ds = br²/2 grows as r², giving (0.60)² = 0.36. Each shell's volume is 4πs² ds, which adds two more powers of r.
C0.13MCorrect The mass within radius r is ∫₀ʳ b s 4πs² ds = πbr⁴, so it grows as r⁴, and the fraction within 0.60R is (0.60)⁴ = 0.13. The inner region holds little of the mass because it has only 0.22 of the volume and the lowest density.
D0.60M A student who takes the enclosed mass as proportional to the radius picks this. The enclosed mass depends on volume and density, here as r⁴, so 0.60 of the radius encloses only 0.13 of the mass.
Working M(r) = ∫₀ʳ b s 4πs² ds = πbr⁴, so the mass within r is proportional to r⁴: M(0.60R)/M(R) = (0.60)⁴ = 0.13. (Uniform density, mass ∝ r³: (0.60)³ = 0.22. ∫ρ dr ∝ r²: (0.60)² = 0.36. Mass ∝ radius: 0.60.) Checked with sympy.
Rods 1, 2 and 3 have the same length L and each lies along the x-axis from x = 0 to x = L. The graph shows the linear mass density λ of each rod as a function of x. x₁, x₂ and x₃ are the x-coordinates of the rods' centers of mass. Which ranking is correct?
Answer and reasoning
Ax₃ > x₂ > x₁Correct xcm = ∫ x λ dx/∫ λ dx. The uniform rod 1 has x₁ = L/2. Rod 2 (λ ∝ x) has x₂ = 2L/3. Rod 3 (λ ∝ x²) has x₃ = 3L/4: its mass is concentrated even more toward x = L. So x₃ > x₂ > x₁.
Bx₁ = x₂ = x₃ A student who puts every rod's center of mass at its geometric middle picks this. Equal lengths do not mean equal mass distributions: rods 2 and 3 have most of their mass toward x = L.
Cx₁ > x₂ > x₃ A student who thinks the center of mass sits toward the lighter end picks this. Each element dm = λ dx is at its own position x, so mass concentrated toward x = L pulls the center of mass toward x = L.
Dx₂ = x₃ > x₁ A student who places the center of mass at the densest point picks this, at x = L for rods 2 and 3 (and the middle for the uniform rod 1). The less dense parts also have mass, so neither center of mass is at the end, and rod 3's mass is more concentrated near x = L than rod 2's.
Working xcm = ∫ x λ dx/∫ λ dx. Rod 1, λ constant: L/2. Rod 2, λ ∝ x: (L³/3)/(L²/2) = 2L/3. Rod 3, λ ∝ x²: (L⁴/4)/(L³/3) = 3L/4. x₃ > x₂ > x₁. (Geometric middle for all: equal. Toward the lighter end: x₁ > x₂ > x₃. At the densest point, x = L for rods 2 and 3, middle for uniform rod 1: x₂ = x₃ > x₁.)
A thin, uniform disk of radius R is centered on the origin of the xy-plane. A circular hole of radius R/3 is cut out of the disk; the hole's center is on the x-axis at x = −2R/3, so the hole touches the disk's edge. What is the x-coordinate of the center of mass of the remaining plate?
Answer and reasoning
A0.07R A student who divides by the original disk's mass M rather than by the remaining mass 8M/9 picks this: (M/9)(2R/3)/M = 2R/27. The system is the plate that remains, so the removed mass comes out of the denominator too.
B0.08RCorrect Model the plate as the full disk, mass M centered at x = 0, plus the hole as a negative mass. The hole has one-ninth of the disk's area, so it removes M/9 from x = −2R/3. Then xcm = [0 − (M/9)(−2R/3)]/(M − M/9) = R/12 ≈ 0.08R, shifted slightly away from the hole.
C0.33R A student who takes the hole's mass to be proportional to its radius, M/3, rather than to its area, picks this: (M/3)(2R/3)/(2M/3) = R/3. A uniform sheet's mass is proportional to area, so a hole of one-third the radius removes one-ninth of the mass.
D0.00R A student who places the center of mass at the center of the plate's circular outline picks this. Material has been removed to the left of the origin, so the mass is no longer symmetric about the y-axis, and the center of mass lies to the right of the origin.
Working Model the plate as the full disk (mass M, center at x = 0) plus the hole as a negative mass. The hole's area is π(R/3)² = πR²/9, one-ninth of the disk's area, so for a uniform sheet it would have mass M/9, centered at x = −2R/3. xcm = [M·0 − (M/9)(−2R/3)]/(M − M/9) = (2MR/27)/(8M/9) = R/12 ≈ 0.08R. (Divided by the original mass M: (M/9)(2R/3)/M = 2R/27 ≈ 0.07R. Hole's mass taken in proportion to its radius, M/3: (M/3)(2R/3)/(2M/3) = R/3 ≈ 0.33R. Center of the circular outline: 0.) Checked with sympy.
A uniform solid hemisphere of radius R rests with its flat face on the xy-plane, centered on the origin, and its curved surface in the region z ≥ 0. At what height z above the flat face is the hemisphere's center of mass?
Answer and reasoning
A0.500R A student who treats the hemisphere's mass per unit height as the same at every z, M/R, as for a uniform rod, picks this: the midpoint of the height. The slices near the flat face are wider and hold more mass per unit height than those near the top, so the center of mass is below R/2.
B0.424R A student who takes each disk slice's mass to be proportional to its radius, √(R² − z²), rather than to its area, picks this: 4R/(3π). A thin disk's mass is its density times its volume, ρπr² dz, so each height is weighted by R² − z².
C0.625R A student who weights each height z by the mass of the slice at the mirror-image height, R − z, picks this: 5R/8, up where the slices are small. Each slice's mass sits at its own height z, so the center of mass is nearer the massive flat face.
D0.375RCorrect Slice the hemisphere into thin disks parallel to its flat face: the disk at height z has mass dm = ρπ(R² − z²) dz. Then M = (2/3)πρR³ and ∫ z dm = πρR⁴/4, so zcm = 3R/8 = 0.375R, below half the height, because the wide slices near the flat face hold most of the mass.
Working Slice the hemisphere into thin disks parallel to the flat face. The disk at height z has radius √(R² − z²) and mass dm = ρπ(R² − z²) dz. M = ∫₀ᴿ ρπ(R² − z²) dz = (2/3)πρR³. ∫ z dm = ρπ∫₀ᴿ (R²z − z³) dz = πρR⁴/4. zcm = (πρR⁴/4)/((2/3)πρR³) = 3R/8 = 0.375R. (Same mass per unit height at every z: R/2 = 0.500R. Slice mass taken proportional to its radius, dm ∝ √(R² − z²) dz: (R³/3)/(πR²/4) = 4R/(3π) ≈ 0.424R. Each height weighted by the slice mass at the mirror height R − z: 5R/8 = 0.625R.) Checked with sympy.
Compiled from the AP Physics C: Mechanics Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
