Corrected exercises: torque, rotation and equilibrium (PHYS 101)

a) One revolution is $2\pi$ rad and one minute is $60$ s, so $1800$ rev/min $= \frac{1800 \times 2\pi}{60} = 60\pi = 188.5$ rad/s. Keep the exact form $60\pi$ as long as you can: every later line stays clean. The rim speed is $v_{0} = r\omega_{0} = 0.12 \times 188.5 = 22.6$ m/s, about $81$ km/h, which is why a shattered grinding wheel is a workshop hazard. The radian earns its place here: had you fed $1800$ rev/min into $v = r\omega$ you would have obtained $216$ m/s, a factor of $9.55$ too large, because the formula counts arcs and only the radian measures an angle as an arc.

b) Uniform braking means constant $\alpha$, so the rotational twin of $v_{f} = v_{i} + a\,\Delta t$ applies: $\alpha = \frac{\omega_{f} - \omega_{0}}{\Delta t} = \frac{0 - 188.5}{12.0} = -15.7$ rad/s$^2$. The minus sign says the wheel is slowing, not that it turns backwards. It will turn backwards only if $\omega$ itself changes sign, which never happens here since the brake releases the wheel at rest.

c) The angular acceleration is constant, so the average angular velocity is the plain average of the endpoints: $\bar{\omega} = \frac{188.5 + 0}{2} = 94.25$ rad/s. Then $\theta = \bar{\omega}\,\Delta t = 94.25 \times 12.0 = 1131$ rad, and $\frac{1131}{2\pi} = 180$ revolutions. Read it without any formula as a check: $1800$ rev/min is $30$ rev/s, the average rate is $15$ rev/s, and $15 \times 12 = 180$. Two independent routes to the same integer is the cheapest verification in the chapter.

d) Tangential acceleration comes from $\alpha$: $a_{t} = r|\alpha| = 0.12 \times 15.7 = 1.88$ m/s$^2$. Centripetal acceleration comes from $\omega$: $a_{c} = r\omega_{0}^{2} = 0.12 \times (188.5)^{2} = 4.26 \times 10^{3}$ m/s$^2$, roughly $435g$. The two are perpendicular, so the total acceleration is $\sqrt{a_{t}^{2} + a_{c}^{2}}$, which here is $4.26 \times 10^{3}$ m/s$^2$ to three figures: the tangential part is utterly negligible. This is the standard trap of the chapter. A spinning body can have zero angular acceleration and still have an enormous acceleration at every point of it, because turning is itself an acceleration.

e) Set $r\alpha = r\omega^{2}$, so $\omega = \sqrt{|\alpha|} = \sqrt{15.7} = 3.96$ rad/s. Working out when the wheel is that slow: $t = \frac{188.5 - 3.96}{15.7} = 11.7$ s, that is, in the last third of a second before it stops. Until then the centripetal term dominates. Note the units in $\omega = \sqrt{|\alpha|}$ do not appear to match, and they do not need to: the radian is a ratio of two lengths, so it carries no dimension and may appear and disappear at will. That is the deepest thing the radian teaches, and it is also why a marker never writes rad in a final answer in metres per second.

a) Use $\tau = rF\sin\phi$, where $\phi$ is the angle between the handle and the force. $\tau_{1} = 0.25 \times 120 \times \sin 90^{\circ} = 30.0$ N m. $\tau_{2} = 0.25 \times 120 \times \sin 40^{\circ} = 30.0 \times 0.643 = 19.3$ N m. $\tau_{3} = 0.15 \times 120 \times \sin 90^{\circ} = 18.0$ N m. $\tau_{4} = 0.25 \times 120 \times \sin 0^{\circ} = 0$. A force pointing straight at the axis, or straight away from it, turns nothing at all, however hard you pull.

b) $\tau_{1} > \tau_{2} > \tau_{3} > \tau_{4}$. Four equal forces, four different torques: the magnitude of the force decides nothing here. What decides is the LEVER ARM $d = r\sin\phi$, the perpendicular distance from the axis to the LINE of the force, extended as far as you need. Read the four lever arms off the figure: $0.25$ m, $0.161$ m, $0.15$ m and $0$. Divide each torque by $120$ and you recover exactly those four numbers.

c) $d_{2} = r\sin 40^{\circ} = 0.25 \times 0.643 = 0.161$ m. To reach $30.0$ N m with that lever arm you need $F = \frac{30.0}{0.161} = 187$ N, which is $56$ percent more force for the same result. This is the whole practical content of the chapter: a mechanic pulls perpendicular to the wrench not out of habit but because any other angle throws away a factor $\sin\phi$. The trap is to compute $rF\cos\phi$ instead. Test it on the two extremes: at $\phi = 90^{\circ}$ the torque must be maximal, and $\cos 90^{\circ} = 0$ would give zero. If the extreme case is absurd, the formula is wrong, and you have caught it before spending the points.

d) Torques about a fixed axis are signed, not vectors to be added by components. Call the sense of $F_{1}$ positive. The fifth force has the same lever arm $0.25$ m and the opposite sense, so $\tau_{5} = -0.25 \times 80 = -20.0$ N m and $\tau_{\text{net}} = 30.0 - 20.0 = +10.0$ N m, turning in the sense of $F_{1}$. Notice what the net FORCE is doing meanwhile: $120 - 80 = 40$ N, which the bolt itself absorbs. Force and torque answer two different questions, and a net torque of $10$ N m tells you nothing about the net force, nor the reverse.

a) Two conditions, in this order. Torques first, about $A$, because that axis kills $N_{A}$ and leaves one unknown: $N_{B}(3.0) = (294)(2.0) + (588)(1.5)$, the plank weight $294$ N acting at its centre $2.0$ m from $A$ and the person weight $588$ N at $1.5$ m. So $N_{B} = \frac{588 + 882}{3.0} = 490$ N. Forces second: $N_{A} = 294 + 588 - 490 = 392$ N. The person is nearer $A$ than $B$, yet $B$ carries more: the PLANK leans on $B$ because its own centre sits closer to $B$ than to $A$.

b) About $B$, distances are measured from $B$: the person is $1.5$ m to its left, the plank centre $1.0$ m to its left, and $A$ is $3.0$ m to its left. Both weights turn the plank one way about $B$ and $N_{A}$ turns it the other: $N_{A}(3.0) = (294)(1.0) + (588)(1.5) = 294 + 882 = 1176$, so $N_{A} = 392$ N. The two routes agree because the body is in equilibrium, and for a body in equilibrium the net torque vanishes about EVERY axis, not merely about the real pivot. That freedom is the most useful tool in the chapter: you pick the axis that deletes the unknown you do not want.

c) About $A$ again: $N_{B}(3.0) = (294)(2.0) + (588)(3.6) = 588 + 2116.8 = 2704.8$, so $N_{B} = 901.6$ N, and $N_{A} = 882 - 901.6 = -19.6$ N. A negative answer here is not an arithmetic slip, it is information. It says the plank would need a force of $19.6$ N pulling DOWN at $A$ to stay level, and a plank simply resting on a support has nothing to supply it. The plank has already tipped about $B$, the left end has lifted, and the whole equilibrium model has stopped applying. Bolt the left end down and the $-19.6$ N becomes a real, physical pull in the bolt.

d) Tipping starts exactly when $N_{A}$ reaches zero, so set $N_{A} = 0$ and take torques about $B$. Only the plank weight holds the left end down, with a lever arm of $1.0$ m, against the person at $(d - 3.0)$ m past $B$: $294(1.0) = 588(d - 3.0)$, hence $d - 3.0 = 0.5$ and $d = 3.5$ m. The person may walk half a metre past the support and no further. Sanity check: the person is twice as heavy as the plank, so their lever arm at the tipping point must be half the plank's, and $0.5 = \frac{1.0}{2}$ exactly. Note what the answer does NOT depend on: not on $g$, which cancels on both sides, and not on how strong the supports are.

a) $I = \frac{1}{2}MR^{2} = \frac{1}{2}(2.0)(0.10)^{2} = 1.0 \times 10^{-2}$ kg m$^2$. Notice the radius is SQUARED: halve the radius at fixed mass and the moment of inertia drops by four. Mass counts once in rotation, distance counts twice.

b) Two equations, one for each object, and one constraint. Block: $mg - T = ma$. Pulley: the only torque about the axle is $TR$, so $TR = I\alpha$. Constraint: the cord does not slip, so $a = R\alpha$. Substituting $\alpha = a/R$ into the second gives $T = \frac{Ia}{R^{2}}$, and then $mg = ma + \frac{Ia}{R^{2}}$, so $a = \frac{mg}{m + I/R^{2}} = \frac{29.4}{3.0 + 1.0} = 7.35$ m/s$^2$ and $\alpha = \frac{7.35}{0.10} = 73.5$ rad/s$^2$. The quantity $I/R^{2} = 1.0$ kg is worth naming: the disk resists exactly as if a further $1.0$ kg, one half of its mass, had been hung on the cord.

c) $T = m(g - a) = 3.0(9.8 - 7.35) = 7.35$ N, while the weight of the block is $29.4$ N. The tension is FOUR times smaller, and writing $T = mg$ is the single most common error on this problem. It is self contradictory: if $T$ equalled $mg$ the block would not accelerate at all, and if it did not accelerate the pulley would not turn. Check the other end: $\tau = TR = 0.735$ N m and $I\alpha = (1.0 \times 10^{-2})(73.5) = 0.735$ N m, so the two equations close on each other.

d) With constant acceleration, $v^{2} = 2a\,\Delta y = 2(7.35)(1.5) = 22.05$, so $v = 4.70$ m/s. Energy check, which is the real point of the question. The block loses $mgh = 3.0(9.8)(1.5) = 44.1$ J of potential energy. It gains $\frac{1}{2}mv^{2} = \frac{1}{2}(3.0)(22.05) = 33.1$ J of translational kinetic energy, and the pulley gains $\frac{1}{2}I\omega^{2}$ with $\omega = v/R = 47.0$ rad/s, that is $\frac{1}{2}(1.0 \times 10^{-2})(2205) = 11.0$ J. The sum is $33.1 + 11.0 = 44.1$ J. The missing quarter of the energy is not lost, it is spinning: a pulley with mass is a place where kinetic energy is stored, and a solution that forgets the $\frac{1}{2}I\omega^{2}$ term returns $v = 5.42$ m/s, too fast by $15$ percent.

a) Rolling without slipping does no work through friction, because the contact point never slides, so mechanical energy is conserved: $mgh = \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}$. Put $I = kmR^{2}$ and $\omega = v/R$: the second term becomes $\frac{1}{2}kmR^{2}\frac{v^{2}}{R^{2}} = \frac{1}{2}kmv^{2}$, and $R$ disappears. So $mgh = \frac{1}{2}mv^{2}(1+k)$ and $v = \sqrt{\frac{2gh}{1+k}}$. Both $m$ and $R$ cancel, and that is not an accident of the algebra: $k$ is a pure number saying how far from the axis the mass sits, in units of $R$.

b) With $2gh = 23.52$ m$^2$/s$^2$. Hoop, $k = 1$: $v = \sqrt{\frac{23.52}{2}} = 3.43$ m/s. Disk, $k = \frac{1}{2}$: $v = \sqrt{\frac{23.52}{1.5}} = 3.96$ m/s. Sphere, $k = \frac{2}{5}$: $v = \sqrt{\frac{23.52}{1.4}} = 4.10$ m/s. Three bodies, one drop, three different speeds.

c) Sphere first, then disk, then hoop: smallest $k$ wins, always. The result does not depend on mass, does not depend on radius, and does not depend on the slope angle either, since $h$ is all that enters. A brass hoop and a plastic hoop of quite different sizes cross the line together, and a marble beats both. This is the experiment to picture when the algebra stops making sense: the hoop carries all its mass at the rim, so at a given $v$ it must spin with more energy, and it can only pay for that out of the same $mgh$.

d) The rotational share is $\frac{\frac{1}{2}kmv^{2}}{\frac{1}{2}mv^{2}(1+k)} = \frac{k}{1+k}$, again a pure number. Hoop: $\frac{1}{2}$, exactly half the energy is spin. Disk: $\frac{1}{3}$. Sphere: $\frac{2}{7} \approx 0.286$. The ranking of the fractions is the ranking of the race, upside down.

e) A sliding block has no rotation to pay for, so $k = 0$ and $v = \sqrt{2gh} = 4.85$ m/s. It beats all three. The lesson to carry into the exam: rolling is SLOWER than sliding, at equal drop, and the deeper the mass sits away from the axis, the slower still. If your answer has the hoop winning, or has a rolling body matching the free slide, you have dropped the $\frac{1}{2}I\omega^{2}$ term, and the whole question goes with it.

a) Vertical forces: the floor pushes up with $N$, and the two weights pull down, $117.6$ N for the ladder and $686$ N for the painter. The wall is smooth, so it contributes nothing vertical. Hence $N = 117.6 + 686 = 803.6 \approx 804$ N, and this is true wherever on the ladder the painter happens to stand. Only the SHARE between wall and friction moves as he climbs.

b) Take torques about the FOOT of the ladder: that axis deletes $N$ and $f$ at a stroke, leaving $N_{w}$ alone. Lever arms are horizontal distances for vertical forces and vertical distances for horizontal ones. The ladder centre is $1.5$ m horizontally from the foot; the painter, $4.0$ m along a ladder whose horizontal run is $\frac{3}{5}$ of its length, is $4.0 \times 0.6 = 2.4$ m horizontally from the foot; the wall force acts $4.0$ m up. So $N_{w}(4.0) = 117.6(1.5) + 686(2.4) = 176.4 + 1646.4 = 1822.8$, giving $N_{w} = 456$ N. Horizontal forces then give $f = N_{w} = 456$ N: the wall pushes the ladder out, friction holds it in, and they are equal because nothing else is horizontal.

c) Static friction obeys $f \le \mu_{s}N$, so the ladder stands only if $\mu_{s} \ge \frac{f}{N} = \frac{455.7}{803.6} = 0.567$. That is a demanding floor: rubber on dry concrete reaches it, a shoe on a dusty tile does not. Note that the condition mixes the two equilibrium conditions, one from forces, one from torques, which is exactly why the chapter insists on both.

d) With $\mu_{s} = 0.40$ the friction can never exceed $f_{\max} = 0.40(803.6) = 321.4$ N, and therefore $N_{w}$ can never exceed $321.4$ N either. Put that ceiling into the torque equation with the painter at horizontal distance $x$: $321.4(4.0) = 176.4 + 686x$, so $x = \frac{1285.8 - 176.4}{686} = 1.617$ m horizontally, which along the ladder is $\frac{1.617}{0.6} = 2.70$ m. He may climb little more than half way. The trap in this part is to hunt for a new $N$: the normal force does not change, only the demand on friction does. And the physical reading is worth a line on the copy: climbing raises the painter's lever arm about the foot, which raises the wall force, which raises the friction needed, until the floor gives out.

a) The joint force is unknown in both magnitude and direction, which would be two unknowns in any force equation. Its line of action passes through the elbow, so about THAT axis its lever arm is zero and its torque is zero: choosing the elbow deletes it from the torque equation and leaves a single unknown. This is the reflex the chapter is really teaching, and it is worth stating on the copy in exactly those words. Torques about the elbow: $F_{b}(0.040) = (17.64)(0.16) + (49.0)(0.32) = 2.822 + 15.68 = 18.50$, so $F_{b} = \frac{18.50}{0.040} = 463$ N.

b) Now the force condition, which was useless a moment ago and is exactly what is needed here. Upward: $F_{b} = 462.6$ N. Downward: the two weights, $17.64 + 49.0 = 66.6$ N. The joint must supply $66.6 - 462.6 = -396$ N, that is $396$ N DOWNWARD on the forearm. The humerus is not holding the forearm up, it is pulling it back down, because the biceps overshoots by a factor of seven. Students who expect a joint always to push up lose the sign here and then the marks; let the algebra decide the direction, and read the sign afterwards.

c) The ball weighs $49.0$ N and the biceps pulls with $463$ N, a factor of $9.4$. The reason is the ratio of lever arms, $\frac{0.32}{0.040} = 8$ for the ball alone, with the forearm's own weight adding the rest. As a MACHINE the arm is terrible: it multiplies the required force by nearly ten. What it buys with that loss is SPEED and RANGE, because the hand sweeps eight times the distance of the tendon in the same time, which is why a human can throw. Every lever trades force against distance, and evolution picked the side of the trade that catches prey.

d) Nothing at all: $F_{b}$ stays $463$ N. Rotate the whole picture by $30^{\circ}$ and every lever arm is multiplied by the same $\cos 30^{\circ}$, on the left of the equation and on the right, so the factor cancels: $F_{b}(0.040\cos 30^{\circ}) = (17.64)(0.16\cos 30^{\circ}) + (49.0)(0.32\cos 30^{\circ})$. Numerically $F_{b}(0.03464) = 16.02$, which returns $463$ N. It is worth verifying rather than asserting, because the cancellation only happens when every force is parallel, here all vertical. Tilt just one of them, hang the ball from a slanted cord for instance, and the angle stops cancelling at once.

a) FALSE. Torque is $\tau = rF\sin\phi$, so a force contributes nothing at all when its line passes through the axis, whatever its size. A $200$ N pull straight along a wrench handle gives $\tau = 0$, while a $100$ N push perpendicular at $0.25$ m gives $25$ N m. Correct statement: at equal lever arm the larger force gives the larger torque, and the lever arm, not the force, is what must be compared first.

b) FALSE, and this is the most expensive misconception in the chapter. For a body in EQUILIBRIUM the net torque vanishes about every axis, real or imaginary, inside the body or outside it. That freedom is a tool: choose the axis through an unknown force and it drops out of the equation, which is why a ladder problem is solved about the foot and a forearm problem about the elbow. The one restriction: the equilibrium must hold first, and once you have chosen an axis, every torque in that equation must be measured about the SAME axis.

c) TRUE. Such a pair is called a couple. Two equal and opposite forces applied at different points, for instance the two hands on a steering wheel, give $\sum \vec{F} = 0$ so the centre of mass does not accelerate, yet $\sum \tau = Fd \ne 0$, where $d$ is the distance between the two lines of action. The torque of a couple is $Fd$ about ANY axis, which is worth checking once by hand. It is the cleanest proof that the two equilibrium conditions are genuinely independent: neither one implies the other.

d) FALSE, though the reason given is not entirely wrong. Mass does cancel for both bodies, and so does radius, and yet they do not arrive together. What decides is $k$ in $I = kmR^{2}$: $v = \sqrt{\frac{2gh}{1+k}}$, with $k = 1$ for the hoop and $k = \frac{1}{2}$ for the disk. From a drop of $1.20$ m the disk reaches $3.96$ m/s and the hoop only $3.43$ m/s. Correct statement: the arrival order depends on how the mass is DISTRIBUTED, not on how much of it there is.

e) FALSE on the second half. No external torque acts about her axis, so $L = I\omega$ is indeed conserved: pulling her arms in cuts $I$, so $\omega$ rises in exactly the same proportion. But $K = \frac{1}{2}I\omega^{2} = \frac{L^{2}}{2I}$, and with $L$ fixed, halving $I$ DOUBLES the kinetic energy. The energy is not created from nothing: her muscles do work pulling the arms inward against the outward pull they feel while spinning. Correct statement: angular momentum is conserved, kinetic energy is not, and the difference is the work her muscles did.

a) $I = \frac{1}{2}MR^{2} = \frac{1}{2}(0.48)(0.10)^{2} = 2.4 \times 10^{-3}$ kg m$^2$. Then $\omega = \frac{6000 \times 2\pi}{60} = 200\pi = 628$ rad/s. Half a kilogram and ten centimetres give a very small moment of inertia, and that is the point of the rig: a light rotor turns quickly and stops quickly, so the bearings show themselves in seconds rather than in hours.

b) Use the average rate rather than a formula: the flywheel averages $50$ rev/s over $8.0$ s, so it makes $400$ revolutions. In radians, $\alpha = \frac{628.3}{8.0} = 78.5$ rad/s$^2$ and $\theta = \frac{1}{2}\alpha t^{2} = \frac{1}{2}(78.5)(64) = 2513$ rad, which divided by $2\pi$ is $400$ again. Agreement of the two routes is the check, and the plain average route is faster on an exam paper than any of the four equations.

c) The net torque is what $I\alpha$ measures, never the motor torque: $\tau_{\text{net}} = I\alpha = (2.4 \times 10^{-3})(78.5) = 0.188$ N m. The motor must beat the bearings as well: $\tau_{\text{motor}} = \tau_{\text{net}} + \tau_{f} = 0.188 + 0.050 = 0.238$ N m. Reading $\sum \tau = I\alpha$ as though the left side were the applied torque alone is the rotational version of writing $T = mg$ for a hanging block, and it costs the same marks for the same reason: $I\alpha$ is a SUM, exactly as $ma$ is.

d) $K = \frac{1}{2}I\omega^{2} = \frac{1}{2}(2.4 \times 10^{-3})(628.3)^{2} = \frac{1}{2}(2.4 \times 10^{-3})(3.948 \times 10^{5}) = 474$ J. The rim moves at $v = R\omega = 0.10 \times 628.3 = 62.8$ m/s, about $226$ km/h. Those two numbers together are why a test rig has a steel guard: half a kilogram carrying $474$ J is the energy of a $1$ kg mass thrown at $31$ m/s. Note also how the energy scales: $K$ goes with $\omega^{2}$, so halving the speed leaves only a quarter of the stored energy.

e) With the motor off, the bearings are the only torque: $\alpha = \frac{-0.050}{2.4 \times 10^{-3}} = -20.8$ rad/s$^2$, and $t = \frac{628.3}{20.8} = 30$ s. A stop in $8.0$ s instead would require $\alpha = -78.5$ rad/s$^2$, that is a friction torque of $(2.4 \times 10^{-3})(78.5) = 0.188$ N m, $3.8$ times the rated $0.050$ N m: the bearing is dry or seizing. Read the ratio directly and the arithmetic almost disappears, since the coast-down time is inversely proportional to the friction torque. This is exactly the measurement a technician makes, and it needs nothing but a stopwatch.

a) No external torque acts about the axle, since the axle is frictionless and both weights act parallel to it, so angular momentum about that axis is conserved. Before: $I_{0} = \frac{1}{2}(180)(1.8)^{2} = 291.6$ kg m$^2$ and $L = (291.6)(0.60) = 175.0$ kg m$^2$/s. The child arrives radially, carrying no angular momentum about the axle. After: $I' = 291.6 + (40)(1.8)^{2} = 291.6 + 129.6 = 421.2$ kg m$^2$, hence $\omega' = \frac{175.0}{421.2} = 0.415$ rad/s. The carousel slows by $31$ percent because the child, sitting at the rim, is expensive: $40$ kg at full radius counts as much as $80$ kg spread over the disk.

b) $K_{0} = \frac{1}{2}(291.6)(0.60)^{2} = 52.5$ J and $K' = \frac{1}{2}(421.2)(0.4154)^{2} = 36.3$ J. Sixteen joules have gone, about $31$ percent. They went into heat and sound under the child's shoes, which slid on the platform for the fraction of a second it took him to be dragged up to rim speed. This is the rotational analogue of a perfectly inelastic collision, and the same warning applies: angular momentum is conserved in it, kinetic energy is NOT, and a solution that conserves energy here to find $\omega'$ gets a wrong answer from a correct looking equation.

c) Walking to the centre changes $I$ back to $291.6$ kg m$^2$, and $L$ is still $175.0$, so $\omega'' = \frac{175.0}{291.6} = 0.60$ rad/s and $K'' = \frac{1}{2}(291.6)(0.60)^{2} = 52.5$ J. The carousel has returned exactly to its original state. The $16.2$ J came from the child's legs: walking inward on a spinning platform means pushing outward against the floor while moving inward, which is work done on the system, joule for joule. This is the skater, in slow motion and with the numbers visible. Angular momentum is conserved throughout; kinetic energy rose because something did work, and fell earlier because something rubbed.

d) No. Just before the child steps on, the system has the child's inward linear momentum; just after, the centre of mass of the whole assembly is moving around the axle. The axle is bolted to the ground and pushes horizontally on the carousel with whatever force is needed, and that force is external to the carousel plus child system. Linear momentum therefore is not conserved, while angular momentum about the axle is, because that same axle force has zero lever arm about the axis it passes through. Choosing the axis on the axle is what saves the problem, and it is the same choice as the elbow in the forearm problem and the foot in the ladder problem.