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

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.

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.

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.

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.

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 quarter of a second before it stops ($12.0 - 11.75 = 0.25$ s). 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.

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.

$\tau_{1} > \tau_{2} > \tau_{3} > \tau_{4}$. 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.

$d_{2} = r\sin 40^{\circ} = 0.25 \times 0.643 = 0.161$ m. To reach $30.0$ N m with that lever arm you need $F = \frac{30.0}{0.161} = 186$ N with the rounded lever arm, or $\frac{120}{\sin 40^{\circ}} = 187$ N without rounding, 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.

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.

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 $A$.

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.

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.

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.

$I = \frac{1}{2}MR^{2} = \frac{1}{2}(2.0)(0.10)^{2} = 1.0 \times 10^{-2}$ 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.

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.

$T = m(g - a) = 3.0(9.8 - 7.35) = 7.35$ 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.

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.

The formula sheet gives $I = \sum m_{i}r_{i}^{2}$, mass by mass, with every $r_{i}$ measured from the NAMED axis.

About the centre, the two inner weights sit at $\frac{s}{2}$ and the two outer ones at $\frac{3s}{2}$:

About one end, the weights sit at $0$, $s$, $2s$ and $3s$:

Dimensions: $ms^{2}$ is a mass times a length squared, kg m$^2$, the unit of a moment of inertia. The ratio is a pure number, as it must be: neither $m$ nor $s$ survives in it, so the factor $2.8$ holds for any mobile built on this pattern.

Sense of variation: moving the axis from the centre to the end pushes three of the four weights farther out, the far one from $\frac{3s}{2}$ to $3s$, so $I_{e} > I_{c}$ was expected before any arithmetic. Limiting case: shrink the spacing, $s \to 0$, and both results vanish, which is right for mass sitting on the axis.

Second route, if you know the parallel axis rule (it is not on the PHYS 101 formula sheet, so the sum is the method to trust): $I_{e} = I_{c} + (4m)\left(\frac{3s}{2}\right)^{2} = 5ms^{2} + 9ms^{2} = 14ms^{2}$.

Pitfall: quoting a moment of inertia without its axis. Same object, same mass, and a factor $2.8$ between the two answers: $I$ belongs to a body AND an axis.

$ms^{2} = (0.25)(0.40)^{2} = 0.040$ kg m$^2$, so:

$I_{c} = 5(0.040) = 0.20$ kg m$^2$ and $I_{e} = 14(0.040) = 0.56$ kg m$^2$

$K_{c} = \frac{1}{2}I_{c}\omega^{2} = \frac{1}{2}(0.20)(3.0)^{2} = 0.90$ J

$K_{e} = \frac{1}{2}I_{e}\omega^{2} = \frac{1}{2}(0.56)(3.0)^{2} = 2.52$ J

At the same $\omega$ the energy follows $I$: the version hung from its end needs $2.8$ times the work, $2.52$ J against $0.90$ J, from the hand that launches it.

Pitfall: computing $\frac{1}{2}Mv^{2}$ with a single speed for the whole mobile. Each weight has its own speed $v_{i} = r_{i}\omega$, and $\frac{1}{2}I\omega^{2}$ is exactly the sum $\sum \frac{1}{2}m_{i}(r_{i}\omega)^{2}$, done once: $\frac{1}{2}(0.25)(3.0)^{2}(0 + 0.16 + 0.64 + 1.44) = 2.52$ J.

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 $R$.

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.

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 $mgh$.

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.

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.

Rolling without slipping: friction does no work, so energy is conserved, with $\omega = \frac{v}{R}$.

$v = \sqrt{\frac{4gh}{3}}$ and $\omega = \frac{v}{R} = \frac{1}{R}\sqrt{\frac{4gh}{3}}$

Shares: $\frac{1}{2}Mv^{2} = \frac{2}{3}Mgh$ goes into translation and $\frac{1}{4}Mv^{2} = \frac{1}{3}Mgh$ into spin, whatever $M$, $R$ and $h$.

Dimensions: $gh$ is in (m/s$^2$)(m) $=$ m$^2$/s$^2$, so $\sqrt{gh}$ is a speed; $\frac{v}{R}$ is in s$^{-1}$, which is rad/s since the radian carries no dimension.

Limiting case: give the body no rotational inertia, $k \to 0$ in a), and the $\frac{1}{4}Mv^{2}$ term disappears: $Mgh = \frac{1}{2}Mv^{2}$ and $v = \sqrt{2gh}$, the frictionless slide. The spool's $\sqrt{\frac{4}{3}} = 1.15$ sits below $\sqrt{2} = 1.41$, as it must.

$v = \sqrt{\frac{4(9.8)(1.10)}{3}} = \sqrt{14.37} = 3.79$ m/s

$\omega = \frac{3.79}{0.30} = 12.6$ rad/s

$K_{\text{rot}} = \frac{1}{3}Mgh = \frac{1}{3}(60)(9.8)(1.10) = 216$ J, out of $Mgh = 647$ J

Notice that $R$ sets $\omega$ but not $v$: a thicker spool turns more slowly and arrives just as fast.

Both start from rest with a constant acceleration, so the average speed over the ramp is half the final speed and $t = \frac{2L}{v}$.

Crate: $v = \sqrt{2gh} = \sqrt{2(9.8)(1.10)} = 4.64$ m/s, so $t = \frac{2(5.5)}{4.64} = 2.37$ s

Spool: $t = \frac{2(5.5)}{3.79} = 2.90$ s

$\frac{t_{\text{spool}}}{t_{\text{crate}}} = \frac{v_{\text{crate}}}{v_{\text{spool}}} = \sqrt{\frac{3}{2}} = 1.22$, so the crate arrives first.

Why the spool turns at all: about its axle the weight and the normal force have zero lever arm, since both lines pass through the axle. The only torque comes from static friction at the contact point, and it must point UP the slope to spin the spool the right way. With $\sin\theta = \frac{h}{L}$:

Along the slope: $Mg\sin\theta - f = Ma$

About the axle: $fR = I\alpha = \frac{1}{2}MR^{2}\cdot\frac{a}{R}$, so $f = \frac{1}{2}Ma$

$a = \frac{2}{3}g\sin\theta$ and $f = \frac{1}{3}Mg\sin\theta$

$\sin\theta = \frac{1.10}{5.5} = 0.20$, so $f = \frac{1}{3}(60)(9.8)(0.20) = 39.2$ N

Cross-check with f): $a = \frac{2}{3}(9.8)(0.20) = 1.31$ m/s$^2$ and $v^{2} = 2aL = 2(1.31)(5.5) = 14.37$, the same $v^{2}$ as the energy route.

Limiting case: on ice, $f \to 0$, there is no torque, the spool does not spin, and it slides down like the crate with $a = g\sin\theta$.

Pitfall: concluding that the friction eats energy, $39.2 \times 5.5 = 216$ J of heat. The contact point is at rest at each instant, so this friction does no net work: it takes $216$ J from the translation and hands exactly those $216$ J to the spin, the $K_{\text{rot}}$ of f). Rolling needs friction; it does not cost energy.

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.

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.

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.

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.