Corrected exercises: rotation of rigid bodies (203-SN1-RE)

$1800$ rev/min $=\dfrac{1800\times 2\pi}{60}=188.5$ rad/s. Starting from rest, $\alpha=\dfrac{\omega-\omega_{0}}{t}=\dfrac{188.5}{15.0}=12.57$ rad/s².

$\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}=0+\frac{1}{2}\times 12.57\times 15.0^{2}=1414$ rad. The number of revolutions is $\dfrac{1414}{2\pi}=225$ revolutions. Check with the average angular velocity: $\overline{\omega}=\dfrac{0+188.5}{2}=94.25$ rad/s, and $94.25\times 15.0=1414$ rad. The two methods agree.

$v=\omega r=188.5\times 0.300=56.6$ m/s, which is over 200 km/h. $a_{t}=\alpha r=12.57\times 0.300=3.77$ m/s². $a_{c}=\omega^{2}r=188.5^{2}\times 0.300=1.07\times 10^{4}$ m/s².

The centripetal acceleration is about 2800 times larger than the tangential acceleration, and it is more than a thousand times $g$. It is the centripetal acceleration that sets the design limit of the wheel: the material has to be held on a circular path, which requires an enormous centripetal force supplied by the internal cohesion of the disc. If the speed exceeds the limit of the material, the wheel BURSTS, with the fragments flying off along the tangent. That is why grinding wheels carry a maximum speed marked on them and are always used behind a guard. Note that $a_{c}$ varies as $\omega^{2}$: doubling the rotation speed quadruples the stress.

$\tau=rF\sin\theta=0.250\times 80.0\times\sin 60.0^{\circ}=0.250\times 80.0\times 0.866=17.3$ N·m.

Applied perpendicularly, $\sin 90^{\circ}=1$ and $\tau=0.250\times 80.0=20.0$ N·m, the MAXIMUM possible value for this force and this arm. At 60.0°, you get only 86.6 % of that maximum, a loss of 13.4 %. This is why you always pull perpendicular to a wrench: any other direction wastes part of the effort pulling the bolt instead of turning it.

We want $\tau=20.0$ N·m with $r=0.750$ m and $\sin\theta=1$: $F=\dfrac{20.0}{0.750}=26.7$ N. Tripling the arm divides the required force by three. That is the whole point of the extension, and also its danger: it becomes easy to exceed the specified tightening torque and strip the thread.

Both forces are tangential, hence perpendicular to the radius. They tend to turn the pulley in opposite directions, so their torques subtract: $\tau_{net}=\left(45.0-30.0\right)\times 0.200=15.0\times 0.200=3.00$ N·m. The pulley turns in the direction imposed by the 45.0 N force, the larger one. Note that the net force on the pulley is not zero either, but the axle takes it up: the two questions, translation and rotation, are treated separately.

The rod is uniform, with linear mass density $\lambda=\frac{M}{L}$. An element of length $dx$ at distance $x$ from the axis has mass $dm=\lambda\,dx$ and contributes $dI=x^{2}\,dm$. Integrating over the whole rod: $I=\int_{-L/2}^{L/2}\lambda x^{2}\,dx=\lambda\left[\frac{x^{3}}{3}\right]_{-L/2}^{L/2}=\lambda\frac{L^{3}}{12}=\frac{ML^{2}}{12}$. The key point of the slicing is that all the points of a given element $dx$ are at the same distance from the axis.

For a disc, you must slice into RINGS and not into strips, because all the points of a ring of radius $r$ are at the same distance from the axis. Areal mass density: $\sigma=\frac{M}{\pi R^{2}}$. A ring of radius $r$ and thickness $dr$ has area $2\pi r\,dr$, so $dm=\sigma\,2\pi r\,dr$ and $dI=r^{2}\,dm$. Thus $I=\int_{0}^{R}2\pi\sigma r^{3}\,dr=2\pi\sigma\frac{R^{4}}{4}=\frac{\pi\sigma R^{4}}{2}=\frac{MR^{2}}{2}$. A thin hoop of the same mass and radius would give $MR^{2}$, twice as much: all of its mass is at the maximum distance.

The disc is flat, so the theorem applies with $z$ the axis of symmetry. By symmetry, two perpendicular diameters are equivalent: $I_{x}=I_{y}$. Hence $I_{z}=2I_{x}$, which gives $I_{diam}=I_{x}=\frac{I_{z}}{2}=\frac{MR^{2}}{4}$. This smaller moment of inertia is the one that matters when a coin tips over on its edge instead of spinning flat.

About the centre: the rod contributes $\frac{1.20(0.800)^{2}}{12}=0.0640$ kg$\cdot$m$^{2}$, and each point mass, at $0.400$ m from the axis, contributes $2.50(0.400)^{2}=0.400$ kg$\cdot$m$^{2}$. Total: $I_{cm}=0.0640+2(0.400)=0.864$ kg$\cdot$m$^{2}$. About one end: the rod gives $\frac{1.20(0.800)^{2}}{3}=0.256$ kg$\cdot$m$^{2}$, the mass located on the axis does not contribute at all, and the other one gives $2.50(0.800)^{2}=1.60$ kg$\cdot$m$^{2}$. Total: $I=1.856$ kg$\cdot$m$^{2}$. Check with the parallel-axis theorem, with a total mass of $6.20$ kg and an offset of $0.400$ m: $I=I_{cm}+Md^{2}=0.864+6.20(0.400)^{2}=0.864+0.992=1.856$ kg$\cdot$m$^{2}$. The two methods agree.

First error: the theorem $I=I_{cm}+Md^{2}$ MUST start from the centre of mass. Let us look for the moment of inertia of the rod alone about an axis a quarter of the way along it, $0.200$ m from the centre. The correct answer is $0.0640+1.20(0.200)^{2}=0.112$ kg$\cdot$m$^{2}$. Starting wrongly from the end, you would write $0.256+1.20(0.200)^{2}=0.304$ kg$\cdot$m$^{2}$, almost triple. The theorem is not a general rule for shifting an axis: it is a relationship that singles out the centre of mass, and it follows that $I_{cm}$ is always the SMALLEST of the moments of inertia for a given direction of axis. Second error: the same rod of $1.20$ kg and $0.800$ m has $0.0640$ kg$\cdot$m$^{2}$ about its centre, $0.256$ kg$\cdot$m$^{2}$ about one end, and practically zero about its own long axis. Speaking of the moment of inertia of a body without naming the axis is meaningless, whereas speaking of its mass always makes sense.

About the centre: $I_{cm}=\frac{1}{12}ML^{2}=\frac{1}{12}\times 2.00\times 1.20^{2}=0.240$ kg·m². About one end: $I=\frac{1}{3}ML^{2}=\frac{1}{3}\times 2.00\times 1.44=0.960$ kg·m². The moment of inertia is FOUR times larger, for the same object and the same mass: it is the distribution relative to the axis that changes everything.

The distance between the centre of mass and the end is $d=\dfrac{L}{2}=0.600$ m. Then $I=I_{cm}+Md^{2}=0.240+2.00\times 0.600^{2}=0.240+0.720=0.960$ kg·m². We recover exactly the value from part a.

The weight acts at the centre of mass, 0.600 m from the axis, and it is perpendicular to the rod when the rod is horizontal. So $\tau=Mg\dfrac{L}{2}=2.00\times 9.80\times 0.600=11.8$ N·m. Then $\alpha=\dfrac{\tau}{I}=\dfrac{11.76}{0.960}=12.25$ rad/s².

The end is at $r=L=1.20$ m, so $a_{t}=\alpha L=12.25\times 1.20=14.7$ m/s². That is 1.5 times $g$: the end of the rod accelerates FASTER than an object in free fall. There is no paradox: the rod is not a free object, it is held by its pivot, and it rotates. The centre of mass, at $\frac{L}{2}$ from the axis, accelerates at $\alpha\frac{L}{2}=7.35$ m/s², which is LESS than $g$: the pivot pushes UP on the rod, with a force $M(g-7.35)=4.90$ N, a quarter of the weight, and so supports part of it. But the rod is rigid: all its points turn with the same angular acceleration, and the acceleration of a point is proportional to its distance from the axis. The end, twice as far as the centre of mass, accelerates twice as much: $2\times 7.35=14.7$ m/s². The weight does not make every point fall freely, it makes the rod rotate, and nothing prevents the end from exceeding $g$.

The coin, on the other hand, is FREE: it can only accelerate at $g=9.80$ m/s². The end of the rod, which descends at 14.7 m/s², drops away faster than the coin falls. The coin is therefore left in the air, separated from the rod, and lands farther along it, or in a cup placed at the right spot if you want to perform the classic magic trick. The crossover point is where $a_{t}=g$, that is at $r=\dfrac{g}{\alpha}=\dfrac{9.80}{12.25}=0.800$ m from the axis: beyond that, the rod outruns free fall.

Rolling without slipping means that the point of contact is instantaneously at rest, which requires $v_{cm}=\omega R$. The total kinetic energy is $K=\frac{1}{2}Mv^{2}+\frac{1}{2}I\omega^{2}$. Substituting $\omega=\dfrac{v}{R}$: $K=\frac{1}{2}Mv^{2}+\frac{1}{2}I\dfrac{v^{2}}{R^{2}}=\frac{1}{2}Mv^{2}\left(1+\dfrac{I}{MR^{2}}\right)$. The factor $\dfrac{I}{MR^{2}}$ is $\frac{2}{5}$ for the sphere, $\frac{1}{2}$ for the disc and 1 for the hoop.

By conservation of energy over a descent of height $h$: $Mgh=\frac{1}{2}Mv^{2}\left(1+\frac{I}{MR^{2}}\right)$. Differentiating, or writing $v^{2}=2ad$ with $h=d\sin\theta$, we obtain $a=\dfrac{g\sin\theta}{1+\frac{I}{MR^{2}}}$. With $g\sin 30^{\circ}=4.90$ m/s²: sphere $\dfrac{4.90}{1.4}=3.50$ m/s²; disc $\dfrac{4.90}{1.5}=3.27$ m/s²; hoop $\dfrac{4.90}{2}=2.45$ m/s².

$v=\sqrt{\dfrac{2gh}{1+\frac{I}{MR^{2}}}}$ with $h=2.00$ m. Sphere: $\sqrt{\dfrac{39.2}{1.4}}=5.29$ m/s. Disc: $\sqrt{\dfrac{39.2}{1.5}}=5.11$ m/s. Hoop: $\sqrt{\dfrac{39.2}{2}}=4.43$ m/s. The sphere wins, and the hoop comes in a clear last.

Because $M$ and $R$ DISAPPEAR from the result. The factor $\dfrac{I}{MR^{2}}$ is a pure number that depends only on the SHAPE: mass and radius cancel in the ratio. Physically, a more massive object has more potential energy but also more inertia, in the same proportion. The only criterion is therefore how the mass is distributed: the hoop has all its mass at the maximum radius, so it has to devote half of its energy to rotation, against only two sevenths for the sphere, which has more mass near the axis. A cannonball and a ball bearing arrive exactly together.

Without friction, nothing can exert a torque about the centre of mass: none of the three objects starts to rotate, and they all SLIDE without rolling. All the potential energy then goes into translational energy, and the three arrive AT THE SAME TIME, with $a=g\sin\theta=4.90$ m/s² and $v=\sqrt{2gh}=6.26$ m/s, faster than any of them reached while rolling. It is friction, often seen as a mere loss, that creates all the richness of this problem.

Solid disc: $I=\frac{1}{2}MR^{2}=\frac{1}{2}(8.00)(0.150)^{2}=0.0900$ kg$\cdot$m$^{2}$. Converting the angular velocity: $\omega=1500\times\frac{2\pi}{60}\approx 157$ rad/s. Rotational kinetic energy: $K=\frac{1}{2}I\omega^{2}=\frac{1}{2}(0.0900)(157.1)^{2}\approx 1.11\times 10^{3}$ J. This is the energy such a flywheel gives back, and it is also what makes a spinning grinding wheel dangerous.

Angular acceleration: $\alpha=\frac{0-157.1}{45.0}\approx -3.49$ rad/s$^{2}$. Friction torque: $\tau=I\left|\alpha\right|=0.0900(3.49)\approx 0.314$ N$\cdot$m. Angle swept: $\theta=\frac{\omega^{2}}{2\left|\alpha\right|}=\frac{(157.1)^{2}}{2(3.49)}\approx 3.53\times 10^{3}$ rad, or $\frac{3534}{2\pi}\approx 563$ revolutions. Check with the average angular velocity: $\theta=\bar{\omega}t=\frac{157.1}{2}(45.0)\approx 3.53\times 10^{3}$ rad, identical. Energy dissipated, from the work done by the torque: $W=\tau\theta=0.314(3534)\approx 1.11\times 10^{3}$ J, exactly the kinetic energy calculated in part a. The work-energy theorem holds in rotation as in translation, provided $Fd$ is replaced by $\tau\theta$.

At constant speed the load is in equilibrium, so the cable tension is $T=mg=50.0(9.80)=490$ N. The torque required is $\tau=TR=490(0.250)=123$ N$\cdot$m. Angular velocity: $\omega=\frac{v}{R}=\frac{0.600}{0.250}=2.40$ rad/s. Power: $P=\tau\omega=122.5(2.40)=294$ W, which the translational calculation $P=Fv=490(0.600)=294$ W confirms. The mass of the drum does NOT come into it: at constant speed the angular acceleration is zero, so its inertia requires no torque.

We now need two equations. Moment of inertia of the drum: $I=\frac{1}{2}MR^{2}=\frac{1}{2}(30.0)(0.250)^{2}=0.938$ kg$\cdot$m$^{2}$. For the load: $T-mg=ma$. For the drum: $\tau-TR=I\alpha=I\frac{a}{R}$. Dividing the second by $R$: $\frac{\tau}{R}-T=\frac{I}{R^{2}}a=15.0a$. Adding to the first: $\frac{\tau}{R}-mg=\left(m+\frac{I}{R^{2}}\right)a$, that is $640-490=65.0a$ and $a\approx 2.31$ m/s$^{2}$. Tension: $T=m(a+g)=50.0(2.31+9.80)\approx 605$ N. Check on the drum: $\tau-TR=160-605.4(0.250)=8.65$ N$\cdot$m, and $I\alpha=0.938\left(\frac{2.31}{0.250}\right)=8.65$ N$\cdot$m. The term $\frac{I}{R^{2}}=15.0$ kg reads as an extra mass to be dragged along: it is the price of the drum's inertia.

Speed: $v=at=2.308(3.00)\approx 6.92$ m/s. Height: $h=\frac{1}{2}at^{2}=\frac{1}{2}(2.308)(9.00)\approx 10.4$ m. Angle turned by the drum: $\theta=\frac{h}{R}=\frac{10.385}{0.250}\approx 41.5$ rad. Work done by the motor: $W=\tau\theta=160(41.54)\approx 6.65\times 10^{3}$ J. Breakdown: potential energy $mgh=490(10.385)\approx 5.09\times 10^{3}$ J; translational kinetic energy of the load $\frac{1}{2}(50.0)(6.923)^{2}\approx 1.20\times 10^{3}$ J; rotational kinetic energy of the drum $\frac{1}{2}(0.9375)\left(\frac{6.923}{0.250}\right)^{2}\approx 3.59\times 10^{2}$ J. Sum: $5088+1198+359\approx 6.65\times 10^{3}$ J, exactly the work done by the motor. The balance closes to the joule, and we see that $76\,\%$ of the energy actually goes into raising the load, the rest being temporarily stored as kinetic energy.

No external torque acts about the vertical axis: the ice is nearly frictionless and the weight is parallel to the axis. So $L$ is conserved: $I_{1}\omega_{1}=I_{2}\omega_{2}$, hence $\omega_{2}=\dfrac{4.50\times 1.50}{1.20}=5.63$ rad/s. The angular velocity is multiplied by 3.75, exactly the ratio of the moments of inertia.

$K_{1}=\frac{1}{2}I_{1}\omega_{1}^{2}=\frac{1}{2}\times 4.50\times 1.50^{2}=5.06$ J. $K_{2}=\frac{1}{2}\times 1.20\times 5.63^{2}=19.0$ J. The kinetic energy has been MULTIPLIED by 3.75 as well.

The 13.9 J difference comes from the skater's MUSCULAR WORK. In pulling her arms in, she has to pull them inward against their tendency to keep moving outward (the apparent centrifugal force in her own frame), and a displacement against a force means positive work done on the system. There is therefore no contradiction: conservation of angular momentum only requires the absence of EXTERNAL torque, whereas conservation of energy would require the absence of work, internal work included. The general result to remember: when $I$ decreases at constant $L$, $\omega$ varies as $\frac{1}{I}$ and $K=\frac{L^{2}}{2I}$ also varies as $\frac{1}{I}$, so the energy always increases.

The person, modelled as a point 2.00 m from the axis, contributes $I_{p}=mr^{2}=60.0\times 2.00^{2}=240$ kg·m². She arrives with no angular momentum of her own about the axis. Conservation gives $250\times 2.00=\left(250+240\right)\omega'$, hence $\omega'=\dfrac{500}{490}=1.02$ rad/s. The merry-go-round slows to roughly half its speed.

$K_{1}=\frac{1}{2}\times 250\times 2.00^{2}=500$ J and $K_{2}=\frac{1}{2}\times 490\times 1.02^{2}=255$ J: 49 % of the energy has been LOST, whereas in part b it was gained. The difference lies in the nature of the process. In part b, an internal agent supplies work in a controlled way. In part d, it is a PERFECTLY INELASTIC collision: the person, initially at rest, has to be brought into rotation by friction against the floor of the merry-go-round, and this friction dissipates energy as heat. It is the exact rotational analogue of a perfectly inelastic collision, in which momentum is conserved but kinetic energy always decreases.

For $m_{1}$, which rises: $T_{1}-m_{1}g=m_{1}a$. For $m_{2}$, which descends: $m_{2}g-T_{2}=m_{2}a$. For the pulley, the two tensions act tangentially and in opposite directions: $\left(T_{2}-T_{1}\right)R=I\alpha$, with $\alpha=\dfrac{a}{R}$ since the string does not slip. If $T_{1}$ were equal to $T_{2}$, the net torque would be zero and the pulley could not accelerate: it is precisely the DIFFERENCE in tension that sets it spinning. Equal tensions are valid only for a pulley of negligible mass.

Adding the three equations, the tensions cancel: $m_{2}g-m_{1}g=m_{1}a+m_{2}a+\dfrac{I}{R^{2}}a$, which gives the stated result. Numerically, $\dfrac{I}{R^{2}}=\dfrac{0.0900}{0.0225}=4.00$ kg, and $a=\dfrac{\left(5.00-3.00\right)\times 9.80}{3.00+5.00+4.00}=\dfrac{19.6}{12.0}=1.63$ m/s².

$T_{1}=m_{1}\left(g+a\right)=3.00\times\left(9.80+1.633\right)=34.3$ N. $T_{2}=m_{2}\left(g-a\right)=5.00\times\left(9.80-1.633\right)=40.8$ N. Indeed $T_{2}>T_{1}$. Check: $\left(T_{2}-T_{1}\right)R=\left(40.83-34.30\right)\times 0.150=0.980$ N·m, and $I\alpha=0.0900\times\dfrac{1.633}{0.150}=0.0900\times 10.89=0.980$ N·m. The two coincide.

With a massless pulley, $I=0$ and $a=\dfrac{19.6}{8.00}=2.45$ m/s², which is 50 % more. The term $\dfrac{I}{R^{2}}$, which has the dimensions of a mass and equals 4.00 kg here, simply adds to the translating masses: it is the pulley's EQUIVALENT MASS, that is, the mass that would have to be added to the translating system to produce the same inertia. A massive pulley therefore slows the system exactly as an extra load would, which gives a very intuitive reading of $I$.

After $t=1.50$ s, $v=at=1.633\times 1.50=2.45$ m/s and $\omega=\dfrac{v}{R}=16.3$ rad/s. The total kinetic energy is $K=\frac{1}{2}\left(m_{1}+m_{2}\right)v^{2}+\frac{1}{2}I\omega^{2}=\frac{1}{2}\times 8.00\times 6.00+\frac{1}{2}\times 0.0900\times 266.9=24.0+12.0=36.0$ J. Check with energy: each mass has moved $d=\frac{1}{2}at^{2}=1.84$ m, so $m_{2}$ has descended 1.84 m and $m_{1}$ has risen the same distance. The change in potential energy is $\left(m_{2}-m_{1}\right)gd=2.00\times 9.80\times 1.838=36.0$ J, exactly the kinetic energy gained. Since no friction was assumed, the balance is perfect.