Torque and Rotational Motion Practice Problems

Practice problems

Work each one before opening the solution. Every step is shown, with the arithmetic left in so you can find where yours diverged.

1. 1. A wrench pulled straight, then pulled at an angle

   You pull on a wrench handle at a point 0.45 m from the center of the bolt, with a force of 60.0 N. (a) Find the torque about the bolt when the pull is perpendicular to the handle. (b) Find the torque when the same 60.0 N is applied at $40^\circ$ to the handle. (c) Find the lever arm in case (b) and use it to reproduce the torque.

   Show the worked solution

   1. The axis is the bolt and $r = 0.45\ \text{m}$ is the distance from that axis to your hand. Use $\tau = rF\sin\theta$, where $\theta$ is the angle between the force and the position vector running from the axis to the point where the force is applied.

   2. (a) Perpendicular means $\theta = 90^\circ$, and $\sin 90^\circ = 1$, so the sine factor disappears: $\tau = (0.45\ \text{m})(60.0\ \text{N})(1) = 27.0\ \text{N} \cdot \text{m}$. That is the largest torque this force can produce from this distance.

   3. (b) At $40^\circ$, $\sin 40^\circ = 0.6428$, so $\tau = (0.45)(60.0)(0.6428) = (27.0)(0.6428) = 17.4\ \text{N} \cdot \text{m}$.

   4. (c) The lever arm is the perpendicular distance from the axis to the line along which the force acts: $r_{\perp} = r\sin\theta = (0.45\ \text{m})(0.6428) = 0.2893\ \text{m}$, which is $0.289\ \text{m}$ to three figures.

   5. Check: $\tau = r_{\perp}F = (0.2893\ \text{m})(60.0\ \text{N}) = 17.36\ \text{N}\cdot\text{m}$, the same $17.4\ \text{N}\cdot\text{m}$ as part (b). The two routes are one calculation with the sine attached to a different factor, which is why the sheet prints them as one line, $\tau = r_{\perp}F = rF\sin\theta$.

   6. Sanity check on the size. $\sin 40^\circ = 0.643$, so the angled pull delivers 64.3 percent of the perpendicular maximum. Tilting your pull $50^\circ$ away from perpendicular threw away more than a third of the torque you were paying for, which is why you instinctively pull square on a stiff bolt.

   (a) $\tau = 27.0\ \text{N}\cdot\text{m}$. (b) $\tau = 17.4\ \text{N}\cdot\text{m}$. (c) The lever arm is $0.289\ \text{m}$, and multiplying it by 60.0 N reproduces the part (b) torque. The angled pull is worth 64.3 percent of the perpendicular one.

2. 2. Two ropes, one wheel, one net torque

   A wheel turns freely about a fixed axle through its center. A rope wound around the rim, at radius 0.30 m, is pulled tangentially with 25 N in the sense that would turn the wheel clockwise. A second rope wound around an inner hub, at radius 0.12 m, is pulled tangentially with 45 N in the counterclockwise sense. Find the net torque on the wheel and the sense in which it starts to turn.

   Show the worked solution

   1. Take counterclockwise as positive. The axis is the axle, and both ropes leave their drums tangentially, so each force is already perpendicular to its own position vector: $\sin\theta = 1$ in both cases and $\tau = rF$ with no trigonometry.

   2. Rim rope: $\tau_1 = -(0.30\ \text{m})(25\ \text{N}) = -7.5\ \text{N}\cdot\text{m}$. Negative because it acts clockwise.

   3. Hub rope: $\tau_2 = +(0.12\ \text{m})(45\ \text{N}) = +5.4\ \text{N}\cdot\text{m}$.

   4. Net torque: $\tau_{\text{net}} = (-7.5) + (+5.4) = -2.1\ \text{N}\cdot\text{m}$, so $2.1\ \text{N}\cdot\text{m}$ clockwise. The wheel starts turning clockwise, the way the smaller force pulls it.

   5. Why the larger force lost, in one line. It is 1.8 times bigger, $45/25 = 1.8$, but it acts at 0.40 of the radius, $0.12/0.30 = 0.40$, and $(1.8)(0.40) = 0.72$, which is less than one. Where a force acts counts exactly as much as how hard it pulls.

   6. Adding magnitudes instead of signed values would give $7.5 + 5.4 = 12.9\ \text{N}\cdot\text{m}$, more than six times the real answer and pointing nowhere. Torques about one axis add like signed numbers, and the sign is the whole of the bookkeeping.

   7. A scope note while the signs are fresh. The Topic 5.3 boundary statement says that while AP Physics 1 expects students to mathematically manipulate the magnitude of torque using vector conventions, the direction of torque is beyond the scope of the course. The useful output here is the magnitude, $2.1\ \text{N}\cdot\text{m}$, together with the sense of rotation it would produce.

   $\tau_{\text{net}} = 2.1\ \text{N}\cdot\text{m}$ clockwise. The 45 N pull is nearly twice the force but acts at 0.40 of the radius, so it produces the smaller torque and loses.

3. 3. Spinning a rotor up from rest

   A centrifuge rotor starts from rest and reaches an angular velocity of 90.0 rad/s in 15.0 s under constant angular acceleration. Find (a) the angular acceleration, (b) the angle it turns through in that time, and (c) the number of complete revolutions.

   Show the worked solution

   1. Take the direction of the spin as positive. You have $\omega_0 = 0$, $\omega = 90.0\ \text{rad/s}$ and $t = 15.0\ \text{s}$, and you want $\alpha$ and $\theta$. The three constant angular acceleration equations are the linear kinematics equations with the symbols swapped, so pick them the same way: choose the one that is missing the quantity you do not care about.

   2. (a) $\omega = \omega_0 + \alpha t$ with $\omega_0 = 0$ rearranges to $\alpha = \omega/t = (90.0\ \text{rad/s})/(15.0\ \text{s}) = 6.00\ \text{rad/s}^2$.

   3. (b) $\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}(6.00)(15.0)^2 = (3.00)(225) = 675\ \text{rad}$.

   4. Check with the average angular velocity, which is only legitimate because $\alpha$ is constant: $\omega_{\text{avg}} = (0 + 90.0)/2 = 45.0\ \text{rad/s}$, and $(45.0)(15.0) = 675\ \text{rad}$.

   5. Check a third way, with the equation that has no time in it: $\omega^2 = \omega_0^2 + 2\alpha\theta$ gives $\theta = (90.0)^2/(2 \times 6.00) = 8100/12.00 = 675\ \text{rad}$. Three routes, one number.

   6. (c) One revolution is $2\pi$ radians, so $675/(2\pi) = 675/6.2832 = 107.4$ revolutions. That is 107 complete turns plus about four tenths of another.

   7. A units note worth carrying. Everything above would have worked in degrees, or in revolutions, as long as every angular quantity used the same unit: the three equations are unit-agnostic that way. Radians are only forced on you once the rotation is connected to something linear, which is what problem 6 does.

   (a) $\alpha = 6.00\ \text{rad/s}^2$. (b) $\theta = 675\ \text{rad}$. (c) $675/(2\pi) = 107.4$ revolutions, so 107 complete ones.

4. 4. A seesaw whose pivot is not in the middle

   A uniform plank 3.6 m long and of mass 8.0 kg rests on a pivot placed 1.2 m from its left end. A 25 kg child sits on the extreme left end. How far to the right of the pivot must a 15 kg child sit for the plank to stay horizontal and still? Use $g = 9.8\ \text{m/s}^2$.

   Show the worked solution

   1. Take the pivot as the axis and counterclockwise as positive. With the plank drawn horizontally, a downward force applied to the left of the pivot turns it counterclockwise, and a downward force applied to the right turns it clockwise.

   2. Locate all three weights relative to the axis. The 25 kg child is 1.2 m to the left. The plank is uniform, so its entire weight acts at its center, 1.8 m from the left end, which is $1.8 - 1.2 = 0.6\ \text{m}$ to the right of the pivot. The 15 kg child sits an unknown distance $d$ to the right.

   3. The plank is not turning, so the net torque about the axis is zero:

      $$(25)(9.8)(1.2) - (8.0)(9.8)(0.6) - (15)(9.8)\,d = 0$$

   4. Every term carries the same factor of 9.8, so $g$ divides straight out before you touch a calculator: $(25)(1.2) - (8.0)(0.6) - 15d = 0$. That happens in any balance where every torque comes from a weight, and it is worth spotting because it removes a whole class of arithmetic slips.

   5. $30 - 4.8 = 15d$, so $25.2 = 15d$ and $d = 1.68\ \text{m}$.

   6. Check with $g$ put back in, as three real torques. Counterclockwise: $(25)(9.8)(1.2) = 294.0\ \text{N}\cdot\text{m}$. Clockwise: $(8.0)(9.8)(0.6) = 47.04\ \text{N}\cdot\text{m}$ from the plank and $(15)(9.8)(1.68) = 246.96\ \text{N}\cdot\text{m}$ from the second child. And $47.04 + 246.96 = 294.00\ \text{N}\cdot\text{m}$ exactly.

   7. Check that the answer fits on the plank. There is $3.6 - 1.2 = 2.4\ \text{m}$ of board to the right of the pivot, and 1.68 m sits comfortably inside it.

   8. Notice what the off-center pivot bought. With the pivot at the plank's center, the plank's own weight would produce no torque about it and the answer would be $(25)(1.2)/15 = 2.0\ \text{m}$. Shifting the pivot left puts more of the board on the right, and that extra clockwise torque lets the second child sit 0.32 m closer in.

   9. One force never appeared: the pivot pushes up with the full $(25 + 8.0 + 15)(9.8) = 470.4\ \text{N}$. It acts at the axis, so its lever arm is zero and it contributes no torque. Choosing the axis at the unknown force is what made this a single equation.

   The 15 kg child must sit $1.68\ \text{m}$ from the pivot on the right, about 1.7 m. Because every torque comes from a weight, $g$ cancels out of the balance entirely, and the plank's own 8.0 kg matters only because the pivot is off center.

5. 5. Rotational inertia of a loaded rod, about two axes

   Two 0.60 kg balls are fixed at the two ends of a uniform rod of mass 0.40 kg and length 0.90 m. Treat the balls as point objects, and take every axis below as perpendicular to the rod. You are given that a uniform rod's rotational inertia about an axis through its center is $I_{\text{cm}} = \frac{1}{12}ML^2$. Find (a) the rotational inertia of the two balls alone about the center axis, (b) the rotational inertia of the whole assembly about the center axis, and (c) the rotational inertia about an axis through one end, first with the parallel axis theorem and then by direct summation.

   Show the worked solution

   1. Each ball sits $r = L/2 = 0.45\ \text{m}$ from the center axis. Point objects are exactly what $I = \sum m_i r_i^2$ is for.

   2. (a) $I_{\text{balls}} = 2(0.60\ \text{kg})(0.45\ \text{m})^2 = 2(0.60)(0.2025) = 0.243\ \text{kg}\cdot\text{m}^2$.

   3. (b) Rotational inertias about the same axis add. The rod contributes $\frac{1}{12}(0.40)(0.90)^2 = \frac{1}{12}(0.40)(0.81) = 0.0270\ \text{kg}\cdot\text{m}^2$, so $I_{\text{cm}} = 0.243 + 0.0270 = 0.270\ \text{kg}\cdot\text{m}^2$.

   4. Look at that split before moving on. The balls carry 1.20 kg between them and supply $0.243/0.270 = 90$ percent of the rotational inertia; the 0.40 kg rod supplies 10 percent. The rod's mass is spread across every distance from 0 to 0.45 m, while the balls sit entirely at the far end of that range, and $r^2$ rewards distance heavily.

   5. (c) Parallel axis theorem, $I' = I_{\text{cm}} + Md^2$. The assembly is symmetric, so its center of mass is at the rod's center, and the new axis is $d = 0.45\ \text{m}$ away. The total mass is $M = 0.60 + 0.60 + 0.40 = 1.60\ \text{kg}$.

   6. $I' = 0.270 + (1.60)(0.45)^2 = 0.270 + (1.60)(0.2025) = 0.270 + 0.324 = 0.594\ \text{kg}\cdot\text{m}^2$.

   7. Now check it without the theorem. About the end axis, one ball sits at $r = 0$ and contributes nothing at all, while the other sits at $r = 0.90\ \text{m}$ and contributes $(0.60)(0.81) = 0.486\ \text{kg}\cdot\text{m}^2$. The rod about its own end is $\frac{1}{12}(0.40)(0.81) + (0.40)(0.45)^2 = 0.0270 + 0.0810 = 0.108\ \text{kg}\cdot\text{m}^2$, the familiar $\frac{1}{3}ML^2$. Total: $0.486 + 0.108 = 0.594\ \text{kg}\cdot\text{m}^2$. Identical.

   8. Moving the axis from the center to the end multiplied the rotational inertia by $0.594/0.270 = 2.2$. Essential knowledge 5.4.B.1 is the general version: a rigid system's rotational inertia in a given plane is at a minimum when the rotational axis passes through the system's center of mass. No axis anywhere gives a smaller value than 0.270 here.

   9. Note where $\frac{1}{12}ML^2$ came from. The problem handed it to you, and the Topic 5.4 boundary statement says that is what the exam does: students do not need to know the rotational inertia of extended rigid systems, as these will be provided within the exam. Read the stem before assuming you were meant to recall one.

   (a) $0.243\ \text{kg}\cdot\text{m}^2$. (b) $0.270\ \text{kg}\cdot\text{m}^2$. (c) $0.594\ \text{kg}\cdot\text{m}^2$, the same from the parallel axis theorem and from summing $m r^2$ term by term. Moving the axis from the center to the end multiplies the rotational inertia by 2.2.

6. 6. A winch, from torque to rope speed

   A rope is wound around a winch drum of radius 0.16 m whose rotational inertia about its axle is $0.24\ \text{kg}\cdot\text{m}^2$. The drum starts at rest. The rope is pulled with a constant 30 N force tangential to the drum, while the bearings exert a constant frictional torque of $0.60\ \text{N}\cdot\text{m}$ opposing the rotation. Find (a) the angular acceleration, (b) the angular velocity after 3.0 s, (c) the speed of the rope at that instant, and (d) the length of rope unwound during the 3.0 s.

   Show the worked solution

   1. Take the sense the rope turns the drum as positive. The axis is the axle.

   2. (a) The rope leaves the drum tangentially, so its torque is $\tau = rF = (0.16\ \text{m})(30\ \text{N}) = 4.8\ \text{N}\cdot\text{m}$. Friction opposes the motion, so it enters as $-0.60\ \text{N}\cdot\text{m}$. Net torque: $4.8 - 0.60 = 4.2\ \text{N}\cdot\text{m}$.

   3. Newton's second law in rotational form: $\alpha = \tau_{\text{net}}/I = (4.2\ \text{N}\cdot\text{m})/(0.24\ \text{kg}\cdot\text{m}^2) = 17.5\ \text{rad/s}^2$.

   4. (b) $\omega = \omega_0 + \alpha t = 0 + (17.5)(3.0) = 52.5\ \text{rad/s}$.

   5. (c) Here the angular unit stops being a formality. $v = r\omega$ holds only when $\omega$ is in radians per second, and it is: $v = (0.16\ \text{m})(52.5\ \text{rad/s}) = 8.4\ \text{m/s}$. Had you carried the angular velocity in revolutions per second, the same multiplication would be wrong by a factor of $2\pi$.

   6. (d) $\theta = \frac{1}{2}\alpha t^2 = \frac{1}{2}(17.5)(3.0)^2 = \frac{1}{2}(17.5)(9.0) = 78.75\ \text{rad}$, and the rope paid out is the arc length that angle sweeps at the drum's surface: $\Delta x = r\Delta\theta = (0.16)(78.75) = 12.6\ \text{m}$.

   7. Check part (d) twice over without touching angles. The rope's linear acceleration is $a_T = r\alpha = (0.16)(17.5) = 2.8\ \text{m/s}^2$, so from rest it covers $\frac{1}{2}(2.8)(3.0)^2 = 12.6\ \text{m}$. Or use the average rope speed: $(0 + 8.4)/2 = 4.2\ \text{m/s}$ for 3.0 s is $12.6\ \text{m}$.

   8. The trap is dropping the friction torque. Using $4.8\ \text{N}\cdot\text{m}$ as the net torque gives $\alpha = 20\ \text{rad/s}^2$, $\theta = 90\ \text{rad}$ and $14.4\ \text{m}$ of rope, about 14 percent too much. Only the net torque goes into $\alpha = \tau_{\text{net}}/I$.

   (a) $\alpha = 17.5\ \text{rad/s}^2$. (b) $\omega = 52.5\ \text{rad/s}$. (c) The rope moves at $8.4\ \text{m/s}$. (d) $12.6\ \text{m}$ of rope comes off the drum.

7. 7. A beam on two supports, with the person near the end

   A uniform beam of mass 24 kg and length 5.0 m lies horizontally across two supports: one at its left end, the other 4.0 m from the left end. A 60 kg person stands 4.5 m from the left end. Find the upward force each support exerts. Use $g = 9.8\ \text{m/s}^2$.

   Show the worked solution

   1. Two unknown forces means two equations: the forces must balance and the torques must balance. Take up as positive for forces and counterclockwise as positive for torques.

   2. Weights first. The beam is uniform, so its $W_b = (24)(9.8) = 235.2\ \text{N}$ acts at its center, 2.5 m from the left end. The person's $W_p = (60)(9.8) = 588\ \text{N}$ acts at 4.5 m. Total downward force: $235.2 + 588 = 823.2\ \text{N}$.

   3. Choose the axis at the left support. $N_1$ acts exactly there, so its lever arm is zero and it drops out of the torque equation, leaving one equation with one unknown. Putting the axis on an unknown force is the whole trick of a two-support problem.

   4. Torques about the left support, with every distance measured from it:

      $$N_2(4.0) - (235.2)(2.5) - (588)(4.5) = 0$$

   5. $(235.2)(2.5) = 588.0\ \text{N}\cdot\text{m}$ and $(588)(4.5) = 2646\ \text{N}\cdot\text{m}$, so $N_2(4.0) = 3234\ \text{N}\cdot\text{m}$ and $N_2 = 3234/4.0 = 808.5\ \text{N}$.

   6. Now the force equation: $N_1 + N_2 = 823.2\ \text{N}$, so $N_1 = 823.2 - 808.5 = 14.7\ \text{N}$.

   7. Check by redoing the torques about the other support, at 4.0 m. The beam's center is 1.5 m to its left and the person 0.5 m to its right, and $N_1$ acts 4.0 m to its left: $N_1(4.0) = (235.2)(1.5) - (588)(0.5) = 352.8 - 294.0 = 58.8$, so $N_1 = 14.7\ \text{N}$. Two independent routes, one answer.

   8. That 14.7 N looks alarmingly small, and it should. Slide the person a little further right and $N_1$ falls to zero: solving $(235.2)(1.5) = 588(x - 4.0)$ gives $x = 4.6\ \text{m}$. Standing at 4.5 m, the person is 0.1 m short of lifting the left end of the beam clean off its support.

   9. A sign check that catches real errors: both support forces came out positive. A support that can only push is not allowed to give a negative answer, so a minus sign there would mean the beam has already tipped and your model no longer describes it.

   The left support pushes up with $14.7\ \text{N}$ and the support at 4.0 m with $808.5\ \text{N}$, together the full 823.2 N of weight. The left force is nearly zero because the person is standing 0.1 m short of the position that would tip the beam off that support.

8. 8. Angular impulse spinning up a flywheel

   A flywheel with rotational inertia $3.2\ \text{kg}\cdot\text{m}^2$ about its axle sits at rest. A motor applies a constant torque of $4.0\ \text{N}\cdot\text{m}$ for 6.0 s, and bearing friction is negligible. Find (a) the angular impulse delivered, (b) the flywheel's angular momentum and angular velocity at the end of the push, and (c) its rotational kinetic energy, checked against the work the torque did.

   Show the worked solution

   1. Take the driven sense as positive. Everything below is about one axis, the axle.

   2. (a) For a constant torque, $\Delta L = \tau\Delta t = (4.0\ \text{N}\cdot\text{m})(6.0\ \text{s}) = 24\ \text{N}\cdot\text{m}\cdot\text{s}$. That is the angular impulse, and $\text{N}\cdot\text{m}\cdot\text{s}$ and $\text{kg}\cdot\text{m}^2/\text{s}$ are the same unit, exactly as $\text{N}\cdot\text{s}$ and $\text{kg}\cdot\text{m/s}$ are the same unit for linear impulse.

   3. (b) The wheel started at rest, so its final angular momentum is the whole of the change: $L = 24\ \text{kg}\cdot\text{m}^2/\text{s}$. Then $L = I\omega$ gives $\omega = L/I = 24/3.2 = 7.5\ \text{rad/s}$.

   4. Cross-check through the second law instead of through impulse: $\alpha = \tau/I = 4.0/3.2 = 1.25\ \text{rad/s}^2$, and $\omega = \alpha t = (1.25)(6.0) = 7.5\ \text{rad/s}$. The two routes have to agree, because $\Delta L = \tau\Delta t$ is just $\tau = I\alpha$ multiplied through by the time.

   5. (c) $K = \frac{1}{2}I\omega^2 = \frac{1}{2}(3.2)(7.5)^2 = (1.6)(56.25) = 90\ \text{J}$.

   6. Check that against the work done. The wheel turned through $\theta = \frac{1}{2}\alpha t^2 = \frac{1}{2}(1.25)(36) = 22.5\ \text{rad}$, and $W = \tau\Delta\theta = (4.0)(22.5) = 90\ \text{J}$. The work the torque did is exactly the kinetic energy the wheel gained, which is the work-energy theorem written in angular symbols.

   7. The two answers are answering different questions about one push, and it is worth keeping them apart. The torque acted for 6.0 s, and that fixes the angular momentum. It acted through 22.5 rad, and that fixes the energy. Run the motor for 12.0 s instead and the angular momentum doubles to $48\ \text{kg}\cdot\text{m}^2/\text{s}$ while the energy quadruples to 360 J.

   8. For the picture: 22.5 rad is $22.5/(2\pi) = 3.58$ revolutions, so the whole spin-up takes less than four turns.

   (a) Angular impulse $= 24\ \text{N}\cdot\text{m}\cdot\text{s}$. (b) $L = 24\ \text{kg}\cdot\text{m}^2/\text{s}$ and $\omega = 7.5\ \text{rad/s}$. (c) $K = 90\ \text{J}$, which matches the work $\tau\Delta\theta$ over the 22.5 rad turned, to the joule.

9. 9. The spinning skater pulls her arms in

   A skater spins on frictionless ice at 1.5 rad/s with her arms and one leg extended, giving a rotational inertia of $5.4\ \text{kg}\cdot\text{m}^2$ about her vertical spin axis. She pulls everything in close to her body, dropping her rotational inertia to $1.8\ \text{kg}\cdot\text{m}^2$. Find (a) her new angular velocity, (b) her rotational kinetic energy before and after, and (c) where the change in kinetic energy came from.

   Show the worked solution

   1. Positive is the sense she is already turning. Take the skater as the system. The ice is frictionless and every force she applies to move her arms is internal to that system, so no net external torque acts about the spin axis and her angular momentum is constant. Essential knowledge 6.4.B.2 says exactly that: if the net external torque exerted on a selected object or rigid system is zero, the total angular momentum of that system is constant.

   2. (a) $L = I_i\omega_i = (5.4\ \text{kg}\cdot\text{m}^2)(1.5\ \text{rad/s}) = 8.1\ \text{kg}\cdot\text{m}^2/\text{s}$, and that number does not move for the rest of the problem. So $\omega_f = L/I_f = (8.1)/(1.8) = 4.5\ \text{rad/s}$.

   3. The ratio is the fast route and the better check: she cut $I$ by a factor of 3, so $\omega$ rises by a factor of 3, from 1.5 to 4.5 rad/s. In periods, one turn took $2\pi/1.5 = 4.19\ \text{s}$ and now takes $2\pi/4.5 = 1.40\ \text{s}$.

   4. (b) $K_i = \frac{1}{2}(5.4)(1.5)^2 = (2.7)(2.25) = 6.075\ \text{J}$, and $K_f = \frac{1}{2}(1.8)(4.5)^2 = (0.9)(20.25) = 18.225\ \text{J}$.

   5. So the kinetic energy tripled while the angular momentum sat perfectly still. The cleanest way to see why is to write the energy in terms of the conserved quantity: $K = L^2/(2I)$. With $L$ fixed, $K$ goes as $1/I$, so cutting $I$ by 3 multiplies $K$ by 3. Check it: $(8.1)^2/(2 \times 5.4) = 65.61/10.8 = 6.075\ \text{J}$ and $(8.1)^2/(2 \times 1.8) = 65.61/3.6 = 18.225\ \text{J}$.

   6. (c) From her muscles. The rise is $18.225 - 6.075 = 12.15\ \text{J}$, and she supplied every joule of it by pulling mass inward while the rotation was trying to carry it outward. That is a force acting through a displacement, which is work.

   7. Nothing external did work and nothing external exerted a torque, and those are two separate facts. The torque statement is why the angular momentum held; the work statement is why the energy did not have to.

   8. Essential knowledge 6.4.A.2.iii names this case directly: the angular speed of a nonrigid system may change without the angular momentum of the system changing, if the system changes shape by moving mass closer to or further from the rotational axis. The word nonrigid is load-bearing. A rigid body cannot do this to itself.

   9. Run it backwards as a final check. If she extends again to $5.4\ \text{kg}\cdot\text{m}^2$, the angular velocity drops back to $8.1/5.4 = 1.5\ \text{rad/s}$ and the rotation does 12.15 J of work on her arms as they swing out, which her muscles absorb. The angular momentum reads $8.1\ \text{kg}\cdot\text{m}^2/\text{s}$ at every instant of both moves.

   (a) $\omega_f = 4.5\ \text{rad/s}$, three times faster. (b) 6.075 J before, 18.225 J after. (c) From the skater: she does 12.15 J of work pulling mass toward the axis. Her angular momentum holds at $8.1\ \text{kg}\cdot\text{m}^2/\text{s}$ throughout, because nothing outside her exerts a torque about the spin axis.

10. 10. A child jumps onto a stationary roundabout

    A playground roundabout is a uniform disk of mass 120 kg and radius 1.50 m, free to turn about a frictionless vertical axle through its center, and initially at rest. Its rotational inertia about that axle is $I = \frac{1}{2}MR^2$. A 30 kg child runs at 4.0 m/s along a line tangent to the rim and jumps on at the very edge, holding on. Find (a) the roundabout's rotational inertia, (b) the child's angular momentum about the axle just before landing, (c) the angular velocity of child and roundabout just after, (d) the child's speed just after, and (e) the kinetic energy lost in the landing.

    Show the worked solution

    1. Take the sense the child is running as positive, and take child plus roundabout as the system. The axle can push on the disk but cannot exert a torque about itself, and the child's weight is vertical and so produces no torque about a vertical axis, so the system's angular momentum about that axle is conserved through the landing.

    2. (a) $I_{\text{disk}} = \frac{1}{2}MR^2 = \frac{1}{2}(120\ \text{kg})(1.50\ \text{m})^2 = \frac{1}{2}(120)(2.25) = 135\ \text{kg}\cdot\text{m}^2$. That formula came from the problem statement, as the Topic 5.4 boundary statement says it will: rotational inertias of extended rigid systems are provided within the exam.

    3. (b) An object moving in a straight line still has angular momentum about a point it is not heading toward, and the sheet prints the expression for it: $L = rmv\sin\theta$. Running tangent to the rim puts the velocity perpendicular to the line from the axle, so $\theta = 90^\circ$ and $\sin\theta = 1$: $L = (1.50\ \text{m})(30\ \text{kg})(4.0\ \text{m/s}) = 180\ \text{kg}\cdot\text{m}^2/\text{s}$.

    4. (c) Once aboard, the child rides at the rim and counts as a point mass at $r = 1.50\ \text{m}$: $I_{\text{child}} = (30)(1.50)^2 = 67.5\ \text{kg}\cdot\text{m}^2$. The system's rotational inertia is $135 + 67.5 = 202.5\ \text{kg}\cdot\text{m}^2$.

    5. Conservation of angular momentum then gives $\omega = L/I_{\text{total}} = (180)/(202.5) = 0.8889\ \text{rad/s}$, which is $0.889\ \text{rad/s}$.

    6. (d) $v = r\omega = (1.50)(0.8889) = 1.33\ \text{m/s}$. The child dropped from 4.0 m/s to a third of that, and the factor of 3 is $I_{\text{total}}/I_{\text{child}} = 202.5/67.5$.

    7. (e) Before: $K_i = \frac{1}{2}(30)(4.0)^2 = 240\ \text{J}$, all of it the child's. After: $K_f = \frac{1}{2}(202.5)(0.8889)^2 = \frac{1}{2}(202.5)(0.7901) = 80.0\ \text{J}$.

    8. Get that final energy a second way, straight from the conserved quantity and never touching the rounded $\omega$: $K = L^2/(2I) = (180)^2/(2 \times 202.5) = 32400/405 = 80.0\ \text{J}$. Exactly the same.

    9. So $240 - 80.0 = 160\ \text{J}$, two thirds of the original kinetic energy, went into the scuff of the landing, the flex of the child's legs, sound and warming. The surviving fraction is $I_{\text{child}}/I_{\text{total}} = 67.5/202.5 = 1/3$, which is the rotational twin of the result for a perfectly inelastic collision with a stationary target.

    10. The trap, priced out. Anyone who sets the child's 240 J equal to $\frac{1}{2}I_{\text{total}}\omega^2$ gets $\omega = \sqrt{2(240)/202.5} = \sqrt{2.370} = 1.54\ \text{rad/s}$, 73 percent too fast. Angular momentum carries you through the landing; energy does not, because the landing is not an energy-conserving event.

    (a) $135\ \text{kg}\cdot\text{m}^2$. (b) $180\ \text{kg}\cdot\text{m}^2/\text{s}$. (c) $\omega = 0.889\ \text{rad/s}$. (d) The child now moves at $1.33\ \text{m/s}$. (e) The system keeps 80.0 J of the original 240 J, so 160 J is lost. Angular momentum is conserved through the landing; kinetic energy is not.

11. 11. Why it matters where you push on a gate

    A garden gate 1.2 m wide is hinged along one edge. Using torque, explain why an 18 N push applied at the latch edge swings it easily, why the same 18 N applied 0.30 m from the hinge barely moves it, and why a push directed along the gate straight at the hinge does nothing at all however hard you push. Support the argument with the three torques.

    Show the worked solution

    1. The axis is the hinge line. Whether a rigid body starts turning about a fixed axis is settled by torque, not by force, and $\tau = rF\sin\theta$ has exactly three factors. The whole answer is a statement about what each of them is doing.

    2. Case 1, at the latch edge, perpendicular to the gate. Here $r = 1.2\ \text{m}$ and $\theta = 90^\circ$, so $\tau = (1.2)(18)(1) = 21.6\ \text{N}\cdot\text{m}$.

    3. Case 2, 0.30 m from the hinge, still perpendicular. $\tau = (0.30)(18)(1) = 5.4\ \text{N}\cdot\text{m}$. Same force, same direction, one quarter the torque, because $r$ fell to a quarter of its value. Torque is linear in the distance from the axis, so where you push is not a detail of the push, it is a factor in the answer.

    4. Case 3, at the latch edge but aimed along the gate toward the hinge. The force is now parallel to the position vector, so $\theta = 0^\circ$, $\sin 0^\circ = 0$, and $\tau = 0$ for any force whatever. Essential knowledge 5.3.A.1 states the reason: torque results only from the force component perpendicular to the position vector from the axis of rotation to the point of application of the force. A push aimed at the hinge has no such component, so it has nothing to contribute.

    5. The lever arm reaches the same verdict from the other side. Essential knowledge 5.3.A.2 defines the lever arm as the perpendicular distance from the axis of rotation to the line of action of the force. In case 3 the line of action runs straight through the hinge, so that perpendicular distance is zero and $\tau = r_{\perp}F = 0$ again.

    6. The general case sits between the two extremes. Push at the latch edge but at $30^\circ$ to the gate and $\tau = (1.2)(18)\sin 30^\circ = (21.6)(0.5) = 10.8\ \text{N}\cdot\text{m}$, exactly half of case 1, because only half of the 18 N is perpendicular to the gate.

    7. Then the consequence for the motion, through $\alpha_{\text{sys}} = \tau_{\text{net}}/I_{\text{sys}}$. The gate's rotational inertia about its hinges is a property of the gate and does not care where you push it, so the angular acceleration is proportional to the torque. Case 1 starts the gate turning four times as briskly as case 2, and case 3 never starts it turning at all.

    8. The same three factors are why door handles are fitted as far from the hinges as the door allows, why a longer wrench beats a shorter one on a seized bolt, and why pushing near the hinge side of a heavy door feels like pushing on a wall.

    Because torque is what starts a rigid body turning, and torque is $rF\sin\theta$. The same 18 N gives $21.6\ \text{N}\cdot\text{m}$ at the latch edge, $5.4\ \text{N}\cdot\text{m}$ at 0.30 m from the hinge (a quarter of the distance, so a quarter of the torque), and exactly zero when aimed along the gate at the hinge, because then the force has no component perpendicular to the position vector and its line of action passes through the axis, making the lever arm zero. Since $\alpha = \tau_{\text{net}}/I$ and the gate's rotational inertia is the same in all three cases, the angular accelerations stand in the same ratio as the torques.