Solved Examples

How to Use This Problem Bank

Everything in this chapter now has to work together. A rolling-cylinder question is a moment-of-inertia question wearing a disguise; an exploding-shell question is a centre-of-mass question; a ladder question is a torque question. These 44 problems are arranged so that the early ones drill one idea each and the later ones need two or three at once.

Work them with a pen, not with your eyes. Cover the solution, try it, then compare — including the checks, because the checks are where the marks are lost.

The four questions to ask before you write anything

1. What is the system? One particle, a rigid body, or a body plus something sitting on it?
2. Which point or axis am I taking moments about? Choose it so that the forces you do not know pass straight through it and vanish.
3. Is anything conserved? No external force gives you momentum; no external torque about a point gives you angular momentum about that point.
4. Does the answer have the right units and the right size? A moment of inertia in kg m$^2$, an angular speed in rad/s, an answer that is bigger than the thing it came from is a mistake.

Key Point: Torque, angular momentum and moment of inertia are all defined about something. An answer that does not say "about which axis" or "about which point" is not an answer. Write the axis down before the number.

The standard moments of inertia used throughout

Everything below is about an axis through the centre of mass unless the row says otherwise. $M$ is the total mass, $R$ the radius, $L$ the length, $k$ the radius of gyration with $I = Mk^2$.

Body	Axis	$I$	$k^2/R^2$
Thin ring / hoop	through centre, perpendicular to plane	$MR^2$	$1$
Thin ring	any diameter	$\frac{1}{2}MR^2$	—
Disc / solid cylinder	central axis	$\frac{1}{2}MR^2$	$\frac{1}{2}$
Disc	any diameter	$\frac{1}{4}MR^2$	—
Hollow cylinder (thin)	central axis	$MR^2$	$1$
Solid sphere	any diameter	$\frac{2}{5}MR^2$	$\frac{2}{5}$
Spherical shell	any diameter	$\frac{2}{3}MR^2$	$\frac{2}{3}$
Thin rod, length $L$	through centre, perpendicular	$\frac{1}{12}ML^2$	—
Thin rod, length $L$	through one end, perpendicular	$\frac{1}{3}ML^2$	—
Rectangular plate, sides $a \times b$	through centre, perpendicular	$\frac{M(a^2+b^2)}{12}$	—

And the two theorems that move an axis for you:

$$I = I_{cm} + Md^2 \qquad \text{(parallel axis, } d \text{ measured from the centre of mass)}$$ $$I_z = I_x + I_y \qquad \text{(perpendicular axis, flat bodies only)}$$

A note on $g$, and on notation

Each problem states the value of $g$ it uses in its own statement, and no problem mixes 9.8 and 10. Several problems need no $g$ at all. Position of the centre of mass is written $\vec{R}$ here; many books write $\vec{r}_{cm}$ and mean exactly the same thing. $\vec{L}$ is the angular momentum of a whole system and $\vec{l}$ that of a single particle.

[Board Important] Every solution below quotes the formula on its own line before substituting numbers into it. Do the same in the exam — a correct formula with an arithmetic slip still earns most of the marks, while a bare number earns none.

Solved Examples

Part 1: Locating the Centre of Mass, and the Vector Toolkit

Example 1: Three particles strung out along a line

Particles of mass 1 kg, 2 kg and 3 kg sit on the $x$-axis at $x = 0$, $x = 2$ m and $x = 5$ m. Find the centre of mass. Then redo the calculation with the origin moved to the 2 kg particle, and show that the two answers describe the same physical point.

Solution:

1. Quote the definition. $$X = \frac{\sum m_ix_i}{\sum m_i}$$

2. Substitute, in the original frame. $$X = \frac{(1)(0) + (2)(2) + (3)(5)}{1 + 2 + 3} = \frac{0 + 4 + 15}{6} = \frac{19}{6} = 3.17 \text{ m}$$

3. Move the origin to the 2 kg particle. The new coordinates are $-2$ m, 0 and 3 m: $$X_{new} = \frac{(1)(-2) + (2)(0) + (3)(3)}{6} = \frac{-2 + 0 + 9}{6} = \frac{7}{6} = 1.17 \text{ m}$$

4. Translate back. The new origin sits at $x = 2$ m in the old frame, so $$1.17 + 2 = 3.17 \text{ m}$$ the same point.

5. Sanity check. Most of the mass is bunched to the right, so the centre of mass must lie to the right of the middle particle at 2 m. It does.

Final Answer: The centre of mass is at $x = 19/6 = 3.17$ m, and it is the same physical point whichever origin you choose.

Takeaway: The centre of mass is a point in space, not a number on a particular ruler. [JEE Tip] Put your origin on the heaviest particle or at a corner where several coordinates are zero — you will kill two or three terms before you start.

Example 2: Symmetric solids, and whether the centre of mass must lie inside

State where the centre of mass lies for a uniform sphere, a uniform solid cylinder, a uniform ring and a uniform cube. Then answer the question that always follows: must the centre of mass of a body lie inside the material of the body?

Solution:

1. The rule that does all four at once. For a body of uniform density, the centre of mass lies at the geometric centre, because every mass element has a partner directly opposite it that cancels its contribution.

2. Apply it.

   Body	Centre of mass	Is it inside the material?
   Uniform sphere	its centre	yes
   Uniform solid cylinder	mid-point of the axis	yes
   Uniform ring	its centre	no
   Uniform cube	intersection of the body diagonals	yes

3. The ring settles the question. All of a ring's mass sits on the rim, yet the centre of mass is at the middle of the hole, where there is nothing at all.

4. More cases of the same kind. A hollow sphere, a bangle, a horseshoe, a boomerang, a hollow pipe — in every one the centre of mass lies in empty space. A high jumper doing a Fosbury flop arches so much that her centre of mass passes under the bar while she goes over it.

Final Answer: All four lie at the geometric centre; and no, the centre of mass need not lie within the material of the body — a ring is the standard counter-example.

Takeaway: "Centre of mass" is a weighted average of positions, and an average need not be one of the things averaged. [NEET Important] This one-liner is asked almost verbatim: the centre of mass of a ring lies at its centre, which contains no mass.

Example 3: The centre of mass of a water molecule

In a water molecule each hydrogen nucleus sits 95.8 pm from the oxygen nucleus, and the angle between the two bonds is $104.5°$. Oxygen is 16 times as massive as hydrogen, and essentially all the mass of an atom is in its nucleus. Locate the centre of mass.

Solution:

1. Choose axes that exploit the symmetry. Put the oxygen nucleus at the origin and let the $y$-axis run along the bisector of the bond angle. Each bond then makes an angle of $104.5°/2 = 52.25°$ with that bisector.

2. Coordinates of the two hydrogens. $$x = \pm\,(95.8)\sin 52.25° = \pm\,75.75 \text{ pm}, \qquad y = (95.8)\cos 52.25° = 58.66 \text{ pm}$$

3. The $x$-coordinate of the centre of mass. The two hydrogens have equal masses at $+75.75$ pm and $-75.75$ pm, so $$X = 0$$ by symmetry, exactly as it had to be.

4. The $y$-coordinate. Take the hydrogen mass as 1 unit and the oxygen as 16: $$Y = \frac{\sum m_iy_i}{\sum m_i} = \frac{(16)(0) + (1)(58.66) + (1)(58.66)}{16 + 1 + 1} = \frac{117.3}{18} = 6.52 \text{ pm}$$

5. Read the answer physically. The centre of mass lies on the bisector, 6.52 pm from the oxygen nucleus, displaced towards the hydrogens.

Final Answer: On the bond-angle bisector, 6.52 pm from the oxygen nucleus, on the hydrogen side.

Takeaway: With a heavy atom and two light ones, the centre of mass barely moves off the heavy nucleus — 6.52 pm out of a 95.8 pm bond is under 7%. [JEE Tip] Always align one axis with a symmetry line first. It turns a two-coordinate problem into a one-coordinate problem.

Example 4: A uniform wire bent into an L

A uniform wire is bent at right angles so that one arm is 0.60 m long and the other 0.80 m. Taking the corner as the origin with the arms along the $x$- and $y$-axes, find the centre of mass of the bent wire.

Solution:

1. Split it into two straight pieces. For a uniform wire the mass of a piece is proportional to its length, so take the masses as 0.60 and 0.80 in any convenient unit, giving a total of 1.40.

2. Locate each piece's own centre of mass. A uniform straight rod has its centre of mass at its midpoint: $$\text{arm along } x: \ (0.30,\ 0) \qquad \text{arm along } y: \ (0,\ 0.40)$$

3. Combine. $$X = \frac{(0.60)(0.30) + (0.80)(0)}{1.40} = \frac{0.18}{1.40} = 0.129 \text{ m}$$ $$Y = \frac{(0.60)(0) + (0.80)(0.40)}{1.40} = \frac{0.32}{1.40} = 0.229 \text{ m}$$

4. Distance from the corner. $$d = \sqrt{0.129^2 + 0.229^2} = 0.262 \text{ m}$$

5. Check the position makes sense. The longer arm lies along $y$, so the centre of mass should sit closer to the $y$-axis than to the $x$-axis. It does: 0.129 across, 0.229 up.

Final Answer: At $(0.129, 0.229)$ m, which is 0.262 m from the corner, in the empty space between the two arms.

Takeaway: A composite body is just a small discrete system whose "particles" are the pieces, each placed at its own centre of mass. [Board Important] Note again that the centre of mass is off the wire entirely — it sits in the air inside the elbow.

Example 5: A square plate with a circular hole punched in it

A uniform square plate of side 0.40 m has its centre at the origin. A circular hole of radius 0.10 m is punched out, the centre of the hole being 0.10 m from the centre of the plate along the $x$-axis. Find the centre of mass of what is left.

Solution:

1. Use the negative-mass idea. Treat the intact plate as a full plate plus a disc of negative mass sitting where the hole is. Masses go as areas for a uniform sheet: $$A_{plate} = (0.40)^2 = 0.1600 \text{ m}^2, \qquad A_{hole} = \pi(0.10)^2 = 0.0314 \text{ m}^2$$ $$A_{remaining} = 0.1600 - 0.0314 = 0.1286 \text{ m}^2$$

2. Write the combination with the hole carrying a minus sign. $$X = \frac{A_{plate}x_{plate} - A_{hole}x_{hole}}{A_{plate} - A_{hole}}$$

3. Substitute. The full plate is centred at $x = 0$ and the hole at $x = +0.10$ m: $$X = \frac{(0.1600)(0) - (0.0314)(0.10)}{0.1286} = \frac{-0.00314}{0.1286} = -0.0244 \text{ m}$$

4. The $y$-coordinate. Both the plate and the hole are centred on $y = 0$, so $Y = 0$.

5. Read the sign. The centre of mass has moved 2.44 cm to the side away from the hole, which is what removing mass from the right-hand side must do.

Final Answer: At $(-0.0244, 0)$ m, that is 2.44 cm from the plate's centre, on the side opposite the hole.

Takeaway: Cutting a hole shifts the centre of mass away from the hole, never towards it. [JEE Tip] The shift is $\dfrac{A_{hole}\,d}{A_{plate} - A_{hole}}$ — memorise that shape and you can do any cavity problem in one line.

Example 6: The centre of mass of a solid hemisphere, from scratch

Derive the position of the centre of mass of a uniform solid hemisphere of radius $R$, and evaluate it for $R = 0.12$ m.

Solution:

1. Set up. Put the flat face on the $xy$-plane with the curved surface above, centre of the flat face at the origin. By symmetry the centre of mass lies on the $z$-axis, so only $Z$ has to be found.

2. Choose the mass element. Slice the hemisphere into thin discs of thickness $dz$ at height $z$. The radius of such a disc is $r = \sqrt{R^2 - z^2}$, so $$dm = \rho\,\pi r^2\,dz = \rho\pi(R^2 - z^2)\,dz$$

3. Quote the definition and integrate the numerator. $$Z = \frac{1}{M}\int z\,dm = \frac{\rho\pi}{M}\int_0^R z(R^2 - z^2)\,dz$$ $$\int_0^R (zR^2 - z^3)\,dz = \frac{R^4}{2} - \frac{R^4}{4} = \frac{R^4}{4}$$

4. The denominator is just the mass. $$M = \rho\,\frac{2}{3}\pi R^3$$

5. Divide. $$Z = \frac{\rho\pi R^4/4}{\rho(2/3)\pi R^3} = \frac{3R}{8}$$

6. Put the number in. $$Z = \frac{3(0.12)}{8} = 0.045 \text{ m} = 4.5 \text{ cm above the flat face}$$

Final Answer: $Z = 3R/8$, which is 0.045 m for $R = 0.12$ m.

Takeaway: $3R/8$ for a solid hemisphere is worth memorising, alongside $R/2$ for a hemispherical shell and $2R/\pi$ for a semicircular wire. [JEE Tip] The integration always follows the same three steps: pick a slice whose points all share the coordinate you are averaging, write $dm$, integrate.

Example 7: A cross product, and the trap in recovering the angle

Given $\vec{A} = 3\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{B} = \hat{i} + 4\hat{j} - 2\hat{k}$, find $\vec{A} \times \vec{B}$, the area of the parallelogram they span, a unit vector perpendicular to both, and the angle between them.

Solution:

1. Set up the determinant. $$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 1 \\ 1 & 4 & -2 \end{vmatrix}$$

2. Expand. $$= \hat{i}\,[(-2)(-2) - (1)(4)] - \hat{j}\,[(3)(-2) - (1)(1)] + \hat{k}\,[(3)(4) - (-2)(1)]$$ $$= \hat{i}(4 - 4) - \hat{j}(-6 - 1) + \hat{k}(12 + 2) = 7\hat{j} + 14\hat{k}$$

3. Check it is perpendicular to both — this takes four seconds and catches sign errors: $$(7\hat{j} + 14\hat{k})\cdot\vec{A} = 7(-2) + 14(1) = 0, \qquad (7\hat{j} + 14\hat{k})\cdot\vec{B} = 7(4) + 14(-2) = 0$$

4. Magnitude, which is the area of the parallelogram. $$|\vec{A} \times \vec{B}| = \sqrt{0 + 49 + 196} = \sqrt{245} = 15.65 \text{ square units}$$

5. Unit vector perpendicular to both. $$\hat{n} = \frac{7\hat{j} + 14\hat{k}}{15.65} = \frac{\hat{j} + 2\hat{k}}{\sqrt{5}}$$ and $-\hat{n}$ is equally valid — there are always two.

6. The angle, done carefully. With $|\vec{A}| = \sqrt{14} = 3.742$ and $|\vec{B}| = \sqrt{21} = 4.583$, $$\sin\theta = \frac{15.65}{(3.742)(4.583)} = 0.9128$$ which alone gives either $65.9°$ or $114.1°$. The dot product decides: $$\vec{A}\cdot\vec{B} = 3 - 8 - 2 = -7 \quad \Rightarrow \quad \cos\theta = \frac{-7}{17.15} = -0.408 \quad \Rightarrow \quad \theta = 114.1°$$

Final Answer: $\vec{A} \times \vec{B} = 7\hat{j} + 14\hat{k}$; area 15.65 square units; $\hat{n} = (\hat{j} + 2\hat{k})/\sqrt{5}$; $\theta = 114.1°$.

Takeaway: The sine alone can never tell an acute angle from its obtuse partner. [JEE Tip] Get the angle from the dot product, and use the cross product only as a check that $|\vec{A}||\vec{B}|\sin\theta$ comes out right.

Example 8: A scalar triple product, a volume, and a test for coplanarity

For $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{c} = 3\hat{i} - \hat{j} + 2\hat{k}$, evaluate $\vec{a}\cdot(\vec{b} \times \vec{c})$ and interpret it. Then decide whether $\hat{i}+\hat{j}$, $\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+\hat{k}$ are coplanar.

Solution:

1. Inner cross product first. $$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & -1 & 2 \end{vmatrix} = \hat{i}(4 + 3) - \hat{j}(2 - 9) + \hat{k}(-1 - 6) = 7\hat{i} + 7\hat{j} - 7\hat{k}$$

2. Now the dot product. $$\vec{a}\cdot(\vec{b} \times \vec{c}) = (2)(7) + (1)(7) + (-1)(-7) = 14 + 7 + 7 = 28$$

3. Interpret it. The scalar triple product is the signed volume of the parallelepiped with the three vectors as edges, so that volume is 28 cubic units. The positive sign says the three form a right-handed set.

4. Check by the single determinant, which is the faster route in an exam: $$\begin{vmatrix} 2 & 1 & -1 \\ 1 & 2 & 3 \\ 3 & -1 & 2 \end{vmatrix} = 2(4 + 3) - 1(2 - 9) + (-1)(-1 - 6) = 14 + 7 + 7 = 28 \ \checkmark$$

5. The coplanarity test. Three vectors are coplanar exactly when their scalar triple product vanishes: $$\begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 1 \end{vmatrix} = 1(1 - 2) - 1(0 - 1) + 0 = -1 + 1 = 0$$ So yes, those three are coplanar — the box they would span has zero volume.

Final Answer: $\vec{a}\cdot(\vec{b} \times \vec{c}) = 28$, a parallelepiped of volume 28 cubic units; and the second set of three vectors is coplanar.

Takeaway: A zero scalar triple product means "flat", a zero cross product means "parallel", a zero dot product means "perpendicular". [JEE Tip] Cyclic swaps leave the value alone, $\vec{a}\cdot(\vec{b}\times\vec{c}) = \vec{b}\cdot(\vec{c}\times\vec{a})$, so rearrange until the easiest cross product is on the inside.

Part 2: Angular Quantities, Torque and the First Equilibrium Problems

Example 9: From the speedometer to the wheel

A car travels at a steady 54 km/h on wheels of radius 0.35 m, which roll without slipping. Find (a) the angular speed of a wheel in rad/s and in rpm, (b) how many turns a wheel makes in 1 km, (c) the centripetal acceleration of a point on the tyre as seen from the wheel's centre, and (d) the angular retardation if the car is brought uniformly to rest in 5.0 s.

Solution:

1. Convert the speed first. $$v = 54 \times \frac{1000}{3600} = 15 \text{ m/s}$$

2. (a) Angular speed, from the rolling relation $v = \omega R$: $$\omega = \frac{v}{R} = \frac{15}{0.35} = 42.86 \text{ rad/s}$$ $$\text{in rpm} = \omega \times \frac{60}{2\pi} = 42.86 \times 9.549 = 409.3 \text{ rpm}$$

3. (b) Turns per kilometre. Each turn lays down one circumference: $$n = \frac{1000}{2\pi R} = \frac{1000}{2\pi(0.35)} = \frac{1000}{2.199} = 454.7 \text{ revolutions}$$

4. (c) Centripetal acceleration of a tyre point about the centre. $$a_c = \omega^2R = (42.86)^2(0.35) = 642.9 \text{ m/s}^2$$ Cross-check with the other form: $a_c = v^2/R = 225/0.35 = 642.9$ m/s$^2$. Agreed.

5. (d) Angular retardation. Uniform braking to rest in 5.0 s: $$\alpha = \frac{\omega_2 - \omega_1}{t} = \frac{0 - 42.86}{5.0} = -8.57 \text{ rad/s}^2$$

Final Answer: 42.86 rad/s, that is 409.3 rpm; 454.7 turns per kilometre; 642.9 m/s$^2$; and a retardation of 8.57 rad/s$^2$.

Takeaway: That 643 m/s$^2$ is about 65 times $g$ — which is why a stone lodged in a tyre tread flies out so violently. [NEET Important] The conversion $\omega\ (\text{rad/s}) \times 9.549 =$ rpm is worth memorising as a single number.

Example 10: A torque in components, and the moment arm hiding in it

A force $\vec{F} = 5\hat{i} - 2\hat{j} + 3\hat{k}$ N acts at the point $\vec{r} = 2\hat{i} + 3\hat{j} - \hat{k}$ m measured from the origin. Find the torque about the origin, its magnitude, and the perpendicular distance from the origin to the line of action of the force.

Solution:

1. Quote the definition. $$\vec{\tau} = \vec{r} \times \vec{F}$$

2. Expand the determinant. $$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 5 & -2 & 3 \end{vmatrix} = \hat{i}(9 - 2) - \hat{j}(6 + 5) + \hat{k}(-4 - 15)$$ $$\vec{\tau} = 7\hat{i} - 11\hat{j} - 19\hat{k} \text{ N m}$$

3. Check perpendicularity both ways. $$\vec{\tau}\cdot\vec{r} = 14 - 33 + 19 = 0, \qquad \vec{\tau}\cdot\vec{F} = 35 + 22 - 57 = 0 \ \checkmark$$

4. Magnitude. $$|\vec{\tau}| = \sqrt{49 + 121 + 361} = \sqrt{531} = 23.04 \text{ N m}$$

5. Extract the moment arm. Since $|\vec{\tau}| = r_\perp |\vec{F}|$ and $|\vec{F}| = \sqrt{25 + 4 + 9} = \sqrt{38} = 6.164$ N, $$r_\perp = \frac{23.04}{6.164} = 3.74 \text{ m}$$

Final Answer: $\vec{\tau} = 7\hat{i} - 11\hat{j} - 19\hat{k}$ N m, of magnitude 23.04 N m; the line of action passes 3.74 m from the origin.

Takeaway: The two dot-product checks cost nothing and catch the single commonest error in this chapter — a sign lost in the middle term of the determinant. [JEE Tip] $r_\perp = |\vec{\tau}|/|\vec{F}|$ is how you get a perpendicular distance without any geometry at all.

Example 11: Three forces on a square plate

A square plate of side 2.0 m has its centre at the origin, so its corners are at $(\pm1, \pm1)$ m. Three forces act in the plane of the plate: 10 N along $+x$ at the corner $(1, 1)$; 15 N along $+y$ at the corner $(1, -1)$; and 20 N along $-x$ at the corner $(-1, -1)$. Find the net force and the net torque about the centre, and then the net torque about the corner $(1, 1)$.

Solution:

1. Net force, added component by component: $$\vec{F}_{net} = (10 - 20)\hat{i} + 15\hat{j} = -10\hat{i} + 15\hat{j} \text{ N}, \qquad |\vec{F}_{net}| = \sqrt{100 + 225} = 18.0 \text{ N}$$ It is not zero, which matters for the last part.

2. Torque about the centre. For forces in the $xy$-plane, only the $z$-component survives, and $$\tau_z = xF_y - yF_x$$

3. Apply it force by force. $$\text{10 N at }(1,1): \ \tau_z = (1)(0) - (1)(10) = -10 \text{ N m}$$ $$\text{15 N at }(1,-1): \ \tau_z = (1)(15) - (-1)(0) = +15 \text{ N m}$$ $$\text{20 N at }(-1,-1): \ \tau_z = (-1)(0) - (-1)(-20) = -20 \text{ N m}$$

4. Add. $$\tau_{net} = -10 + 15 - 20 = -15 \text{ N m}$$ The minus sign means clockwise, seen from the $+z$ side.

5. Now about the corner $(1,1)$. Shift every position vector by $-(\hat{i} + \hat{j})$: $$\text{10 N: } \vec{r} = 0 \Rightarrow \tau_z = 0$$ $$\text{15 N: } \vec{r} = (0,-2) \Rightarrow \tau_z = (0)(15) - (-2)(0) = 0$$ $$\text{20 N: } \vec{r} = (-2,-2) \Rightarrow \tau_z = (-2)(0) - (-2)(-20) = -40 \text{ N m}$$ $$\tau_{net} = -40 \text{ N m}$$

6. Why the two differ. Shifting the reference point by $\vec{d}$ changes the total torque by $-\vec{d} \times \vec{F}_{net}$. Here $\vec{d} = \hat{i}+\hat{j}$, and $-[(1)(15) - (1)(-10)] = -25$ N m, which is exactly $-40 - (-15)$.

Final Answer: Net force $-10\hat{i} + 15\hat{j}$ N; net torque $-15\hat{k}$ N m about the centre and $-40\hat{k}$ N m about the corner $(1,1)$.

Takeaway: Net torque depends on the reference point unless the net force is zero — and when the net force is zero the system is a couple, whose moment is the same about every point. [JEE Tip] If a question gives you a torque without naming the point, look for a couple.

Example 12: The Earth's two angular momenta

Treat the Earth as a uniform sphere of mass $5.97 \times 10^{24}$ kg and radius $6.37 \times 10^6$ m, spinning once in 24 hours, and orbiting the Sun in a circle of radius $1.496 \times 10^{11}$ m once in $3.156 \times 10^7$ s. Find its spin angular momentum about its own axis and its orbital angular momentum about the Sun, and compare them.

Solution:

1. Spin: moment of inertia first. $$I = \frac{2}{5}MR^2 = 0.4 \times (5.97 \times 10^{24})(6.37 \times 10^6)^2 = 9.69 \times 10^{37} \text{ kg m}^2$$

2. Spin angular speed. $$\omega_{spin} = \frac{2\pi}{T} = \frac{2\pi}{86400} = 7.272 \times 10^{-5} \text{ rad/s}$$

3. Spin angular momentum. $$L_{spin} = I\omega_{spin} = (9.69 \times 10^{37})(7.272 \times 10^{-5}) = 7.05 \times 10^{33} \text{ kg m}^2\text{/s}$$

4. Orbit: treat the Earth as a point mass at distance $r$, so $I_{orbit} = Mr^2$: $$\omega_{orb} = \frac{2\pi}{3.156 \times 10^7} = 1.991 \times 10^{-7} \text{ rad/s}$$ $$L_{orb} = Mr^2\omega_{orb} = (5.97 \times 10^{24})(1.496 \times 10^{11})^2 (1.991 \times 10^{-7}) = 2.66 \times 10^{40} \text{ kg m}^2\text{/s}$$

5. Compare. $$\frac{L_{orb}}{L_{spin}} = \frac{2.66 \times 10^{40}}{7.05 \times 10^{33}} = 3.8 \times 10^{6}$$

Final Answer: $L_{spin} = 7.05 \times 10^{33}$ kg m$^2$/s and $L_{orb} = 2.66 \times 10^{40}$ kg m$^2$/s; the orbital value is about 3.8 million times the spin value.

Takeaway: Angular momentum is dominated by how far the mass is from the axis, not by how fast the body spins — the Earth turns 365 times faster about its own axis, yet the orbit still wins by a factor of millions. [JEE Tip] Treating an orbiting body as a point mass, $L = Mvr = Mr^2\omega$, is legitimate whenever the body is tiny compared with its orbit.

Example 13: A non-uniform bar hung from two slanting strings

A bar of weight $W$ and length 2.0 m hangs at rest from two light strings tied to its two ends. The left string makes $36.9°$ with the vertical and the right string $53.1°$. The bar is not uniform. Find the distance $d$ of its centre of gravity from the left end, and the two tensions if the bar weighs 40 N. No value of $g$ is needed.

Solution:

1. Name the forces. Tension $T_1$ along the left string, $T_2$ along the right string, and the weight $W$ acting vertically down through the centre of gravity, a distance $d$ from the left end.

2. Horizontal equilibrium. The horizontal components of the two tensions must cancel: $$T_1\sin 36.9° = T_2\sin 53.1° \quad \Rightarrow \quad 0.6004\,T_1 = 0.7997\,T_2$$

3. Vertical equilibrium. $$T_1\cos 36.9° + T_2\cos 53.1° = W \quad \Rightarrow \quad 0.7997\,T_1 + 0.6004\,T_2 = W$$

4. Solve the pair. From step 2, $T_1 = 1.332\,T_2$. Substituting, $$0.7997(1.332\,T_2) + 0.6004\,T_2 = W \quad \Rightarrow \quad 1.6656\,T_2 = W$$ $$T_2 = 0.6004\,W, \qquad T_1 = 0.7997\,W$$

5. Take moments about the left end, where $T_1$ acts and therefore drops out. The horizontal component of $T_2$ acts along the line of the bar, so its moment arm about the left end is zero too. Only the vertical component of $T_2$ and the weight survive: $$(T_2\cos 53.1°)(2.0) = W d$$ $$d = \frac{(0.6004W)(0.6004)(2.0)}{W} = 0.72 \text{ m}$$

6. Numbers for $W = 40$ N. $$T_1 = 0.7997 \times 40 = 32.0 \text{ N}, \qquad T_2 = 0.6004 \times 40 = 24.0 \text{ N}$$

7. Check. The left string is more nearly vertical, so it should carry more load — and it does, 32 N against 24 N. The centre of gravity must therefore lie nearer the left end: 0.72 m out of 2.0 m. Consistent.

Final Answer: $d = 0.72$ m from the left end; $T_1 = 32.0$ N and $T_2 = 24.0$ N for a 40 N bar.

Takeaway: The weight cancels out of $d$ entirely — the geometry of the strings alone fixes where the centre of gravity is. [Board Important] Since $36.9° + 53.1° = 90°$, the two strings are perpendicular, and $d = L\cos^2 53.1°$ falls out in one line. Spotting that saves two minutes.

Example 14: The load a car puts on each of its wheels

A car of mass 1800 kg has its front and rear axles 1.8 m apart, and its centre of gravity lies 1.05 m behind the front axle on the line joining them. Find the force the level ground exerts on each front wheel and on each back wheel. Take $g = 9.8$ m/s$^2$.

Solution:

1. Weight of the car. $$W = mg = 1800 \times 9.8 = 17640 \text{ N}$$

2. Set up the two equilibrium conditions. Let $N_f$ be the total upward force on the front axle and $N_r$ that on the rear axle. $$N_f + N_r = W = 17640 \text{ N}$$

3. Take moments about the rear axle, which removes $N_r$. The centre of gravity is $1.8 - 1.05 = 0.75$ m in front of the rear axle: $$N_f \times 1.8 = W \times 0.75$$ $$N_f = \frac{17640 \times 0.75}{1.8} = 7350 \text{ N}$$

4. Back-substitute. $$N_r = 17640 - 7350 = 10290 \text{ N}$$

5. Split each axle between its two wheels, which is legitimate because the car is symmetric side to side: $$\text{each front wheel: } \frac{7350}{2} = 3675 \text{ N}, \qquad \text{each back wheel: } \frac{10290}{2} = 5145 \text{ N}$$

6. Check. The centre of gravity is nearer the front axle (1.05 m against 0.75 m from the rear)… and yet the rear carries more. Look again: the load on an axle goes with the distance from the other axle, so being 1.05 m from the front means the front carries the smaller share. $4 \times$ the wheel loads gives $2(3675) + 2(5145) = 17640$ N. Correct.

Final Answer: 3675 N on each front wheel and 5145 N on each back wheel.

Takeaway: A support always carries a share proportional to the distance from the opposite support — the far support takes the larger load. [Board Important] This is the same principle-of-moments arithmetic as a see-saw; only the words change.

Example 15: Lifting a wheel over a kerb

A wheel of radius 0.60 m and mass 20 kg rests on level ground against a step 0.20 m high. Find the least horizontal force, applied at the wheel's centre, that will just lift it onto the step. Take $g = 10$ m/s$^2$.

Solution:

1. Identify the pivot. As the wheel begins to lift, it loses contact with the flat ground and turns about the edge $E$ of the step. Take moments about $E$; the ground's normal reaction and the friction there both pass through $E$ and disappear from the equation.

2. Moment arm of the weight. The centre $C$ is a horizontal distance $x$ from $E$, where by Pythagoras on the right triangle with hypotenuse $R$ and vertical side $R - h$: $$x = \sqrt{R^2 - (R-h)^2} = \sqrt{2Rh - h^2}$$ $$x = \sqrt{2(0.60)(0.20) - (0.20)^2} = \sqrt{0.24 - 0.04} = \sqrt{0.20} = 0.447 \text{ m}$$

3. Moment arm of the applied force. $F$ is horizontal and acts at $C$, which is a height $R - h = 0.60 - 0.20 = 0.40$ m above $E$.

4. Balance the moments at the point of lifting. $$F(R - h) = Mg\sqrt{2Rh - h^2}$$ $$F = \frac{(20)(10)(0.447)}{0.40} = \frac{89.44}{0.40} = 223.6 \text{ N}$$

5. Sanity check on the limit. If the step were as tall as the radius, $h = R$, the arm of $F$ would shrink to zero and no horizontal force at the centre could ever lift the wheel. The formula agrees: the denominator goes to zero and $F \to \infty$.

Final Answer: About 224 N, which is more than the wheel's own weight of 200 N.

Takeaway: Choose the pivot so that the forces you do not know pass through it — here that killed two unknowns at once. [JEE Tip] If the same force is applied at the top of the wheel instead, its arm becomes $2R - h$ and the force needed drops sharply. Questions love that comparison.

Example 16: A shopkeeper's balance with unequal arms

A shopkeeper's beam balance has arms of slightly different lengths. A packet placed in the left pan appears to weigh 9.0 kg; placed in the right pan the same packet appears to weigh 4.0 kg. Find the true mass of the packet and the ratio of the arm lengths. The beam itself is balanced when empty.

Solution:

1. Set up. Let the arms be $a$ (left) and $b$ (right), the true mass $m$, and the two readings $m_1 = 9.0$ kg and $m_2 = 4.0$ kg.

2. First weighing — packet on the left, standard masses on the right. By the principle of moments, $$m\,a = m_1\,b$$

3. Second weighing — packet on the right, standards on the left: $$m\,b = m_2\,a$$

4. Multiply the two equations. The arm lengths cancel completely: $$m^2ab = m_1m_2\,ab \quad \Rightarrow \quad m = \sqrt{m_1m_2}$$

5. Substitute. $$m = \sqrt{(9.0)(4.0)} = \sqrt{36} = 6.0 \text{ kg}$$

6. Arm ratio, by dividing the two equations instead of multiplying: $$\frac{a}{b} = \sqrt{\frac{m_1}{m_2}} = \sqrt{\frac{9}{4}} = 1.5$$

7. The trap. The arithmetic mean $(9 + 4)/2 = 6.5$ kg is not the answer, and it is always an offered option.

Final Answer: The true mass is 6.0 kg, and the left arm is 1.5 times the right arm.

Takeaway: The true value is the geometric mean of the two readings, $\sqrt{m_1m_2}$, never the arithmetic mean. [NEET Important] Notice also that $\sqrt{m_1m_2} < (m_1+m_2)/2$ always, so a false balance of this kind always over-reads on average.

Part 3: Equilibrium Finished, and Moments of Inertia Built From Scratch

Example 17: How far can he walk before the plank tips?

A uniform plank of mass 30 kg and length 4.0 m rests on two supports: support $A$ at the left end, and support $B$ 1.0 m in from the right end. A boy of mass 50 kg starts at $A$ and walks to the right. Find (a) the reactions at the two supports when he stands at the middle of the plank, and (b) how far past $B$ he can walk before the plank tips. Take $g = 10$ m/s$^2$.

Solution:

1. Set up coordinates. Measure $x$ from the left end. Then $A$ is at $x = 0$, $B$ is at $x = 3.0$ m, the plank's weight of 300 N acts at $x = 2.0$ m, and the boy's weight is 500 N.

2. (a) Moments about $A$, with the boy at $x = 2.0$ m, which removes $R_A$: $$R_B(3.0) = (300)(2.0) + (500)(2.0) = 600 + 1000 = 1600$$ $$R_B = \frac{1600}{3.0} = 533.3 \text{ N}$$

3. Vertical equilibrium gives the other one. $$R_A = (300 + 500) - 533.3 = 266.7 \text{ N}$$

4. (b) What "about to tip" means. The plank tips about $B$ the instant the reaction at $A$ falls to zero. Take moments about $B$ with $R_A = 0$. The plank's weight acts 1.0 m to the left of $B$ and holds it down; the boy, a distance $(x - 3.0)$ to the right, lifts that end. $$500(x - 3.0) = 300(1.0)$$

5. Solve. $$x - 3.0 = \frac{300}{500} = 0.6 \text{ m} \quad \Rightarrow \quad x = 3.6 \text{ m}$$

6. State it usefully. He can walk 0.6 m past support $B$, reaching a point 0.4 m short of the right-hand end.

Final Answer: $R_A = 266.7$ N and $R_B = 533.3$ N with the boy at the middle; he may walk 0.6 m beyond $B$, that is up to $x = 3.6$ m.

Takeaway: Toppling always begins the moment a reaction reaches zero, not the moment it becomes negative — a support can push, but it cannot pull. [JEE Tip] Set the reaction you are abandoning to zero, take moments about the other support, and the tipping condition falls out in one line.

Example 18: A ladder with someone standing on it

A uniform ladder of mass 20 kg and length 5.0 m leans against a smooth vertical wall, its foot resting on rough ground 3.0 m from the wall. A person of mass 60 kg stands on the ladder 4.0 m from its foot, measured along the ladder. Find the reaction of the wall, the friction at the floor, the normal reaction at the floor, and the least coefficient of friction that will keep the ladder from slipping. Take $g = 10$ m/s$^2$.

Solution:

1. Get the geometry. Base 3.0 m, ladder 5.0 m, so the top rests $$h = \sqrt{5.0^2 - 3.0^2} = 4.0 \text{ m}$$ up the wall. The ladder makes $\tan^{-1}(4/3) = 53.13°$ with the ground.

2. List the forces. Weight of ladder 200 N at its midpoint; weight of person 600 N; normal reaction $N$ and friction $f$ at the floor; and, because the wall is smooth, a purely horizontal reaction $N_w$ at the top.

3. Vertical equilibrium. $$N = 200 + 600 = 800 \text{ N}$$

4. Horizontal equilibrium. $$f = N_w$$

5. Moments about the foot, which removes both $N$ and $f$ at once. Convert distances along the ladder into horizontal distances by multiplying by $3/5$: $$\text{ladder's weight: } 2.5 \times \tfrac{3}{5} = 1.5 \text{ m}, \qquad \text{person: } 4.0 \times \tfrac{3}{5} = 2.4 \text{ m}$$ $$N_w(4.0) = (200)(1.5) + (600)(2.4) = 300 + 1440 = 1740$$

6. Solve. $$N_w = \frac{1740}{4.0} = 435 \text{ N} \quad \Rightarrow \quad f = 435 \text{ N}$$

7. Least coefficient of friction. Slipping is prevented as long as $f \le \mu N$: $$\mu_{min} = \frac{f}{N} = \frac{435}{800} = 0.544$$

Final Answer: $N_w = f = 435$ N, $N = 800$ N, and the floor must supply $\mu \ge 0.544$.

Takeaway: A smooth wall can only push horizontally, so the whole of the ladder's tendency to slip has to be met by friction at the floor. [Board Important] Distances measured along a ladder are not the moment arms. Project them onto the horizontal before you use them.

Example 19: A step ladder with a tie rope

A step ladder is made of two light legs each 2.0 m long, hinged at the top and standing on a smooth floor with their feet 2.4 m apart. A rope joins the mid-points of the two legs. A load of 400 N hangs from the hinge at the top. Find the reaction under each foot, the tension in the rope, and the force at the hinge on one leg. Treat the legs as weightless.

Solution:

1. Geometry first. Each leg is 2.0 m long and reaches out 1.2 m horizontally, so the hinge stands $$\sqrt{2.0^2 - 1.2^2} = \sqrt{4.00 - 1.44} = 1.6 \text{ m}$$ above the floor. The rope joins the mid-points, which are 0.6 m out horizontally and 0.8 m up.

2. The whole ladder, by symmetry. The floor is smooth so it pushes straight up, and the two legs are identical: $$N = \frac{400}{2} = 200 \text{ N under each foot}$$

3. Now isolate ONE leg — this is the step everyone skips, and the one that makes the problem solvable. The forces on the right-hand leg are: $N = 200$ N up at its foot, the rope tension $T$ pulling inwards (horizontally, towards the other leg) at its mid-point, and whatever force the hinge transmits at the top.

4. Take moments about the hinge $A$. The hinge force vanishes because it acts there, and the 400 N load vanishes too because it also hangs from $A$. Measure moment arms about $A$: the foot is 1.2 m horizontally from $A$, and the mid-point of the leg is 0.8 m vertically below $A$. $$N(1.2) = T(0.8)$$

5. Solve for the tension. $$T = \frac{200 \times 1.2}{0.8} = 300 \text{ N}$$

6. The hinge force on this leg, from the equilibrium of the leg itself: $$\text{horizontal: } H_x = T = 300 \text{ N} \qquad \text{vertical: } H_y = N = 200 \text{ N (downward)}$$ $$|\vec{H}| = \sqrt{300^2 + 200^2} = \sqrt{130000} = 360.6 \text{ N}$$

Final Answer: 200 N under each foot, a rope tension of 300 N, and a hinge force of magnitude 360.6 N on each leg (300 N horizontal, 200 N vertical).

Takeaway: A hinged structure is never solved as one object — cut it at the hinge and write the equilibrium of one piece. [JEE Tip] Take moments about the hinge and both the hinge reaction and any load hanging from the hinge vanish together, leaving one equation in one unknown.

Example 20: A rod, about an axis a quarter of the way along

A uniform rod of mass 3.0 kg and length 1.6 m is free to turn about an axis perpendicular to it, passing through a point 0.40 m from one end. Find its moment of inertia by direct integration, then confirm the result using the parallel axis theorem, and find the radius of gyration.

Solution:

1. Set up the integral. Measure $s$ along the rod from the end nearest the axis, so the axis is at $s = L/4$. The linear mass density is $$\lambda = \frac{M}{L} = \frac{3.0}{1.6} = 1.875 \text{ kg/m}$$ and an element $ds$ at position $s$ lies a distance $\left(s - \frac{L}{4}\right)$ from the axis.

2. Quote the definition and integrate. $$I = \int r^2\,dm = \lambda\int_0^{L}\left(s - \frac{L}{4}\right)^2 ds$$ Substituting $u = s - L/4$, the limits run from $-L/4$ to $3L/4$: $$I = \lambda\left[\frac{u^3}{3}\right]_{-L/4}^{3L/4} = \frac{\lambda}{3}\left(\frac{27L^3}{64} + \frac{L^3}{64}\right) = \frac{\lambda}{3}\cdot\frac{28L^3}{64} = \frac{7\lambda L^3}{48} = \frac{7ML^2}{48}$$

3. Put the numbers in. $$I = \frac{7(3.0)(1.6)^2}{48} = \frac{7 \times 3.0 \times 2.56}{48} = \frac{53.76}{48} = 1.12 \text{ kg m}^2$$

4. Independent confirmation by the parallel axis theorem. The axis is $d = \frac{L}{2} - \frac{L}{4} = \frac{L}{4} = 0.40$ m from the centre of mass: $$I = I_{cm} + Md^2 = \frac{ML^2}{12} + M\left(\frac{L}{4}\right)^2 = \frac{3.0(2.56)}{12} + 3.0(0.16) = 0.64 + 0.48 = 1.12 \text{ kg m}^2 \ \checkmark$$

5. Radius of gyration. $$k = \sqrt{\frac{I}{M}} = \sqrt{\frac{1.12}{3.0}} = 0.611 \text{ m}$$

Final Answer: $I = \frac{7ML^2}{48} = 1.12$ kg m$^2$, and $k = 0.611$ m.

Takeaway: Two completely different routes landing on 1.12 kg m$^2$ is what a verified answer looks like. [JEE Tip] Note that $k = 0.611$ m is larger than the distance from the axis to either end's midpoint — the radius of gyration is a root-mean-square distance, and squaring always favours the far mass.

Example 21: An annular disc, and both axis theorems on one body

A flat annulus (a washer) has inner radius 0.10 m, outer radius 0.20 m and mass 2.4 kg, uniformly distributed. Find its moment of inertia (a) about the central axis perpendicular to its plane, by integration, (b) about a diameter, and (c) about a tangent lying in its plane. Also find the radius of gyration for the central axis.

Solution:

1. (a) Set up the integration. The surface density is $$\sigma = \frac{M}{\pi(R_2^2 - R_1^2)} = \frac{2.4}{\pi(0.04 - 0.01)} = \frac{2.4}{0.09425} = 25.46 \text{ kg/m}^2$$ Take a thin ring of radius $r$ and width $dr$: every point on it is exactly $r$ from the axis, so $$dm = \sigma(2\pi r\,dr), \qquad dI = r^2\,dm = 2\pi\sigma r^3\,dr$$

2. Integrate between the two radii. $$I_z = 2\pi\sigma\int_{R_1}^{R_2} r^3\,dr = 2\pi\sigma\,\frac{R_2^4 - R_1^4}{4} = \frac{\pi\sigma}{2}(R_2^2 - R_1^2)(R_2^2 + R_1^2)$$ and since $M = \pi\sigma(R_2^2 - R_1^2)$, this collapses to $$I_z = \frac{1}{2}M(R_1^2 + R_2^2)$$

3. Evaluate. $$I_z = \frac{1}{2}(2.4)(0.01 + 0.04) = 1.2 \times 0.05 = 0.06 \text{ kg m}^2$$

4. (b) About a diameter, by the perpendicular axis theorem. The annulus is flat, and by symmetry the two in-plane axes give the same value, $I_x = I_y$: $$I_z = I_x + I_y = 2I_x \quad \Rightarrow \quad I_{diameter} = \frac{I_z}{2} = 0.03 \text{ kg m}^2$$

5. (c) About a tangent in its plane, by the parallel axis theorem. That tangent is parallel to a diameter and $R_2 = 0.20$ m away from it: $$I_{tangent} = I_{diameter} + MR_2^2 = 0.03 + (2.4)(0.04) = 0.03 + 0.096 = 0.126 \text{ kg m}^2$$

6. Radius of gyration for the central axis. $$k = \sqrt{\frac{0.06}{2.4}} = \sqrt{0.025} = 0.158 \text{ m}$$ which sits between 0.10 m and 0.20 m, as it must.

Final Answer: 0.06 kg m$^2$ about the central axis, 0.03 kg m$^2$ about a diameter, 0.126 kg m$^2$ about an in-plane tangent, and $k = 0.158$ m.

Takeaway: The perpendicular axis theorem moves you from the face-on axis to an in-plane one; the parallel axis theorem then slides that axis sideways. Used in that order they reach almost any axis of a flat body. [JEE Tip] Set $R_1 = 0$ and $I_z = \frac{1}{2}MR_2^2$ reappears — always test a general result by collapsing it to a case you know.

Example 22: The solid sphere, derived by stacking discs

Derive $I = \frac{2}{5}MR^2$ for a uniform solid sphere about a diameter, and evaluate it for a sphere of mass 5.0 kg and radius 0.20 m. Then find its moment of inertia about a tangent line.

Solution:

1. Choose the element. Slice the sphere perpendicular to the chosen diameter (the $z$-axis) into discs of thickness $dz$ at height $z$. Such a disc has radius $r = \sqrt{R^2 - z^2}$ and mass $$dm = \rho\pi(R^2 - z^2)\,dz$$

2. Use the disc result for each slice. Every disc shares the $z$-axis as its own central axis, so $$dI = \frac{1}{2}r^2\,dm = \frac{1}{2}\rho\pi(R^2 - z^2)^2\,dz$$

3. Integrate from $-R$ to $R$. $$I = \frac{\rho\pi}{2}\int_{-R}^{R}(R^4 - 2R^2z^2 + z^4)\,dz = \frac{\rho\pi}{2}\left[2R^5 - \frac{4R^5}{3} + \frac{2R^5}{5}\right] = \frac{\rho\pi}{2}\cdot\frac{16R^5}{15} = \frac{8\pi\rho R^5}{15}$$

4. Replace $\rho$ using $M = \frac{4}{3}\pi R^3\rho$. $$I = \frac{8\pi R^5}{15}\cdot\frac{3M}{4\pi R^3} = \frac{2}{5}MR^2$$

5. Evaluate. $$I = \frac{2}{5}(5.0)(0.20)^2 = 0.4 \times 5.0 \times 0.04 = 0.08 \text{ kg m}^2$$

6. About a tangent line. A tangent is parallel to a diameter and a distance $R$ from it: $$I_{tangent} = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2 = 1.4 \times 5.0 \times 0.04 = 0.28 \text{ kg m}^2$$

Final Answer: $I = \frac{2}{5}MR^2 = 0.08$ kg m$^2$ about a diameter, and $\frac{7}{5}MR^2 = 0.28$ kg m$^2$ about a tangent.

Takeaway: Building a sphere from discs works because a disc's central axis and the sphere's diameter are the same line. [JEE Tip] Slice a body into pieces whose moment of inertia you already know, and the "integration" is one line of algebra plus a standard integral.

Example 23: A ceiling fan switched off

A ceiling fan turning at 900 rpm is switched off and comes uniformly to rest after 75 revolutions. Find (a) the initial angular speed in rad/s, (b) the angular retardation, (c) the time it takes to stop, and (d) how many revolutions it makes in the first 5.0 s.

Solution:

1. (a) Convert to rad/s. $$\omega_0 = 900 \times \frac{2\pi}{60} = 30\pi = 94.25 \text{ rad/s}$$

2. Convert the angle too. $$\theta = 75 \times 2\pi = 471.2 \text{ rad}$$

3. (b) Use the equation with no time in it. $$\omega^2 = \omega_0^2 + 2\alpha\theta$$ $$0 = (94.25)^2 + 2\alpha(471.2) \quad \Rightarrow \quad \alpha = -\frac{8883}{942.5} = -9.425 \text{ rad/s}^2$$

4. (c) Time, from $\omega = \omega_0 + \alpha t$. $$t = \frac{0 - 94.25}{-9.425} = 10.0 \text{ s}$$

5. Cross-check with the average. For uniform retardation the mean angular speed is $\omega_0/2 = 47.12$ rad/s, and $$\frac{\theta}{t} = \frac{471.2}{10.0} = 47.12 \text{ rad/s} \ \checkmark$$

6. (d) The first 5.0 s. $$\theta_5 = \omega_0t + \tfrac{1}{2}\alpha t^2 = (94.25)(5.0) - \tfrac{1}{2}(9.425)(25) = 471.2 - 117.8 = 353.4 \text{ rad}$$ $$n = \frac{353.4}{2\pi} = 56.25 \text{ revolutions}$$

Final Answer: 94.25 rad/s; a retardation of 9.425 rad/s$^2$; 10.0 s to stop; and 56.25 of the 75 revolutions happen in the first half of that time.

Takeaway: 75% of the turning happens in the first half of the stopping time — a decelerating body always covers more ground early. [NEET Important] The three rotational equations are the linear ones with $s \to \theta$, $u \to \omega_0$, $a \to \alpha$. Learn one set, not two.

Example 24: The angle turned during one particular second

A wheel starts from rest and reaches 300 rpm in 10 s under constant angular acceleration. Find (a) the angular acceleration, (b) the total number of revolutions in those 10 s, and (c) the angle turned during the fifth second alone.

Solution:

1. (a) Final angular speed and $\alpha$. $$\omega = 300 \times \frac{2\pi}{60} = 10\pi = 31.42 \text{ rad/s}$$ $$\alpha = \frac{\omega - \omega_0}{t} = \frac{31.42 - 0}{10} = 3.142 \text{ rad/s}^2$$

2. (b) Total angle. $$\theta = \omega_0t + \tfrac{1}{2}\alpha t^2 = 0 + \tfrac{1}{2}(3.142)(100) = 157.1 \text{ rad}$$ $$n = \frac{157.1}{2\pi} = 25.0 \text{ revolutions}$$

3. (c) The fifth second means from $t = 4$ s to $t = 5$ s. Take the difference of two totals: $$\theta_{5th} = \tfrac{1}{2}\alpha(5)^2 - \tfrac{1}{2}\alpha(4)^2 = \tfrac{1}{2}(3.142)(25 - 16) = \tfrac{1}{2}(3.142)(9) = 14.14 \text{ rad}$$

4. The standard shortcut, for the same answer. For a body starting from rest, the angle in the $n$-th second is $$\theta_n = \frac{\alpha(2n - 1)}{2} = \frac{3.142(9)}{2} = 14.14 \text{ rad} \ \checkmark$$

5. Read it as a pattern. The angles in successive seconds go as $1 : 3 : 5 : 7 : 9 : \ldots$, exactly as distances do in uniformly accelerated straight-line motion.

Final Answer: $\alpha = 3.142$ rad/s$^2$; 25.0 revolutions in 10 s; and 14.14 rad, that is 2.25 revolutions, during the fifth second.

Takeaway: "In the $n$-th second" always means between $t = n-1$ and $t = n$, so subtract two totals rather than substituting $t = n$. [NEET Important] The odd-number pattern $1:3:5:7$ is a free way to check any answer of this type.