Unit 5: Rotational Mechanics

Rotational mechanics is the single highest-yield topic in JEE Advanced mechanics, often supplying a multi-part problem worth 8–12 marks. JEE Main typically has 2–3 questions on moment of inertia, rolling motion, angular momentum conservation. The unit synthesizes everything from earlier units (forces, energy, momentum) with the new concept of rotation.

Typical question types:

• JEE Main: moment of inertia of standard bodies (with parallel/perpendicular axis theorems), pure rolling down an incline, angular momentum conservation (skater, planet).
• JEE Advanced: rotational collisions (bullet stuck in hinged rod), rolling with slipping, instantaneous axis of rotation, toppling vs sliding, combined translation + rotation problems.

Concept Map

• Centre of Mass
  • Discrete systems
  • Continuous bodies (rod, arc, plate, hemisphere)
  • Properties (motion of CM, external forces only)
• Rotational Kinematics
  • $\omega, \alpha$; analogues of linear motion
  • Rotation about fixed axis
• Moment of Inertia (MoI)
  • Definition, standard bodies (with derivations)
  • Parallel & perpendicular axis theorems
  • Radius of gyration
• Torque and Angular Momentum
  • Definitions; relation $\vec\tau = d\vec L/dt$
  • Conservation of angular momentum
• Combined Translation + Rotation
  • Pure rolling condition; KE decomposition
  • Rolling on incline; slipping vs rolling
  • Direction of friction in rolling
• Instantaneous Axis of Rotation
• Angular Momentum about a Point (for a moving particle)
• Toppling vs Sliding
• Rotational Collisions

Topic 1: Centre of Mass

Sub-topic A: Discrete systems

For $n$ particles at positions $\vec r_i$ with masses $m_i$:

$\vec R_{cm} = \frac{\sum m_i \vec r_i}{M},\quad M = \sum m_i.$

If $\vec F_{ext}$ is the net external force, $\vec F_{ext} = M\vec a_{cm}$. Internal forces (e.g. those in a collision) cancel pairwise (Newton's third law).

Sub-topic B: Continuous bodies — derivations

Uniform rod (length $L$, mass $M$). Choose origin at one end. Linear mass density $\lambda = M/L$.

$x_{cm} = \frac{1}{M}\int_0^L x\,\lambda\,dx = \frac{1}{M}\cdot\lambda\frac{L^2}{2} = \frac{L}{2}.$

So CM at the midpoint. (For a rod of varying density $\lambda(x)$, the integral gives a different location.)

Semicircular wire (radius $R$, uniform). Parametrize $\theta\in[0,\pi]$, $x=R\cos\theta$, $y=R\sin\theta$, $ds = R\,d\theta$. Linear density $\lambda = M/(\pi R)$.

$y_{cm} = \frac{1}{M}\int_0^\pi R\sin\theta\cdot \lambda R\,d\theta = \frac{R^2\lambda}{M}\int_0^\pi \sin\theta\,d\theta = \frac{R^2}{\pi R}\cdot 2 = \frac{2R}{\pi}.$

$x_{cm} = 0$ by symmetry. So CM is at $(0, 2R/\pi)$.

Semicircular disc (radius $R$, mass $M$). $\sigma = M/(\tfrac12\pi R^2)$. Use strips parallel to the diameter at height $y$ width $dy$, length $2\sqrt{R^2-y^2}$:

$y_{cm} = \frac{1}{M}\int_0^R y\cdot \sigma\cdot 2\sqrt{R^2-y^2}\,dy.$

Let $u = R^2-y^2$, $du = -2y\,dy$, $y\,dy = -du/2$:

$y_{cm} = \frac{2\sigma}{M}\int_{R^2}^0 \sqrt u\cdot(-\tfrac12)du = \frac{\sigma}{M}\int_0^{R^2}\sqrt u\,du = \frac{\sigma}{M}\cdot\frac{2}{3}R^3 = \frac{2R^3 \sigma}{3M}.$

With $\sigma = 2M/(\pi R^2)$:

$y_{cm} = \frac{2R^3}{3M}\cdot\frac{2M}{\pi R^2} = \frac{4R}{3\pi}.$

Triangular plate (uniform). CM at the centroid: $(x_1+x_2+x_3)/3, (y_1+y_2+y_3)/3$.

Hemispherical solid (radius $R$, mass $M$). Use shells of radius $r$, thickness $dr$... or disks of radius $\sqrt{R^2-y^2}$ at height $y$.

Volume element: disk of radius $\sqrt{R^2-y^2}$, area $\pi(R^2-y^2)$, thickness $dy$. $\rho = M/(\tfrac23\pi R^3)$.

$y_{cm} = \frac{1}{M}\int_0^R y\cdot \rho\pi(R^2-y^2)\,dy = \frac{\rho\pi}{M}\int_0^R(R^2 y - y^3)dy = \frac{\rho\pi}{M}\cdot\frac{R^4}{4}.$

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

Hemispherical shell (thin, radius $R$, mass $M$). Use rings at height $y$ width $R\,d\theta$ at polar angle $\theta$:

$y = R\cos\theta$, ring radius $R\sin\theta$, area = $2\pi R\sin\theta\cdot R\,d\theta = 2\pi R^2\sin\theta\,d\theta$.

$\sigma = M/(2\pi R^2)$.

$y_{cm} = \frac{1}{M}\int_0^{\pi/2} R\cos\theta\cdot \sigma\cdot 2\pi R^2\sin\theta\,d\theta = \frac{2\pi R^3\sigma}{M}\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta = \frac{2\pi R^3\sigma}{M}\cdot \frac{1}{2}=\frac{\pi R^3\sigma}{M}.$

$= \pi R^3\cdot M/(2\pi R^2 M) = R/2$.

Sub-topic C: Centre of mass with cavity

If a body of mass $M$ has a cavity (small piece removed), treat the cavity as a negative mass and use the formula for the CM of two objects.

$\vec R_{cm}^{(\text{body with cavity})} = \frac{M\vec R_{cm}^{(\text{full})} - m\vec R_{cm}^{(\text{cavity})}}{M - m}.$

Sub-topic D: Motion of CM

$\vec F_{ext} = M\vec a_{cm},\quad \vec P_{sys} = M\vec v_{cm}.$

Internal forces don't affect CM motion. Example: when a shell explodes in mid-air, its CM continues on the original projectile trajectory.

Worked Examples (JEE Main level)

Example 1.1. Find CM of a system of three particles at $(0,0)$, $(1,0)$, $(0,1)$ m, masses $1, 2, 3$ kg.

$x_{cm} = (0+2+0)/6 = 1/3$. $y_{cm} = (0+0+3)/6 = 1/2$. So $(1/3, 1/2)$ m.

Example 1.2. A uniform disc of radius $R$ has a circular hole of radius $R/2$ centred at $R/2$ from the centre. Find the CM of the remainder.

Treat hole as negative mass. Original CM at $0$ with mass $M$. Hole mass $M/4$ (area ratio $\tfrac14$). Hole CM at $R/2$.

$x_{cm} = \frac{M\cdot 0 - (M/4)(R/2)}{M - M/4} = \frac{-MR/8}{3M/4} = -\frac{R}{6}.$

(Opposite side of the hole, distance $R/6$ from centre.)

Worked Examples (JEE Advanced level)

Example 1.A1. A boat of mass $M$ and length $L$ is at rest on still water. A man of mass $m$ walks from one end to the other. Displacement of the boat (no friction with water)?

CM stays put (no external horizontal force). Let boat shift by $x$. Man's displacement in ground frame: $L - x$ (in opposite direction effectively, but let's set up coords: man moves $+\hat\imath$ in the boat by $L$, boat moves $-\hat\imath$ by $x$, so man's ground displacement is $L - x$.)

$0 = m(L - x) + M(-x)\Rightarrow x = \dfrac{mL}{m+M}.$

Topic 2: Rotational Kinematics

Sub-topic A: Definitions

For rotation about a fixed axis:

• $\theta(t)$ = angular position, $\omega = d\theta/dt$, $\alpha = d\omega/dt = d^2\theta/dt^2$.
• Linear quantities for a point at distance $r$ from axis: $v = r\omega$, $a_t = r\alpha$, $a_c = r\omega^2$.

Sub-topic B: Constant $\alpha$ equations

$\omega = \omega_0 + \alpha t,\quad \theta = \omega_0 t + \tfrac12\alpha t^2,\quad \omega^2 = \omega_0^2 + 2\alpha\theta.$

Analogous to linear kinematics.

Worked Examples (JEE Main level)

Example 2.1. A wheel starts from rest with $\alpha = 2$ rad/s². Angular speed after $5$ s and angle covered?

$\omega = 10$ rad/s. $\theta = \tfrac12\cdot 2\cdot 25 = 25$ rad.

Topic 3: Moment of Inertia

Sub-topic A: Definition

For a system of particles, the moment of inertia about an axis is

$I = \sum m_i r_i^2,\quad\text{or}\quad I = \int r^2\,dm,$

where $r_i$ is the perpendicular distance from particle $i$ to the axis.

Sub-topic B: Derivations for standard bodies

1. Thin rod, axis through centre perpendicular to length.

$\lambda = M/L$. $I = \int_{-L/2}^{L/2} x^2 \lambda\,dx = \lambda\cdot \tfrac{2}{3}(L/2)^3 = \dfrac{ML^2}{12}.$

2. Thin rod, axis through one end perpendicular.

$I = \int_0^L x^2\lambda\,dx = \dfrac{ML^2}{3}$. (Or via parallel axis: $ML^2/12 + M(L/2)^2 = ML^2/3$.)

3. Thin ring, axis through centre perpendicular.

All mass at distance $R$: $I = MR^2$.

4. Disc, axis through centre perpendicular.

Use rings of radius $r$ width $dr$. $dm = \sigma\cdot 2\pi r\,dr$, $\sigma = M/(\pi R^2)$. $I = \int_0^R r^2\cdot \sigma 2\pi r\,dr = 2\pi\sigma R^4/4 = \dfrac{MR^2}{2}.$

5. Disc, axis along a diameter.

By perpendicular axis theorem: $I_z = I_x + I_y$. By symmetry $I_x = I_y$. So $I_d = I_z/2 = MR^2/4$.

6. Solid sphere, axis through centre.

Use spherical shells of radius $r$ thickness $dr$. Mass $dm = 4\pi r^2 \rho\,dr$. MoI of a thin spherical shell about its diameter is $\tfrac23 m r^2$.

Total: $I = \int_0^R \tfrac23\cdot 4\pi r^2\rho\cdot r^2\,dr = \tfrac{8\pi\rho}{3}\cdot R^5/5 = \tfrac{8\pi R^5}{15}\rho$.

With $\rho = 3M/(4\pi R^3)$: $I = \tfrac{8\pi R^5}{15}\cdot \tfrac{3M}{4\pi R^3} = \dfrac{2MR^2}{5}.$

7. Thin spherical shell, axis through centre.

$I = \tfrac{2}{3}MR^2$.

(Quick derivation: use rings at polar angle $\theta$, radius $R\sin\theta$, mass $dm = \sigma\cdot 2\pi R\sin\theta\cdot R\,d\theta$. MoI of ring about axis $= dm\cdot R^2\sin^2\theta$. Integrate over $\theta\in[0,\pi]$.)

8. Cylinder (solid, length $L$, radius $R$), axis along length.

Like a disc: $I = MR^2/2$.

9. Hollow cylinder (thin, radius $R$), axis along length.

$I = MR^2$.

10. Rectangular plate ($a\times b$, mass $M$), axis through CM perpendicular to plane.

By perp. axis: $I_z = I_x + I_y = M b^2/12 + M a^2/12 = M(a^2+b^2)/12$.

Sub-topic C: Theorems

Parallel axis theorem. If $I_{cm}$ is MoI about an axis through the CM, then MoI about a parallel axis at distance $d$ is

$I = I_{cm} + Md^2.$

Perpendicular axis theorem (for plane laminas). For a plane lamina, MoI about an axis perpendicular to the plane equals the sum of MoIs about two perpendicular axes in the plane intersecting at the same point:

$I_z = I_x + I_y.$

Sub-topic D: Radius of gyration

$I = M k^2$ defines the radius of gyration $k$. For a solid sphere, $k = R\sqrt{2/5}$.

Worked Examples (JEE Main level)

Example 3.1. MoI of a rod ($L=1$ m, $M=2$ kg) about an axis through one end perpendicular?

$I = ML^2/3 = 2/3$ kg·m².

Example 3.2. MoI of a uniform disc of mass $M$ and radius $R$ about an axis tangent to the disc in its plane?

Through diameter: $MR^2/4$. Parallel axis (distance $R$): $I = MR^2/4 + MR^2 = 5MR^2/4$.

Example 3.3. MoI of a thin ring about a tangent in its plane?

Diameter: $MR^2/2$ (by perp. axis, $I_z=MR^2=I_x+I_y=2I_x\Rightarrow I_x = MR^2/2$). Tangent: $MR^2/2 + MR^2 = 3MR^2/2$.

Worked Examples (JEE Advanced level)

Example 3.A1. A uniform rod of length $L$ and mass $M$ has MoI $I_0$ about a perpendicular axis through one end. A small mass $m$ is added at the other end. New MoI?

$I_0 = ML^2/3$. With added mass: $I = ML^2/3 + mL^2$.

Example 3.A2. MoI of a uniform solid cone (semi-vertical angle $\alpha$, height $h$, mass $M$) about its axis.

Volume element: disc at height $z$ from apex, radius $z\tan\alpha$, thickness $dz$.

$dV = \pi z^2\tan^2\alpha\,dz$. $\rho = M/V = M/(\tfrac13\pi h^3\tan^2\alpha)$.

MoI of disc about axis: $\tfrac12 dm\cdot r^2 = \tfrac12\rho\pi z^2\tan^2\alpha\cdot z^2\tan^2\alpha\,dz = \tfrac12\rho\pi z^4\tan^4\alpha\,dz$.

Integrate $0$ to $h$: $I = \tfrac{\rho\pi\tan^4\alpha}{2}\cdot \tfrac{h^5}{5} = \tfrac{\rho\pi h^5\tan^4\alpha}{10}$.

Substituting $\rho$: $I = \dfrac{3 M h^2\tan^2\alpha}{10} = \dfrac{3MR^2}{10}$, where $R = h\tan\alpha$ is the base radius.

Topic 4: Torque and Angular Momentum

Sub-topic A: Torque

About a point $O$, the torque of a force $\vec F$ applied at position $\vec r$ (from $O$) is

$\vec\tau = \vec r\times\vec F,\quad |\tau| = rF\sin\theta = F\cdot r_\perp.$

For rotation about a fixed axis with MoI $I$:

$\tau_{ext} = I\alpha.$

Sub-topic B: Angular momentum

For a single particle, $\vec L = \vec r\times\vec p = m\vec r\times\vec v$.

For a rigid body rotating about a fixed axis with $\omega$: $L = I\omega$.

Theorem: $\vec\tau_{ext} = d\vec L/dt$ (Newton's 2nd law for rotation).

Sub-topic C: Conservation of angular momentum

If $\vec\tau_{ext} = \vec 0$, then $\vec L =$ const.

Classic examples:

• Ice skater pulling arms in: $I$ decreases, $\omega$ increases.
• Planet in elliptic orbit: gravity is central → torque about Sun is zero → $L$ conserved → Kepler's 2nd law (equal areas in equal times).
• Top with no friction: spin axis precession is via gravity torque.

Sub-topic D: Angular momentum about a point (moving particle)

For a particle moving with velocity $\vec v$ at position $\vec r$ relative to $O$,

$\vec L = m\vec r\times\vec v.$

Magnitude: $L = mvr\sin\theta = mv\cdot d$ where $d$ is the perpendicular distance from $O$ to the line of $\vec v$.

For a particle moving in a straight line (constant $\vec v$): the perpendicular distance from a fixed point to the line is constant → $L$ is constant about that point (even though the particle moves).

Worked Examples (JEE Main level)

Example 4.1. A force $\vec F = (3\hat\imath + 4\hat\jmath)$ N acts at position $\vec r = (2\hat\imath - \hat\jmath)$ m. Torque about origin?

$\vec\tau = \vec r\times\vec F = (2\hat\imath - \hat\jmath)\times(3\hat\imath+4\hat\jmath) = 8\hat k + 3\hat k = 11\hat k$ N·m.

Example 4.2. A flywheel of MoI $0.5$ kg·m² spins at $20$ rad/s. Torque to stop in $4$ s?

$\tau = I\alpha = 0.5\cdot(20/4) = 2.5$ N·m.

Worked Examples (JEE Advanced level)

Example 4.A1. A bullet of mass $m$ moving at speed $v$ hits and embeds in the free end of a rod (mass $M$, length $L$) pivoted at the other end. Angular speed of the system just after collision?

Conserve angular momentum about pivot (the impulsive reaction at pivot has zero moment arm):

$mvL = I\omega$, where $I = ML^2/3 + mL^2$.

$\omega = \frac{mvL}{(M/3 + m)L^2} = \frac{mv}{(M/3 + m)L}.$

Example 4.A2. A planet moves in an elliptic orbit; at perihelion distance $r_1$ its speed is $v_1$. Find speed at aphelion ($r_2$).

$L$ conserved: $m v_1 r_1 = m v_2 r_2\Rightarrow v_2 = v_1 r_1/r_2$ (with both velocities perpendicular to position at these extremes).

Topic 5: Rolling Motion

Sub-topic A: Pure rolling condition

For a body of radius $R$ rolling without slipping on a surface:

$v_{cm} = \omega R,\quad a_{cm} = \alpha R.$

The point of contact is instantaneously at rest.

Sub-topic B: KE decomposition

Total KE = translational KE of CM + rotational KE about CM:

$KE = \tfrac12 M v_{cm}^2 + \tfrac12 I_{cm}\omega^2.$

For pure rolling: $KE = \tfrac12 M v^2(1 + I_{cm}/(MR^2))$.

For a solid sphere ($I=2MR^2/5$): $KE = (7/10)Mv^2$. For a disc/cylinder ($I=MR^2/2$): $KE = (3/4)Mv^2$. For a ring ($I=MR^2$): $KE = Mv^2$.

Sub-topic C: Rolling on an incline

A body of mass $M$, radius $R$, MoI $I = \beta M R^2$ (so $\beta = 2/5, 1/2, 1$ for solid sphere, disc, ring).

Newton's 2nd law along incline: $Mg\sin\theta - f = Ma$.

Torque about CM (friction at the contact point): $f R = I\alpha = \beta M R^2\cdot a/R = \beta M R a\Rightarrow f = \beta M a$.

So $Mg\sin\theta - \beta M a = M a\Rightarrow a = \dfrac{g\sin\theta}{1+\beta}.$

Friction needed: $f = \beta M a = \dfrac{\beta M g\sin\theta}{1+\beta}$.

For no slipping, need $f\le \mu_s M g\cos\theta$:

$\boxed{\mu_s \ge \frac{\beta\tan\theta}{1+\beta}.}$

If incline is steeper, body rolls AND slips, and friction becomes kinetic $\mu_k Mg\cos\theta$.

Race down an incline. From $a = g\sin\theta/(1+\beta)$, smaller $\beta$ wins: solid sphere $>$ disc $>$ ring. Speed at the bottom (from $v^2 = 2a L$, where $L$ = incline length, $h=L\sin\theta$):

$v = \sqrt{\frac{2gh}{1+\beta}}.$

Sub-topic D: Direction of friction in rolling

• Body initially rolling on smooth surface: no friction needed; rolls forever.
• Body rolling on rough surface, no other force: no slipping tendency, so no friction.
• Pure translation on rough surface (e.g. a ball placed on belt or kicked): friction acts on the bottom backward (opposing slip), provides torque to start rolling.
• Body rolling down incline: friction acts up the incline (provides torque to keep angular acceleration positive).
• Body being pushed/pulled at the centre on rough surface: friction acts backward (opposes slip tendency at contact).
• Body being pushed/pulled at the top: depends on geometry. Trick: write the torque equation about the bottom contact point to find net rotational effect; friction adjusts accordingly.

Sub-topic E: Slipping → rolling transition

A ball is given pure translation $v_0$ on a rough horizontal surface ($\mu_k$). Find time at which pure rolling begins.

Friction $\mu_k mg$ backward decelerates: $v(t) = v_0 - \mu_k g t$.

Torque about CM (friction at contact, lever arm $R$): $\tau = \mu_k mg R = I\alpha = (2/5) mR^2\alpha\Rightarrow \alpha = (5\mu_k g)/(2R)$.

$\omega(t) = \alpha t = 5\mu_k g t/(2R)$.

Pure rolling when $v = \omega R$:

$v_0 - \mu_k g t = (5/2)\mu_k g t\Rightarrow t = \frac{2 v_0}{7\mu_k g}.$

Speed at this instant: $v = v_0 - \mu_k g\cdot 2v_0/(7\mu_k g) = 5 v_0/7$.

Worked Examples (JEE Main level)

Example 5.1. A solid sphere rolls down a $30^\circ$ incline. Acceleration ($g=10$)?

$a = g\sin\theta/(1+\beta) = 10\cdot 0.5/(1+2/5) = 5/(7/5) = 25/7\approx 3.57$ m/s².

Example 5.2. Min $\mu$ for a disc to roll without slipping on a $45^\circ$ incline?

$\mu\ge \beta\tan\theta/(1+\beta) = (1/2)\cdot 1/(3/2) = 1/3$.

Worked Examples (JEE Advanced level)

Example 5.A1. A solid sphere is set spinning at $\omega_0$ in place (no translation) on a rough horizontal surface ($\mu$). Find time and position when pure rolling starts.

Friction acts on the bottom contact; bottom is moving backward (since sphere spins forward), so friction is forward. This accelerates the CM: $a = \mu g$ (forward). Torque about CM (friction acts at contact, magnitude $\mu mg$, lever arm $R$): $\tau = \mu mg R$, opposite to spin → decelerates: $\alpha = -5\mu g/(2R)$.

$v(t) = \mu g t$. $\omega(t) = \omega_0 - 5\mu g t/(2R)$.

Pure rolling: $v = \omega R\Rightarrow \mu g t = \omega_0 R - 5\mu g t/2\Rightarrow t = \dfrac{2\omega_0 R}{7\mu g}$.

Distance covered $= \tfrac12 \mu g t^2 = \mu g\cdot 2\omega_0^2 R^2/(49\mu^2 g^2) = 2\omega_0^2 R^2/(49\mu g)$.

Final speed $v = 2\omega_0 R/7$.

Example 5.A2. A sphere of radius $r$ rolls in the inside of a hemisphere of radius $R$. Period of small oscillations?

The CM moves on a circle of radius $R-r$. Equation of motion (SHM with effective "$g$" depending on rolling): $T = 2\pi\sqrt{(R-r)/g_{\text{eff}}}$, where $g_{\text{eff}} = g/(1+\beta) = 5g/7$. So $T = 2\pi\sqrt{7(R-r)/(5g)}$.

Topic 6: Instantaneous Axis of Rotation (IAR)

Sub-topic A: Definition

In a rigid body undergoing combined translation and rotation, at any instant there is an axis (possibly external to the body) about which the body appears to be in pure rotation. The point on the body coinciding with this axis has zero velocity.

For pure rolling, the contact point is the IAR.

Sub-topic B: KE about IAR

If $I_{IAR}$ is MoI about the instantaneous axis,

$KE = \tfrac12 I_{IAR}\omega^2.$

For a wheel rolling on the ground with $v_{cm} = \omega R$: $I_{IAR} = I_{cm} + MR^2$. $KE = \tfrac12 (I_{cm}+MR^2)\omega^2 = \tfrac12 I_{cm}\omega^2 + \tfrac12 M v^2$. ✓

Sub-topic C: Useful for finding velocities of various points

Velocity of point $P$ on a rolling body: $\vec v_P = \vec\omega\times\vec r_P$ where $\vec r_P$ is from IAR. Magnitude $= \omega\cdot \vert r_P\vert$.

• Top of wheel: $2v_{cm}$.
• Bottom: $0$.
• Any point: depends on its distance from contact point.

Worked Examples

Example 6.1. A wheel of radius $0.5$ m rolls at $v_{cm}=10$ m/s. Velocity of the topmost point?

$2v_{cm} = 20$ m/s.

Example 6.2. A ladder of length $L$ slides such that its bottom moves on the floor with velocity $v$ and top moves down the wall. Find angular velocity when the ladder makes angle $\theta$ with the floor.

The IAR is the point such that horizontal velocity of the bottom matches and vertical velocity of the top matches. Bottom: $(x_B, 0)$, velocity $(v, 0)$. Top: $(0, y_T)$, velocity $(0, -v_T)$.

The IAR is at $(x_B, y_T)$ — i.e. the corner of the rectangle. $\omega = v/y_T = v/(L\sin\theta)$.

Topic 7: Toppling vs Sliding

A block of width $w$, height $h$ on a rough floor pushed horizontally at the top.

Sliding condition: $F > \mu mg$.

Toppling condition: torque about the front edge (front edge is pivot) exceeds the restoring torque from gravity. $F h > mg(w/2)$, i.e. $F > mgw/(2h)$.

Whichever condition is satisfied first as $F$ increases determines whether the block slides or topples.

• If $\mu mg < mgw/(2h)$, i.e. $\mu < w/(2h)$: sliding first.
• If $\mu > w/(2h)$: toppling first.

(For force applied at height $h_0$ instead of top, replace $h$ by $h_0$ in the toppling condition.)

Worked Examples (JEE Advanced level)

Example 7.A1. A cube of side $a$ on a rough floor is pushed by a horizontal force at the top edge. $\mu = 0.5$. What happens?

Toppling at $F = mga/(2a) = mg/2$. Sliding at $F = \mu mg = mg/2$. They're equal — both occur simultaneously.

Topic 8: Rotational Collisions

Sub-topic A: Impulse and angular impulse

For a collision with a hinged body, angular impulse = change in angular momentum:

$\int \tau\,dt = \Delta L.$

About the pivot, the impulsive hinge force has zero moment arm, so doesn't contribute. Use this to find $\omega$ after collision.

Sub-topic B: Examples

Bullet–rod (hinged at one end): $mvL = I\omega = (ML^2/3 + mL^2)\omega$.

Bullet–free rod: Conserve linear momentum AND angular momentum about CM. Bullet embeds at end → final body has CM somewhere between original rod CM and the embedding point. Find $v_{cm}$ from linear momentum, $\omega$ from angular momentum about CM.

Worked Examples (JEE Advanced level)

Example 8.A1. A rod of mass $M$, length $L$ lies on a smooth horizontal table. A ball of mass $m$ moves perpendicular to the rod with speed $v$ and hits the end of the rod, sticking. Find CM speed and angular speed of the system after collision.

Linear momentum: $mv = (M+m)v_{cm}\Rightarrow v_{cm} = mv/(M+m)$.

Locate new CM: original rod CM at centre, mass $M$; ball at end (distance $L/2$ from rod CM), mass $m$. New CM at distance $mL/(2(M+m))$ from original rod CM, toward the end where the ball stuck.

MoI about new CM: $I_{new} = I_{rod,cm} + M d_1^2 + m d_2^2$, where $d_1 = mL/(2(M+m))$ and $d_2 = ML/(2(M+m))$.

$I_{new} = ML^2/12 + M\left(\dfrac{mL}{2(M+m)}\right)^2 + m\left(\dfrac{ML}{2(M+m)}\right)^2$.

Simplify: $= ML^2/12 + \dfrac{Mm L^2 (m + M)}{4(M+m)^2}\cdot \dfrac{1}{1} = ML^2/12 + \dfrac{Mm L^2}{4(M+m)}$.

Angular momentum about new CM: the ball had momentum $mv$ at distance $d_2 = ML/(2(M+m))$ from CM perpendicular to its motion.

$L_0 = mv\cdot d_2 = \dfrac{m M v L}{2(M+m)}$.

After: $L_0 = I_{new}\omega\Rightarrow \omega = L_0/I_{new}$.

Problem-Solving Heuristics

1. For CM problems, use the formula straightforwardly; for bodies with cavities, use negative mass trick.
2. Choose the axis of MoI carefully — usually the CM or the contact point. Use parallel axis to shift.
3. For a plane lamina, perpendicular axis theorem is your friend ($I_z = I_x+I_y$).
4. For combined translation + rotation, always write two equations: $\sum F = Ma_{cm}$ and $\sum\tau_{cm} = I_{cm}\alpha$. Then use the rolling constraint $a = \alpha R$.
5. Friction in rolling: not always at $\mu N$ — only the maximum. For pure rolling, friction is whatever is needed (static); compute it, compare with $\mu N$ to check.
6. For rotational collisions with a hinged body: conserve angular momentum about the hinge; the impulsive hinge force is unknown but has zero moment arm.
7. Angular momentum of a moving particle about a point: $L = mv\cdot d_\perp$ (perpendicular distance from the point to the line of motion).
8. For two-body rotational problems on a smooth floor, conserve linear momentum, angular momentum about the new CM, and (if elastic) KE.
9. Sphere rolling down: solid sphere wins the race, then disc/cylinder, then ring. Order is by $\beta$ — smaller $\beta$, faster.
10. Toppling vs sliding: compute the force needed for each; the smaller decides the outcome.

Common Traps & Mistakes

• For a disc, the MoI about a diameter is $MR^2/4$, not $MR^2/2$ (that's about the perpendicular axis through centre).
• For a rod, $I = ML^2/12$ about centre, $ML^2/3$ about end — students often invert.
• Rolling without slipping is a constraint: $v_{cm}=\omega R$. You can't assume this if there's slipping.
• Friction in pure rolling is static, not kinetic. So $f$ can be any value up to $\mu_s N$.
• When a body slides and rolls (e.g. on a slippery incline), use kinetic friction; the constraint $v=\omega R$ does NOT hold.
• Sphere on smooth surface kicked at centre: it slides without rolling (no friction → no torque). On rough surface, it eventually rolls.
• A particle moving in a straight line at constant velocity has constant angular momentum about a fixed point (the perpendicular distance is constant).
• The torque on a planet about the Sun is zero, so $L$ is conserved — but NOT the angular velocity (since the distance changes).
• Negative mass for cavities: don't double-count!

Quick Revision Card

• $\vec R_{cm} = \sum m_i\vec r_i/\sum m_i$; for continuous bodies use integrals.
• $I = \sum m_i r_i^2$; key results:
  • Rod about centre $= ML^2/12$, about end $= ML^2/3$
  • Ring about axis $= MR^2$, about diameter $= MR^2/2$
  • Disc about axis $= MR^2/2$, about diameter $= MR^2/4$
  • Solid sphere $= 2MR^2/5$, shell $= 2MR^2/3$
• Parallel axis: $I = I_{cm} + Md^2$.
• Perp. axis (lamina): $I_z = I_x + I_y$.
• $\tau = I\alpha$, $L = I\omega$, $\tau_{ext} = dL/dt$.
• Pure rolling: $v = \omega R$, $KE = \tfrac12 Mv^2(1+\beta)$.
• Rolling down incline: $a = g\sin\theta/(1+\beta)$, $v_{bot} = \sqrt{2gh/(1+\beta)}$.
• Min $\mu$ for rolling: $\mu \ge \beta\tan\theta/(1+\beta)$.
• Bullet–hinged rod: $mvL = (ML^2/3 + mL^2)\omega$.

Formula Sheet