Rotational Motion — Easy Practice (Set 1)

Quick Recap — Rotation Basics

• Moment of inertia $I = \sum m_i r_i^2$ (unit kg m$^2$) — the rotational analogue of mass; radius of gyration $k$ satisfies $I = mk^2$.
• Torque $\vec{\tau} = \vec{r}\times\vec{F} = I\vec{\alpha}$ (unit N m).
• Angular momentum $L = I\omega$ (unit kg m$^2$/s); it is conserved when the net external torque is zero.
• Rotational kinetic energy $= \tfrac12 I\omega^2$; and $v = \omega r$ links linear and angular speed.
• Standard moments of inertia (mass $M$, radius $R$): ring $MR^2$, disc $\tfrac12 MR^2$, solid sphere $\tfrac25 MR^2$, hollow sphere $\tfrac23 MR^2$.

Beyond-NCERT JEE Formulae

Moment of inertia — the six to know cold (mass $M$, radius $R$, length $L$; about the symmetry axis unless stated):

• Ring / thin hoop (perpendicular axis): $I=MR^2$
• Disc / solid cylinder (perpendicular axis): $I=\tfrac12 MR^2$
• Solid sphere (about a diameter): $I=\tfrac25 MR^2$
• Hollow (thin) sphere (about a diameter): $I=\tfrac23 MR^2$
• Rod about its centre (perpendicular): $I=\tfrac{1}{12}ML^2$
• Rod about one end (perpendicular): $I=\tfrac13 ML^2$

When to use: these are the building blocks of every composite-body and rolling question. [JEE Tip] Store each body by its shape factor $\dfrac{k^2}{R^2}$ — ring $1$, disc $\tfrac12$, solid sphere $\tfrac25$, hollow sphere $\tfrac23$. Most rolling MCQs collapse to a single line once you read off this number.

Parallel-axis theorem: $I=I_{cm}+Md^2$ — shift from the centre-of-mass axis to any parallel axis a distance $d$ away. You always ADD the $Md^2$ term, never subtract it.

Perpendicular-axis theorem (flat laminae only): for a planar body lying in the $xy$-plane, $I_z=I_x+I_y$.

When to use: discs, rings and square plates — never a solid 3-D body. [JEE Tip] For a disc, symmetry forces $I_x=I_y$, so the diameter value is $I_{diam}=\tfrac12 I_z=\tfrac14 MR^2$ at a glance; for a ring it comes out as $\tfrac12 MR^2$.

Radius of gyration: $k=\sqrt{I/M}$, equivalently $I=Mk^2$. It is the single distance at which the entire mass could sit and reproduce the same $I$.

Rolling without slipping (so that $v=R\omega$) — total kinetic energy:

$$KE=\tfrac12 Mv^2\left(1+\dfrac{k^2}{R^2}\right)$$

The rotational share of that energy is $\dfrac{k^2/R^2}{1+k^2/R^2}$, which is $\tfrac27$ for a solid sphere and $\tfrac12$ for a ring.

Rolling down a rough incline of angle $\theta$:

$$a=\dfrac{g\sin\theta}{1+k^2/R^2},\qquad v_{bottom}=\sqrt{\dfrac{2gh}{1+k^2/R^2}}$$

Friction supplies the spin-up torque, so the friction actually required and the minimum coefficient for pure rolling are

$$f=\dfrac{Mg\sin\theta\,(k^2/R^2)}{1+k^2/R^2},\qquad \mu_{min}=\dfrac{(k^2/R^2)\tan\theta}{1+k^2/R^2}$$

When to use: any "which body wins the race" or "which rolls faster" problem. [JEE Tip] A smaller $\dfrac{k^2}{R^2}$ always wins — solid sphere beats disc beats hollow sphere beats ring, regardless of mass or radius. If the surface can only offer $\mu<\mu_{min}$ the body slips, and you must switch to kinetic friction $\mu N$.

Torque, angular momentum and Newton's second law for rotation:

$$\tau=I\alpha,\qquad L=I\omega,\qquad \tau=\dfrac{dL}{dt}$$

Conservation of angular momentum (zero external torque): $I_1\omega_1=I_2\omega_2$.

When to use: a skater pulling her arms in, a disc dropped coaxially onto another, clay landing on a turntable. [JEE Tip] Kinetic energy is NOT conserved in these problems — track it with $KE=\dfrac{L^2}{2I}$. Energy falls whenever two bodies merge (the shared $I$ grows) and rises when a body draws itself inward ($I$ shrinks while internal muscles do the work).

Other beyond-NCERT results worth carrying:

• Rotational work-energy and power: $W=\tau\,\theta$ and $P=\tau\omega$.
• Angular impulse: $\displaystyle\int\tau\,dt=\Delta L$ — the tool to reach for when the torque varies with time and $\alpha$ is not constant.
• Kinetic energy from angular momentum: $KE=\dfrac{L^2}{2I}$.
• Compound (physical) pendulum: $T=2\pi\sqrt{\dfrac{I}{Mgd}}$, where $d$ is the pivot-to-centre-of-mass distance.
• Toppling versus sliding of a block (base width $b$, height $h$) on a tilting surface: it topples about its lower edge if $\tan\theta>\dfrac{b}{h}$, but slides first if $\mu<\dfrac{b}{h}$. [JEE Tip] Whichever threshold is crossed at the smaller tilt happens first, so the block topples rather than slides precisely when $\mu>\dfrac{b}{h}$.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Parallel-axis theorem (rod). A uniform rod of mass $2$ kg and length $1.2$ m turns about a perpendicular axis through a point one-third of the way along it from one end. Find its moment of inertia about that axis.

Step 1 — central value. About a perpendicular axis through the centre, $I_{cm}=\tfrac{1}{12}ML^2=\tfrac{1}{12}(2)(1.2)^2=0.24$ kg m$^2$.

Step 2 — shift distance. The axis lies $\tfrac{L}{3}=0.4$ m from the end, so its distance from the centre is $d=\tfrac{L}{2}-\tfrac{L}{3}=0.6-0.4=0.2$ m.

Step 3 — apply the theorem. $I=I_{cm}+Md^2=0.24+2(0.2)^2=0.24+0.08=0.32$ kg m$^2$.

Answer: $I=0.32$ kg m$^2$. The shift adds only $0.08$, since the axis still sits fairly close to the mass.

Example 2 — Perpendicular-axis theorem and radius of gyration (disc). A uniform disc has mass $3$ kg and radius $0.2$ m. Find its moment of inertia about a diameter, and the matching radius of gyration.

Step 1 — central perpendicular axis. $I_z=\tfrac12 MR^2=\tfrac12(3)(0.2)^2=0.06$ kg m$^2$.

Step 2 — perpendicular-axis theorem. The disc is planar, so $I_z=I_x+I_y$; by symmetry $I_x=I_y=I_{diam}$, hence $I_{diam}=\tfrac12 I_z=0.03$ kg m$^2$, which is the familiar $\tfrac14 MR^2$.

Step 3 — radius of gyration. $k=\sqrt{I_{diam}/M}=\sqrt{0.03/3}=\sqrt{0.01}=0.1$ m, exactly $\tfrac{R}{2}$.

Answer: $I_{diam}=0.03$ kg m$^2$ and $k=0.1$ m.

Example 3 — Which rolls faster: solid sphere versus ring. A solid sphere and a ring are each released from rest and roll without slipping from the same height $h=1.4$ m. Take $g=10$ m/s$^2$. Compare their speeds at the bottom.

Step 1 — the formula. Energy conservation for a rolling body gives $v_{bottom}=\sqrt{\dfrac{2gh}{1+k^2/R^2}}$.

Step 2 — solid sphere, for which $\dfrac{k^2}{R^2}=\tfrac25$. Then $v=\sqrt{\dfrac{2(10)(1.4)}{1+2/5}}=\sqrt{\dfrac{28}{1.4}}=\sqrt{20}=4.47$ m/s.

Step 3 — ring, for which $\dfrac{k^2}{R^2}=1$. Then $v=\sqrt{\dfrac{28}{1+1}}=\sqrt{14}=3.74$ m/s.

Step 4 — compare. The sphere is faster by a factor $\sqrt{20/14}=1.20$, because it locks up a smaller share of its energy as spin.

Answer: solid sphere $4.47$ m/s, ring $3.74$ m/s. The outcome is independent of the masses and radii chosen.

Example 4 — Rolling acceleration and the friction it demands. A solid sphere of mass $2$ kg rolls without slipping down an incline of $30^\circ$; take $g=10$ m/s$^2$. Find its acceleration, the friction force acting on it, and the minimum coefficient of friction required.

Step 1 — acceleration. $a=\dfrac{g\sin\theta}{1+k^2/R^2}=\dfrac{10\times 0.5}{1+2/5}=\dfrac{5}{1.4}=3.57$ m/s$^2$.

Step 2 — friction force. $f=\dfrac{Mg\sin\theta\,(k^2/R^2)}{1+k^2/R^2}=\dfrac{(2)(10)(0.5)(2/5)}{1.4}=\dfrac{4}{1.4}=2.86$ N, which is simply $\tfrac27 Mg\sin\theta$.

Step 3 — minimum coefficient. $\mu_{min}=\dfrac{(k^2/R^2)\tan\theta}{1+k^2/R^2}=\tfrac27\tan 30^\circ=0.165$.

Answer: $a=3.57$ m/s$^2$, $f=2.86$ N and $\mu_{min}=0.165$. Should the real $\mu$ fall below $0.165$, the sphere slips instead of rolling purely.

Example 5 — Angular momentum conservation (spinning skater). A skater on frictionless ice spins with her arms out at $\omega_1=2$ rad/s, her moment of inertia being $I_1=6$ kg m$^2$. She pulls her arms in, cutting it to $I_2=2$ kg m$^2$. Find her new angular speed and the change in kinetic energy.

Step 1 — conserve angular momentum. There is no external torque, so $I_1\omega_1=I_2\omega_2$, giving $\omega_2=\dfrac{I_1\omega_1}{I_2}=\dfrac{(6)(2)}{2}=6$ rad/s.

Step 2 — the two energies. $KE_1=\tfrac12 I_1\omega_1^2=\tfrac12(6)(2)^2=12$ J, while $KE_2=\tfrac12 I_2\omega_2^2=\tfrac12(2)(6)^2=36$ J.

Step 3 — interpret. The kinetic energy triples, since $\dfrac{KE_2}{KE_1}=\dfrac{I_1}{I_2}=3$; the extra $24$ J is the work her muscles do drawing the arms inward.

Answer: $\omega_2=6$ rad/s, with the energy rising from $12$ J to $36$ J.

Example 6 — Disc dropped onto a spinning disc. A disc of moment of inertia $I_1=4$ kg m$^2$ spins at $\omega_0=9$ rad/s about a vertical axis. A second, non-rotating disc of $I_2=2$ kg m$^2$ is lowered gently and coaxially onto it, and the two then turn together. Find the common angular speed and the energy lost.

Step 1 — conserve angular momentum. The drop applies no torque about the axis, so $I_1\omega_0=(I_1+I_2)\omega$, giving $\omega=\dfrac{I_1\omega_0}{I_1+I_2}=\dfrac{(4)(9)}{6}=6$ rad/s.

Step 2 — the two energies. $KE_i=\tfrac12 I_1\omega_0^2=\tfrac12(4)(9)^2=162$ J, and $KE_f=\tfrac12(I_1+I_2)\omega^2=\tfrac12(6)(6)^2=108$ J.

Step 3 — the loss. A total of $\Delta KE=162-108=54$ J is dissipated as the surfaces slip into step — a fraction $\dfrac{I_2}{I_1+I_2}=\tfrac13$ of the original energy.

Answer: $\omega=6$ rad/s, with $54$ J (one-third) lost. Angular momentum is conserved; kinetic energy is not.

Example 7 — Topple or slide? (block on a tilting surface). A uniform block of base width $b=0.4$ m and height $h=1.0$ m rests on a plank whose tilt is slowly raised. The coefficient of static friction is $\mu=0.5$. Determine whether the block topples or slides first.

Step 1 — toppling threshold. The block tips about its lower edge once the vertical line through its centre of mass reaches that edge, that is when $\tan\theta=\dfrac{b}{h}=0.4$, so $\theta_{topple}=21.8^\circ$.

Step 2 — sliding threshold. It begins to slide once $\tan\theta=\mu=0.5$, so $\theta_{slide}=26.6^\circ$.

Step 3 — compare. The smaller angle is reached first, and $21.8^\circ<26.6^\circ$, so the block TOPPLES before it slides — equivalently, toppling wins because $\mu>\dfrac{b}{h}$.

Answer: the block topples first, at roughly $22^\circ$. A wider or shorter block (a larger $\dfrac{b}{h}$) would instead slide.