**Ex. A cockroach of mass ‘m’is start moving, with velocity v on the circumference of a disc of mass’ ‘M” and radius’R’what will be angular velocity of disc.**Initially total angular momentum $=$ Final total angular momentum. $$ 0+0=m v R+\frac{m R^{2}}{2} \omega $$ $\omega=(-) \frac{2 m v}{M R} \quad$ -ive angular velocity for opposite direction.\

**Ex. A rotating table has angular velocity ‘$\omega$’ and moment of inertia I. Aperson of mass’m’ stands of centre of rotating table. If the person moves rfrom the centre what will be angular velocity of rotating table.**$\mathrm{I}_{1} \omega=\left(\mathrm{I}_{1}+\mathrm{mr}^{2}\right) \omega_{2} \quad$ or $\quad \omega_{2}=\frac{\mathrm{I}_{1} \omega}{\mathrm{I}_{1}+\mathrm{mr}^{2}}$

**Ex. A horizontal disc is rotating about a vertical axis passing through its centre at the rate of 100 rev $\min .$ A blob of wax, of mass $20 \mathrm{gm},$ falls on the disc and sticks to it at a distance of $5 \mathrm{cm}$ from the axis. If the moment of inertia of the disc about the given axis is $2 \times 10^{-4} \mathrm{kg}-\mathrm{m}^{2},$ find the new frequency of rotation of the disc.**$\begin{array}{ll}{\text { The M.I. of the disc, }} & {I_{1}=2 \times 10^{-4} \mathrm{kg}-\mathrm{m}^{2}} \\ {\text { The M.I. of the blob of wax, }} & {\mathrm{I}_{2}=20 \times 10^{-3} \times(0.05)^{2}=0.5 \times 10^{-4} \mathrm{kg}-\mathrm{m}^{2}}\end{array}$ Let the initial angular speed of the disc be $\omega=2 \pi n$ and let the final angular speed of the disc and blob of wax be $\omega^{\prime}=2 \pi n^{\prime}$\text { Then, } \quad \mathrm{I}_{1} \omega=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega^{\prime}$\mathrm{I}_{1} \times 2 \pi \mathrm{n}=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) 2 \pi \mathrm{n’}$ (The law of conservation of angular momentum)$\therefore 2 \times 10^{-4} \times 100=\left(2 \times 10^{-4}+0.5 \times 10^{-4}\right) \times n’$so, $n^{\prime}=\frac{2}{2.5} \times 10^{2}=80 \mathrm{rev} / \mathrm{min}$

**Ex. A solid cylinder of mass ‘M’kg and radius ‘R’ is rotating along its axis with angular velocity $\omega$ without friction. A particle of mass’ ‘m’moving at v m/sec collide against the cylinder and sticks to it. Then calculate angular velocity and angular momentum of cylinder and initial and final kinetic energy of system?**Intial momentum of cylinder $=\mathrm{I} \omega \quad$ Intial momentum of particle $=\mathrm{m} \mathrm{v} \mathrm{R}$Before sticking total angular momentum $\mathrm{J}_{1}=\mathrm{I} \omega+\mathrm{mvR}$ After sticking total angular momentum $\quad \mathrm{J}_{2}=\left(\mathrm{I}+\mathrm{mR}^{2}\right) \omega^{\prime}$if $\tau=0 \quad$ then $\quad J_{1}=J_{2}$Angular velocity $\quad \omega^{\prime}=\frac{I \omega+m v R}{I+m R^{2}}$Initial kinetic energy of system $=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{mv}^{2}$Final kinetic energy of system $\quad=\frac{1}{2}\left(\mathrm{I}+\mathrm{mR}^{2}\right) \omega^{2}$

**Ex. If the earth were suddenly to shrink to half its size (its mass remaining const) what would be the length of a day.**Sol. $\quad \mathrm{I}_{1} \omega_{1}=\mathrm{I}_{2} \omega_{2} \quad$ so $\quad \frac{2}{5} \mathrm{MR}^{2} \times \frac{2 \pi}{\mathrm{T}_{1}}=\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^{2} \frac{2 \pi}{\mathrm{T}_{2}}$hence $\mathrm{T}_{1}=24 \mathrm{hr} \quad \mathrm{T}_{2}=6 \mathrm{hr}$

**Ex. In which of the following cases, it is most difficult to rotate the rod?**Sol. $\quad(\mathrm{c})$ as $\mathrm{I}$ is maximum.

**Ex. A thin meter scale is kept vertical by placing its one end on floor keeping the end in contact stationary it is allowed to fall. Find the velocity of its upper end when it hit the floor.**Sol. $\quad \mathrm{mgh}=\frac{1}{2} \mathrm{I} \omega^{2} \quad$ where $\mathrm{h}=\frac{\ell}{2}, \mathrm{I}=\frac{\mathrm{m} \ell^{2}}{3}, \omega=\frac{\mathrm{v}}{\ell}$[Loss is P.E. = Rotational K.E.]$\frac{\mathrm{mg} \ell}{2}=\frac{1}{2} \times \frac{\mathrm{m} \ell^{2}}{3} \times \frac{\mathrm{v}^{2}}{\ell^{2}} \quad \mathrm{v}=\sqrt{3 \mathrm{g} \ell}$

**Ex. A particle is moving in $x-y$ plane and the components of its velocity along $x$ and $y$ axis are $v_{x}$ and $v_{y}$. Find the angular momentum about the origin.**We know that angular momentum of a particle $\vec{J}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{r}} \times \mathrm{m} \overrightarrow{\mathrm{v}}=\mathrm{m}(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{v}})$$=\mathrm{m}\left|\begin{array}{ccc}{\mathrm{i}} & {\mathrm{j}} & {\mathrm{k}} \\ {\mathrm{x}} & {\mathrm{y}} & {\mathrm{z}} \\ {\mathrm{V}_{\mathrm{x}}} & {\mathrm{V}_{\mathrm{y}}} & {\mathrm{V}_{\mathrm{z}}}\end{array}\right|=\mathrm{m}\left|\begin{array}{ccc}{\mathrm{i}} & {\mathrm{j}} & {\mathrm{k}} \\ {\mathrm{x}} & {\mathrm{y}} & {\mathrm{o}} \\ {\mathrm{V}_{\mathrm{x}}} & {\mathrm{V}_{\mathrm{y}}} & {\mathrm{o}}\end{array}\right|=\mathrm{m}\left(\mathrm{x} \vee_{\mathrm{y}}-\mathrm{y} \mathrm{v}_{\mathrm{x}}\right) \hat{\mathrm{k}}$

**Ex. A ring of mass $10 \mathrm{kg}$ and diameter 0.4 $\mathrm{m}$ is rotating about its geometrical axis at 1200 rot./min. Find its moment of inertia and angular momentum.**M.I. of a ring about its geometrical axis $=$ M.I. of ring an axis passing through its centre and $\quad$ perpendicular to its plane.$=M R^{2}=10(0.2)^{2}=10 \times 0.04=0.4 \mathrm{kg}-\mathrm{m}^{2}$Now Angular momentum, $J=$ I. $\omega=I \frac{2 \pi n}{t}=0.4 \times \frac{2 \pi \times 1200}{60}=50.24 \mathrm{J-sec}$

**Ex. A homogeneous rod $\mathrm{AB}$ of length $\mathrm{L}$ and mass $\mathrm{M}$ is pivoted at the centre 0 in such a way that it can rotate freely in a vertical plane. The rod $\mathrm{AB}$ is initially in horizontal position. An insect $\mathrm{S}$ of same mass falls vertically with a speed $\mathrm{V}$ at the point c, midway between points 0 and $\mathrm{B}$. Immediately after the insect falls, the angular velocity of the system is**Sol. By Law of conservation of angular momentum $\left(\sum \mathrm{mvr}\right)_{\mathrm{about} 0}=\left(\begin{array}{c}{\mathrm{I}_{\mathrm{system}}} \\ {\mathrm{about} 0}\end{array}\right) \omega$$\Rightarrow M V \frac{L}{4}=\left[\frac{M L^{2}}{12}+M\left(\frac{L}{4}\right)^{2}\right] \omega \Rightarrow \quad M V \frac{L}{4}=M L^{2}\left(\frac{1}{12}+\frac{1}{16}\right) \omega$ $\Rightarrow \mathrm{MV} \frac{\mathrm{L}}{4}=\mathrm{ML}^{2}\left(\frac{4+3}{48}\right) \omega \quad \Rightarrow \quad \omega=\frac{12}{7} \frac{\mathrm{V}}{\mathrm{L}}$

Introduction to Rotational Dynamics Moment of InertiaMoment of Inertia: Perpendicular and Parallel axis theoremRadius of GyrationLaw of Conservation of Angular MomentumConservation of Angular Momentum ExamplesKinetic Energy of a Rotating BodyWork done in rotatory MotionRotational PowerCombine Translational and Rotational MotionRolling without slippingRolling on a plane surfaceRolling on a Inclined PlaneFor Latest updates, Subscribe our Youtube Channel

A puzzle, concerning the conservation of angular momentum. A mass ‘m’ is rotating horizontally at a given velocity ‘v’ at a given radius ‘r’. The mass has energy of J = 1/2*m*v^2 Now let the radius gradually reduce by one half. At the new radius the velocity is a factor of two faster. This gives the mass a larger energy by a factor of four. Where did the mass get the additional energy from?