Rotational Motion – JEE Main PhysicsPractice Questions & PYQs

For rolling without slipping on an inclined plane, we can write $R m g \sin \theta =(m K^{2}+m R^{2}) \alpha$ $\Rightarrow \;\; R m g \sin \theta =m(K^{2}+R^{2}) \alpha$ $\Rightarrow \;\; \alpha =\;\frac{R g \sin \theta }{K^{2}+R^{2}} \Rightarrow \;\frac{\alpha }{R}=\;{g \sin \theta }/{1+\;\frac{K^{2}}{R^{2}}}$ $\Rightarrow \;\; a=\;{g \sin \theta }/{1+\;\frac{K^{2}}{R^{2}}}$ $[ \because a=\;\frac{\alpha }{R}] . . .$ (i) Time period, $t=\sqrt{\;\frac{2 s}{a}}$ ...(ii) From Eqs. (i) and (ii), we get $t=\sqrt{\;\frac{2 s}{g \sin \theta }(1+\;\frac{k^{2}}{R^{2}})}$ For least time acceleration a should be maximum and $K$ should be minimum and we know that $K$ is least for solid sphere. So, time will be least for sphere. It means the body which will reach first at the bottom of the inclined plane is 4 , i.e. solid sphere.

Given, mass of uniform bar, $m=6 {kg}$ Length of bar, $I=2.4 {m}$ Side of hexagon, ${a}=\;\frac{2.4}{6}=0.4 {m}$ Mass of each side, $m^{'}=6 / 6=1 {kg}$ and $O P=\sqrt{a^{2}-a^{2} / 4}=a \sqrt{3} / 2$ $=0.2 \sqrt{3} {m}$ Now, by using parallel axes theorem, $I_{O P} \;\; =\;\frac{m^{'} a^{2}}{12}+m^{'}(\;\frac{\sqrt{3} a}{2})^{2}$ $\;\; =\;\frac{a^{2}}{12}+\;\frac{3 a^{2}}{4}=\;\frac{10 a^{2}}{12}=\;\frac{5 a^{2}}{6}$ $\Rightarrow \;\; I_{O P} \;\; =\;\frac{5}{6} \times 0.4 \times 0.4$ and $I_{\;\text{net }\;}$ (net moment of inertia at $O$ ) $=6 \times \;\frac{5}{6} \times 0.4 \times 0.4=8 \times 10^{-1} {kg}- {m}^{2}$

Given, $M=20 {kg}, R=0.2 {m}$, $F=20 {N}, \omega =50 {rad} {} {s}^{-1}$ $\theta =6.28 {rad}$ (for one revolution). We know that, Torque $=$ Moment of inertia $\times$ Angular acceleration $\Rightarrow \;\; \tau =I \times \alpha$ $\Rightarrow \;\; \alpha =\;\frac{\tau }{l}=\;{F \times R}/{\frac{M R^{2}}{2}}=\;\frac{2 F}{M R} \;\; [ \because \tau =F \times R$ and $I_{ {disc}}=\;\frac{1}{2} M R^{2}]$ $\Rightarrow \;\; \alpha =\;\frac{2 \times 20}{20 \times (0.2)}=10 {rad} / {s}^{2}$ In angular terms, third equation of motion, $\omega ^{2}=\omega _{0}^{2}+2 \alpha \Delta \theta$ $\Rightarrow \;(50)^{2}=(0)^{2}+2 \times 10 \times \Delta \theta$ $\Rightarrow \;\Delta \theta =\;\frac{2500}{20}$ $\Rightarrow \;\Delta \theta =125 {rad}$ $\therefore$ Number of revolution $=\;\frac{\Delta \theta }{\theta }=\;\frac{125}{6.28} \approx 20$

Let $M$ and $R$ be the mass and radius of four bodies. Then, as per question, their moment of inertia are $I_{1}=\;\frac{M R^{2}}{2}, I_{2}=\;\frac{M R^{2}}{2}, I_{3}=\;\frac{M R^{2}}{2}$ $I_{4}=\;\frac{2}{5} M R^{2}$ Clearly, $I_{1}=I_{2}=I_{3} > I_{4}$

Given, $F=4 \hat{i}-3 \hat{j}$ Resolving the components of position vector in horizontal and vertical direction. $\therefore$ Horizontal component, $r_{x}=r \cos \theta$ $=10 \cos 60^{\circ}$ $[ \because \cos 60^{\circ}=\;\frac{1}{2}]$ $=10 \times \;\frac{1}{2}=5$ units Vertical component, $r_{y}=r \sin \theta$ $=10 \sin 60^{\circ}$ $[ \because \sin 60^{\circ}=\;\frac{\sqrt{3}}{2}]$ $=10 \times \;\frac{\sqrt{3}}{2}=5 \sqrt{3}$ units $\therefore$ Position vector w.r.t. $O, r_{O}=5 \hat{i}+5 \sqrt{3} \hat{j}$ and position vector w.r.t. Q, $r_{Q}=-5 \hat{i}+5 \sqrt{3} j$ Torque at point $P$ w.r.t. $O$, $\tau _{O}=r_{O} \times F=(5 \hat{i}+5 \sqrt{3} \hat{j}) \times (4 \hat{i}-3 \hat{j})$ $=(-15-20 \sqrt{3}) {k}^{∧}=(15+20 \sqrt{3})(-{k}^{∧})$ Torque at point $P$ w.r.t. $Q$, $\tau _{Q}=r_{Q} \times F=(4 \hat{i}-3 {j}^{∧}) \times (-5 {i}^{∧}+5 \sqrt{3} {j}^{∧})$ $=(-15+20 \sqrt{3}) \hat{k}=(15-20 \sqrt{3})(-\hat{k})$

The standard moments of inertia for a uniform rod of mass $m$ and length $l$ are $I = \frac{ml^2}{12}$ (about the midpoint) and $I = \frac{ml^2}{3}$ (about an end). Applying these:

(a) Mass $M$, length $L$, midpoint: $I = \frac{ML^2}{12}$ $\rightarrow$ (iii).
(b) Mass $2M$, length $L$, end: $I = \frac{(2M)L^2}{3} = \frac{2ML^2}{3}$ $\rightarrow$ (iv).
(c) Mass $M$, length $2L$, midpoint: $I = \frac{M(2L)^2}{12} = \frac{4ML^2}{12} = \frac{ML^2}{3}$ $\rightarrow$ (ii).
(d) Mass $2M$, length $2L$, end: $I = \frac{(2M)(2L)^2}{3} = \frac{(2M)(4L^2)}{3} = \frac{8ML^2}{3}$ $\rightarrow$ (i).

The correct mapping is (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i).

Angular momentum, $L=I \omega$ The moment of inertia of the circular ring, $I= {Mr}^{2}$ When masses are attached at diametrically opposite ends, the moment of inertia, $I^{'}=M r^{2}+2 m r^{2}$ The new angular speed $=\omega ^{'}$ So, the new angular momentum, $L^{'}=I^{'} \omega ^{'}$ Using the law of conservation of angular momentum $I \omega =I^{'} \omega ^{'}$ $\Rightarrow \;\; M r^{2} \omega =(M r^{2}+2 m r^{2}) \omega ^{'}$ $\Rightarrow \;\; \omega ^{'}=\;\frac{M}{M+2 m} \omega$

${I}_{{z}}={I}_{{x}}+{I}_{{y}}$ (using perpendicular axis theorem) $\& \ {I}={mk}^{2}({K}:$ radius of gyration $)$ so ${mK}_{{z}}^{2}={m} {K}_{{x}}^{2}+{mK}_{{y}}^{2}$ ${K}_{{z}}^{2}={K}_{{x}}^{2}+{K}_{{y}}^{2}$ so radius of gyration about axes $x, y\ \&\ z$ won't be same hense asseration ${A}$ is not correct reason ${R}$ is correct statement (property of a rigid body)

Given, Force, $F=4 \hat{i}+3 \hat{j}+4 \hat{k}$ We know that Torque, $\tau =r \times F$ where, $r$ is the perpendicular distance. $r=(2 \hat{i})-(2 \hat{i}+3 \hat{j}+4 \hat{k})=-3 \hat{j}-4 \hat{k}$ $\Rightarrow \;\; \tau =r \times F=\begin{vmatrix}i^{∧} & j^{∧} & k^{∧} \\ 0 & -3 & -4 \\ 4 & 3 & 4\end{vmatrix}$ $\Rightarrow \;\; \tau =\hat{i}(-12+12)-\hat{j}(0+16)+\hat{k}(0+12)$ $\Rightarrow \;\; \tau =-16 \hat{j}+12 \hat{k}$ $\therefore$ Magnitude of torque $|\tau |=\sqrt{(16)^{2}+(12)^{2}}$ $=\sqrt{256+144}=\sqrt{400}$ $\Rightarrow \;\; |\tau |=20$

For no slipping ${v}_{0}=\omega {R}$ Now $v_{{A}}=v_{B}=\\sqrt{v_{0}^{2}+(\omega R)^{2}}$ $=\\sqrt{2} {v}_{0}$ \Rightarrow ${x}=2$

Given, the mass of the steel rod, m = 2 kg The length of the steel rod, I = 0.6 m According to question, the diagram can be as shown below By using the energy conservation, $m g l=\frac{1}{2} \text{l} \omega ^{2}$ $m g l=\frac{1}{2} \times \frac{m l^{2}}{3} \times \omega ^{2}$ \Rightarrow $\omega = \frac{\sqrt6g}{\text{l}}$ As we know the relation between the linear speed and angularspeed, $v=\omega r=\omega l=\\sqrt{6 g l}$ Substituting the values in the above equation, we get $\text{v}=\\sqrt{6 \times 10 \times 0.6}=6$ m / s Hence, the speed of the free end of the rod when it passes throughits lowest position is 6 m/s.

The uniform wire whose mass is $M$ and length is $L$ and bent into a semicircle of radius $r$. The length of the semicircle wire, $L=\pi r$ $\Rightarrow \;\; r=\;\frac{L}{\pi }$ The moment of inertia, $I=M r^{2}$ $\Rightarrow \;\; I=M(\;\frac{L}{\pi })^{2} \Rightarrow I=\;\frac{M L^{2}}{\pi ^{2}}$

If the force $P$ of magnitude $100 {N}$ is resolved parallel to the arms $A B$ and $A C$ of the frame, the above figure will be represented as followsComponent of force along $A C=100 \cos 35^{\circ} {N}$ $=100 \times 0.819 {N}=81.9 {N} \approx 82 {N}$ This is the required magnitude of the resolved component along the arm $A C$. Compare with given in question, $x=82$

It is given that the system is assumed to be in steady circular motion with constant angular velocity $\omega$. We know that $L=m(r \times v)$ where, $L=$ angular momentum, $v=$ velocity of the particle, $m=$ mass of the particle and $r=$ radius of the circular path traced by the particle. Since, $L_{A}$ is the angular momentum of $M$ about point $A$ which lies in positive $z$-direction. Therefore, with respect to point $A$, we will get the direction of $L$ along positive $Z$-axis and of constant magnitude of mvr. Since, $L_{B}$ is the angular momentum of $M$ about point $B$, so with respect to point $B$, we will get the constant magnitude of $L$ but its direction will be continuously changing. Hence, option (d) is correct.

$\tau _{ {B}}=0$ (torque about point ${B}$ is zero) $({T}_{ {A}} ) \times 100-( {mg})$ $\times 50-(2 {mg}) \times 25=0$ $100 \ {T}_{ {A}}=100 \ {mg}$ $\ {T}_{ {A}}=1 \ {mg}$

Let side of triangle is a and mass is $\ {m}$MOI of plate ABC about centroid ${I}_{0}=\frac{{m}}{3} ( (\frac{{a}}{2 \\sqrt{3}} )^{2} \\times 3\ )=\frac{{ma}^{2}}{12}$ triangle ADE is also an equilateral triangle of side a/2. Let moment of inertia of triangular plate ADE about it's centroid (G') is I $_{1}$ and mass is ${m}_{1}$ ${m}_{1}={{m}}/{\frac{\sqrt{3} {a}^{2}}{4}} \\times \frac{\sqrt{3}}{4}(\frac{{a}}{2} )^{2}=\frac{{m}}{4}$ ${I}_{1}=\frac{{m}_{1}}{12}(\frac{{a}}{2} )^{2}$ $=\frac{{m}}{4 \\times 12}\frac{{a}^{2}}{4}=\frac{{ma}^{2}}{192}$distance ${GG}'=\frac{{a}}{\\sqrt{3}}-\frac{{a}}{2 \sqrt{3}}=\frac{{a}}{2 \\sqrt{3}}$ so MOI of part ADE about centroid G is ${I}_{2}= {I}_{1}+ {m}_{1}(\frac{{a}}{2 \\sqrt{3}} )^{2}$ $=\frac{{ma}^{2}}{192}+\frac{{m}}{4} .\frac{{a}^{2}}{12}$ $= \frac{5 \ {ma}^{2}}{192}$ now MOI of remaining part $=\frac{{ma}^{2}}{12}- \frac{5 {ma}^{2}}{192}$ $= \frac{11 {ma}^{2}}{12 \\times 16}=\ \frac{11 {I}_{0}}{16}$ \Rightarrow $N=11$

Let angular velocity o f the system after collision be $\omega$ . By conservation of angular momentum about the hinge :m(v/\sqrt2)(l/2)=\[{4ml^2}/12+{ml^2}/4]$\omega $On solving$\omega ={3\sqrt2}/7v/l$

${\vec{L}}_{ {i}}={\vec{L}}_{ {f}}$ ${mvL}= {I} \omega$ ${mvL}=(\frac{{ML}^{2}}{3}+{mL}^{2}) \omega$ $0.1 \\times 80 \\times 1$ $=(\frac{0.9 \\times 1^{2}}{3}+0.1 \\times 1^{2} ) \omega$ $8=(\frac{3}{10}+ \frac{1}{10} ) \omega$ $8= \frac{4}{10} \omega$ $\omega =20 {\text{rad}} {{\text{rad}}}/{{\sec}}$

$mgh=1/2mv^{2+}1/2\times 1/2mr^2\times v^2/r^2$ $=3/4mv^2$ $u=\sqrt{4/3gh}$ $\omega =v/r$

Let moment of inertia of bigger disc is $I =\frac{{MR}^{2}}{2}$ $\Rightarrow$ MOI of small disc $\ {I}_{2}={{M}(\frac{{R}}{2} )^{2}}/{2}= \frac{{I}}{4}$ by angular momentum conservation ${I} \omega _{1}+\frac{{I}}{4}(0)= {I} \omega _{2}+\frac{{I}}{4} \omega _{2}$ $\Rightarrow \omega _{2}= \frac{4 \omega _{1}}{5}$ initial kinetic energy ${K}_{1}= \frac{1}{2} {I} \omega _{1}^{2}$ final kinetic energy $\ {K}_{2}$ $= \frac{1}{2}({I}+\frac{{I}}{4})(\frac{4 \omega _{1}}{5} )^{2}$ $= \frac{1}{2} {I} \omega _{1}^{2}(\frac{4}{5} )$ ${P} \%=\frac{{K}_{1}- {K}_{2}}{ {K}_{1}} \\times 100 \%$ $= \frac{1-4 / 5}{1} \\times 100=20 \%$