Circular Motion Practice Questions

To move together$\omega ^2 R \le \mu g$ $\Rightarrow R \le \;\frac{\mu g}{\omega ^2}$

$\vec{v}={\sqrt{u^{2-}2 g L}} \hat{j}$ $\vec{u}=u \hat{i}$ $\therefore |\vec{v}-\vec{u}|={\sqrt{ (u^{2-}2 g L) +u^2}}$ $={\sqrt{2 u-2 g L}}$ $\therefore x=2$

Distance travelled $=60 {\text{m}}$ $\Rightarrow$ Angle covered $=135^{\circ}$Displacement $=2 R \sin (\;\frac{135^{\circ}}{2})$ =2(\;\frac{60}{135} \times \;\frac{180}{\pi })\[;\frac{1-\cos (135^{\circ})}{2}]^{1 / 2}$$=2(;\frac{80}{π})(0.85)^{1 / 2}$$≈ 47 {\text{m}}$

$a_r=k^2 r t^2=\;\frac{v^2}{r}$ $\Rightarrow v^2=k^2 r^2 t^2$ or $v=k r t$and $\;\frac{d|v|}{d t}=k r$ $\Rightarrow a_t=k r$ $\Rightarrow |\overline{F} \cdot \overline{v}|=(m k r)(k r t)$ $=m k^2 r^2 t=$ power delivered

As the particle in uniform circular motion experiences only centripetal acceleration of magnitude $\omega ^2 {\text{R}}$ or $\;\frac{v^2}{R}$ directed towards centre so from diagram,$\vec{a}=\;\frac{v^2}{R} \cos \theta (-\hat{i})+\;\frac{v^2}{R} \sin (-\hat{j})$

$\because v=\sqrt{\mu g r}$ $\Rightarrow \;\frac{v_1}{v_2}=\sqrt{\;\frac{r_1}{r_2}}$ $\Rightarrow \;\frac{30}{v_2}=\sqrt{\;\frac{75}{48}}=\sqrt{\;\frac{25}{16}}=\;\frac{5}{4}$ $\Rightarrow V_2=24 {\text{m}} / {\text{s}}$

$V_{a v}=\;{S_{\;\text{total }\;}}/{T_{\;\text{total }\;}}=\;\frac{200+200+\frac{2 \pi \times 200}{4}}{238}$ $=\;\frac{400+100 \pi }{238}=\;\frac{714}{238}=3 {\text{m}} / {\ s}$

$v_{b o b}={\sqrt{5 g l}}={\sqrt{5 \times 10 \times 2}}=10 {\text{m}} / {\ s}$Conserving momentum :$75 \times v=75 \times \;\frac{v}{3}+50 \times 10$ $\Rightarrow 50 v=50 \times 10$ $\Rightarrow v=10 {\text{m}} / {\ s}$

Given, radius of horizontal circle, $r=\text{L} /\sqrt{2}$ Figure illustrating the particle of mass m moving in a horizontal circle, while suspended from a ceiling is shown In equilibrium condition at point A, $\text{T} \cos \theta =m g$ ...(i) $\text{T} \sin \theta =\frac{m v^{2}}{r}$ ...(ii) Divide Eq. (ii) by Eq. (i), $\tan \theta =v^{2} / r g$ $\Rightarrow v=\\sqrt{r g \tan \theta }$ ...(iii) Now, from figure, we can write $\sin \theta =\frac{\text{L} / \sqrt{2}}{L}=\frac{1}{\\sqrt{2}}$ \Rightarrow θ = 45° Substituting the value of θ in Eq. (iii), we get $v=\\sqrt{r g \tan 45°}=\\sqrt{r g}$ Thus, the value of speed of particle is $v=\\sqrt{r g}$.

Given, radius of circle $=R$ Central fictitious force is, $F \propto \;\frac{1}{R^{3}}$. Let $T$ be the time period of revolution, $m, \omega$ be the mass and angular velocity of Earth. $\because \;\; F=m \omega ^{2} R \propto \;\frac{1}{R^{3}}$ $\Rightarrow \;\; \omega ^{2} \propto \;\frac{1}{R^{4}} \Rightarrow \omega =\;\frac{2 \pi }{T} \propto \;\frac{1}{R^{2}}$ $\Rightarrow \;\; T \propto R^{2}$

At ${v}_{\max }, {f}$ will be limiting in nature. $\therefore$ Balancing force in vertical direction, $N \cos 30°-m g-\muN \cos 60°=0$ $\RightarrowN[\cos 30°-\mu\cos 60°]={mg}$ $\therefore {N}=\frac{800 \\times 10}{(0.87-0.1)} \approx 10.2 \\times 10^{3} {kg} {m} / {s}^{2}$ Hence option 1 .

The normal force by the side walls of the groove will be equal to the centripetal force acting on it. i.e. $N=\;\frac{m v^{2}}{r}$.....(i) where, $r \;\; =20 {cm}=0.2 {m}$ $m \;\; =200 {g}=200 \times 10^{-3} {kg}$ and $v=r \omega =\;\frac{2 \pi r}{T}=\;\frac{2 \pi \times 0.2}{40} \frac{m}{s}$ Substituting the given values in Eq. (i), we get $N \;\; =\;\frac{(200 \times 10^{-3}) \times (\frac{2 \pi \times 0.2}{40})^{2}}{0.2}$ $\;\; \approx 9.859 \times 10^{-4} {N}$

The central potential field when particle moves in circular orbit, $U(r)=-\;\frac{C}{r}$ We know that, $F=-\;\frac{d U}{d r}$ $\Rightarrow \;\; F=-\;\frac{d}{d r}(-\;\frac{C}{r}) \Rightarrow {|F|}=-\;\frac{C}{r^{2}}$ $\Rightarrow \;\; \;\frac{m v^{2}}{r}=-\;\frac{C}{r^{2}} \Rightarrow v^{2} \propto \;\frac{1}{r}$ The graph between velocity and radius is hyperbolic.

Length of ${\text{arc}}, \text{I}=4.4 {\text{ly}}=4.4 \times 9.46 \times 10^{15}$ $=4.1624 \times 10^{16} {\text{m}}$ $[\because 1 \text{Iy}=9.46 \times 10^{15} {\text{m}}]$ Angle subtended by arc, θ = 4 s Number of revolution, n = 4 Speed of body, $v=8 {\text{AUs}}^{-1}=1.5 \times 10^{11} \times 8=1.2 \times 10^{12} {\text{ms}}^{-1} [\because 1 {\text{AU}}=1.5 \times 10^{11} {\text{m}}]$ We know that, length of arc subtended by an angle, $\text{I}=\text{R} \theta$ Here, R = radius of arc. Substituting the values in above expression, we get $4.1624 \times 10^{16}=\text{R} \times 4 \times (\frac{1°}{3600}) \times (\frac{\pi }{180°})$ $(\because 1 {\text{s}}=\frac{1}{3600} {\text{degree}})$ $\text{R}=2.146 \times 10^{21} {\text{m}}$ We know that, time = distance/speed $t=\frac{n \times 2 \pi \text{R}}{\text{V}}$ (\because Distance = n × length of circular path ) Substituting the values in above expression, we get $t=\frac{4 \times 2 \pi \times 2.146 \times 10^{21}}{1.2 \times 10^{12}}$ $=4.5 \times 10^{10} {\text{s}}$ Thus, the time required is $4.5 \times 10^{10} {\text{s}}$.

Given, length of string, $l=1 {m}$ $T_{\max }$ and $T_{\min }$ be the tension in string and $v_{1}$ and $v_{2}$ be the velocities of bob at bottom and top in vertical circle.$\;T_{\max } \;\; =m g+m v_{1}^{2} / l$ $\;\;\text{ and }\;\;T_{\min } \;\; =m v_{2}^{2} / l-m g$ $\; \because \;\;\frac{T_{\max }}{T_{\min }} \;\; =\;\frac{m g+m v_{1}^{2} / l}{m v_{2}^{2} / l-m g}=\;\frac{5}{1}$ $\Rightarrow \;m g+m v_{1}^{2} / l \;\; =5 m v_{2}^{2} / l-5 m g$ $\;\;\text{ Here, }\;\;v_{1} \;\; =\sqrt{v_{2}^{2}+4 g l}$ $\Rightarrow \;m g+\;\frac{m}{l}(v_{2}^{2}+4 g l) \;\; =\;\frac{5 m v_{2}^{2}}{l}-5 m g$ $\Rightarrow \;g+\;\frac{v_{2}^{2}+4 g l}{l} \;\; =\;\frac{5 v_{2}^{2}}{l}-5 g$ $\Rightarrow \;6 g l \;\; =5 v_{2}^{2}-v_{2}^{2}-4 g l \Rightarrow 10 g l=4 v_{2}^{2}$ $\Rightarrow \;v_{2} \;\; =\sqrt{\;\frac{10 g l}{4}}=\sqrt{\;\frac{10 \times 10 \times 1}{4}}$ $\; \;\; =\sqrt{25}=5 {m} / {s}$

We know that, static friction force, $f_{s}=\mu _{s} N$ where, $\mu _{s}$ is the coefficient of static friction and $N$ is the normal force acting on the body. As the car is travelling on a circular track, so centripetal force is also acting on it. $\Rightarrow \;\; f_{C}=\;\frac{m v^{2}}{R}$ In limiting condition, $f_{s}=f_{C} \Rightarrow \mu _{s} N=\;\frac{m v^{2}}{R}$ $\Rightarrow \;\; N=\;\frac{m v^{2}}{\mu _{s} R}$....(i) The magnitude of negative lift $f_{L}$ acting downwards on the car is given by $f_{L}=m g-N$ $=m g-\;\frac{m v^{2}}{\mu _{s} R}$ $\Rightarrow \;\; f_{L}=m(g-\;\frac{v^{2}}{\mu _{s} R}) \Rightarrow f_{L}=-m(\;\frac{v^{2}}{\mu _{s} R}-g)$ $\Rightarrow \;\; {|f_{L}|}=m(\;\frac{v^{2}}{\mu _{s} R}-g)$

The maximum speed of cyclist on turn of unbanked road without slipping is given as $v_{\max }=\sqrt{\mu g R}=\sqrt{0.2 \times 10 \times 2}=2 {ms}^{-1} \;\; [ \because \mu =0.2$ (given) $]$ Given, speed $=7 \frac{km}{h}$ $=\;\frac{7000}{3600} {ms}^{-1}=\;\frac{70}{36}=1.94 {ms}^{-1}$ As given speed is lesser than $v_{\max }$, so the cyclist will not slip. Therefore, Statement I is true. As per Statement II, angle of banking, $\theta =45^{\circ}$ We know that, for banked road, $V_{\max }=\sqrt{\;\frac{g R(\mu +\tan \theta )}{(1-\mu \tan \theta )}}$ and $\;\; v_{\min }=\sqrt{\;\frac{g R(\tan \theta -\mu )}{1+\mu \tan \theta }}$ $\Rightarrow \;\; v_{\max }=\sqrt{\;\frac{10 \times 2(0.2+\tan 45^{\circ})}{1-0.2 \tan 45^{\circ}}}$ and $\;\; v_{\min }=\sqrt{\;\frac{10 \times 2(1-0.2)}{1+0.2}}$ and $\;\; v_{\min }=\sqrt{\;\frac{10 \times 2(1-0.2)}{1+0.2}}$ $\Rightarrow \;\; v_{\max }=5.47 {ms}^{-1}$ and $v_{\min }=3.65 {ms}^{-1}$ $\because \;\; v=18.5 {km} / {h}=\;\frac{18.5 \times 1000}{5600}=5.13 {ms}^{-1}$ $[ \because$ given speed $=18.5 {kmh}^{-1}]$ As, $v_{\min } < v < v_{\max }$, so the cyclist will not slip. $\therefore$ Statement II is also true. Hence, option (d) is the correct.

${mx} \omega ^{2} \cos \theta = {mg} \sin \theta$ ${x} \omega ^{2}= {g} \tan \theta$ ${x} \omega ^{2}= {g} .\frac{{dy}}{ {dx}}$ ${X} \omega ^{2}= {g}.(8 {cx})$ $\omega ^{2}=8 {gc}$ $\omega =2 \\sqrt{2 {gc}}$