Circular Motion Practice Questions

Given ${\text{m}}_1= {\text{m}}_2$ and $\;{ {\text{r}}_1}/{ {\text{r}}_2}=\;\frac{3}{4}$As centripetal force $F=\;\frac{m^2}{r}$In order to have constant (same in this question) centripetal force$\; {\text{F}}_1= {\text{F}}_2$ $\;\;{ {\text{m}}_1 {\text{v}}_1^2}/{ {\text{r}}_1}=\;{ {\text{m}}_2 {\text{v}}_2^2}/{ {\text{r}}_2}$ $\;\Rightarrow \;{ {\text{v}}_1}/{ {\text{v}}_2}=\sqrt{\;{ {\text{r}}_1}/{ {\text{r}}_2}}=\;{{\sqrt{3}}}/{2}$

Natural length $= {\text{L}}_0$Extension $=\text{bo} {x}$ $\; {\text{kx}}= {\text{m}} ( {\text{L}}_0+ {\text{x}}) \omega ^2$ $\;\Rightarrow 12.5 {\text{x}}=\;\frac{1}{5} ( {\text{L}}_0+ {\text{x}}) 25 \Rightarrow 1.5 {\text{x}}= {\text{L}}_0$ $\;\Rightarrow \;{ {\text{x}}}/{ {\text{L}}_0}=\;\frac{2}{3}$

$f_s=\;{m v^2}/{ {\text{r}}}$For maximum speed in safe turning,$\; {\text{f}}_5= {\text{f}}_5 \max =\mu {\text{mg}}$ $\; {\text{v}}_{\max } \;\text{ (for safe turning }\;=\sqrt{\mu {\mu g}}$ $\;={\sqrt{0.34 \times 50 \times 10}} \approx 13 {\text{m}} / {\ s}$

${\text{T}} \cos \theta = {\text{mg}}$ ${\text{T}} sin \;\theta =\;{ {\ mv}^2}/{ {\text{R}}}$ $\tan \theta =\;{ {\text{v}}^2}/{ {\text{Rg}}}$ $\;\tan \theta =\;\frac{20^2}{40 \times 10}$ $\;\tan \theta =1$ $\;\Rightarrow \theta =\;\frac{\pi }{4}$

$\;F_c=m \omega ^2 r=200 \times (0.2)^2 \times 70$ $\;=560 N$

$\;k x=m \omega ^2 r$ $\;\Rightarrow k x=0.1 \times 25 \times (0.2+x)$ $\;\Rightarrow 7.5 x=2.5(0.2+x)$ $\;\Rightarrow 3 x=0.2+x$ $\;\Rightarrow 2 x=0.2$ $\;\Rightarrow x=0.1 {\text{m}}$Now, tension in the spring $=k x=7.5 \times 0.1 {\text{N}}=0.75 {\text{N}}$

$\;a=\omega ^2 {\text{R}}= (\;\frac{28 \times 2 \pi }{60}) ^2 \times 1.8$ $\;= (\;\frac{56}{60} \times \;\frac{22}{7}) ^2 \times 1.8$ $\;=\;\frac{(44)^2}{225} \times 1.8$ $\;=\;\frac{1936 \times 1.8}{225}$ $\;x=125$

Given speed ${\text{v}}=\pi {\text{m}} / {\ s}$or $R \omega =\pi$or $\omega =\;\frac{\pi }{R} {\text{rad}} / {\ s}$angular displacement $\theta =120^{\circ}$ or $\;\frac{2 \pi }{3}$uising $\theta =\omega {\text{t}}$ ${\text{t}}=\;\frac{\theta }{\omega }=\;{2 \pi / 3}/{\pi / {\text{R}}}=\;{2 {\text{R}}}/{3}$linear displacement $d=2 {\text{R}} sin \;(\theta / 2)$ ${\text{d}}=2 {\text{R}} sin \; (\;\frac{120}{2})$ $=2 {\text{R}} \times sin \;60=2 {\text{R}} \times \;{{\sqrt{3}}}/{2}$ $= {\text{R}} {\sqrt{3}}$average velocity $=\;{ {\text{d}}}/{ {\text{t}}}=\;{ {\text{R}} {\sqrt{3}}}/{2 {\text{R}} / 3}=\;{3 {\sqrt{3}}}/{2}$

Given, ${\text{m}}=5 {\text{kg}}, {\text{R}}=2 {\text{m}}$ time ${\text{t}}$ for $1 {\text{rev}}=3.14 {\ sec}$ or $\pi {\ sec}$ $\theta$ for $1 {\text{rev}}=2 \pi {\text{rad}}$ Therefore $\omega =\;{\theta }/{ {\text{t}}}=\;\frac{2 \pi }{\pi }=2 {\text{rad}} / {\ s}$ centrifugal force ${\text{F}}= {\text{mR}} \omega ^2$ or ${\text{F}}=5 \times 2 \times 2^2=40 {\text{N}}$

V_A=v {𝑗}^{\^}And V_B=-v {𝑖}^{\^}Time to reach from ${\text{A}}$ to ${\text{B}}=\;\frac{2 \pi R}{4} \times \;\frac{1}{v}=\;\frac{\pi R}{2 v}$Displacement from ${\text{A}}$ to ${\text{B}}=R {\sqrt{2}}$Now, Average velocity from ${\text{A}}$ to ${\text{B}}=\;\frac{\;\text{ Displacement }\;}{\;\text{ Time }\;}=\;{R {\sqrt{2}}}/{\frac{\pi R}{2 v}}=\;{2 {\sqrt{2}} v}/{\pi }$Instantaneous velocity at $B$ is -v {𝑖}^{\^}According to question, $\;\frac{\;\text{ instantaneous velocity }\;}{\;\text{ average velocity }\;}=\;{\pi }/{x {\sqrt{2}}}$ $\;\;{v}/{{2 {\sqrt{2}} v}/{\pi }}=\;{\pi }/{x {\sqrt{2}}}$ $\;\Rightarrow \;{\pi }/{2 {\sqrt{2}}}=\;{\pi }/{x {\sqrt{2}}}$ $\;\Rightarrow x=2$

$T sin \;\theta =\;\frac{m v^2}{l sin \;\theta }$ $\cos \theta =\;\frac{m g}{T} . . . . . . . . .(i)$ $sin \;^2 \theta =\;\frac{m v^2}{T l} . . . . . . . . . (ii)$From (i) and (ii), $1= (\;\frac{m g}{T}) ^{2+}\;\frac{m v^2}{T l}$ $\Rightarrow 1= (\;\frac{10}{400}) ^{2+}\;\frac{v^2}{400}$ $\Rightarrow v^2=399.78$ $\Rightarrow v=20 {\text{m}} / {\ s}$

\; {\text{v}} \;{ {\text{dv}}}/{ {\text{dx}}}=\;{ {\text{v}}^2}/{ {\text{R}}} \Rightarrow ∫_{15}^{ {\text{v}}} \;{ {\text{dv}}}/{ {\text{v}}}=\;{1}/{ {\text{R}}} ∫_{0^{ {\text{x}}} {\text{dx}} $\;v=15 e^{x / R}$ $\;\;\frac{d x}{d t}=15 e^{x / R}$ $∫_{0^{\;\frac{\pi R}{2}} e}^{-x / R} \text{dx}=15 ∫_{0}^{t_0} \text{dt}$ $t_0=40 (1-e^{-\pi / 2})$

$\omega =\;{2 \pi }/{ {\text{T}}}=\;\frac{2 \pi }{4}=\;\frac{\pi }{2} {\text{rad}} / {\ s}$ $\theta =\omega {\text{t}}$ $\theta =\;\frac{\pi }{2} \times 3$ $\theta =\;\frac{3 \pi }{2} {\text{rad}}$ $r=10 {\text{m}}$ $T=4 {\ sec}$ $d={\sqrt{2}}(10) {\text{m}}$

${fr}= {\text{m}\omega ^2r}$ $\;\mu {\text{mg}}= {\text{m}} \omega ^2 {\text{r}}= {\text{const}}$ $\;\omega ^2 {\text{r}}= {\text{const}}$ $\;\omega _1^2 {\text{r}}_1=\omega _2^2 {\text{r}}_2$ $\;\omega ^2(1)= (\;\frac{\omega }{2}) ^2 r_2$ $\;r_2=4 {\text{cm}}$

$\; {\text{a}}_{ {\text{c}}}=\;{ {\text{V}}^2}/{ {\text{r}}}=\;\frac{4^2}{2}=\;\frac{16}{2}=8 {\text{m}} / {\ s}^2$ \;{ \vec{\text{V}}=4 \;{ {\text{j}}}^{\^} \;{ {\text{a}}_{ \vec{\text{c}}}=8 \;{ {\text{i}}}^{\^}

$\theta _1=\;\frac{1}{2} \alpha (2 \times 1-1)=5 {\text{rad}}$ $\Rightarrow \alpha =10 {\text{rad}} / {\sec}^2$So $\theta _2=\;\frac{1}{2} \times \alpha (2 \times 2-1)=15 {\text{rad}}$

$N=\;\frac{m v^2}{r}$ $\Rightarrow$ The graph given in option A suits the best for the above relation.

At any $\theta : T-m g \cos \theta =\;\frac{m v^2}{R}$ $\Rightarrow T=m g \cos \theta +\;\frac{m v^2}{R}$Since $v$ is constant,$\Rightarrow {\text{T}}$ will be minimum when $\cos \theta$ is minimum.$\Rightarrow \theta =180^{\circ}$ corresponds to ${\text{T}}_{\;\text{minimum }\;}$