Circular Motion Practice Questions

For the mass $M$ to remain stuck to the inner wall, it must undergo circular motion, where the normal force $N$ provides the centripetal force: $N = M \omega^2 R$. To prevent the mass from sliding down due to gravity, the upward static friction force $f$ must balance the weight: $f = Mg$. Given that friction must satisfy $f \le \mu N$, we substitute the expressions for $f$ and $N$: $Mg \le \mu (M \omega^2 R)$. Canceling $M$ from both sides gives $g \le \mu \omega^2 R$, which simplifies to $\omega \ge \sqrt{\frac{g}{\mu R}}$. Thus, the minimum angular velocity is $\omega = \sqrt{\frac{g}{\mu R}}$.

First, convert the speeds from km/h to m/s:
$v_A = 72 \times \frac{5}{18} = 20 \text{ m/s}, \quad v_B = 36 \times \frac{5}{18} = 10 \text{ m/s}$
The relative velocity of car $A$ with respect to car $B$ is:
$v_{rel} = v_A - v_B = 20 - 10 = 10 \text{ m/s}$
The magnitude of angular momentum $L$ of car $A$ with respect to car $B$ is given by $L = m_A \cdot v_{rel} \cdot r$, where $r$ is the perpendicular distance ($10 \text{ m}$) between the parallel paths:
$L = 10^3 \text{ kg} \times 10 \text{ m/s} \times 10 \text{ m} = 10^5 \text{ J}\cdot\text{s}$

For a wheel rolling without slipping with center velocity $v_{cm}$ and angular velocity $\omega$, the condition is $v_{cm} = \omega R$.
The speed at the highest point is the sum of translational and rotational velocities: $v_{top} = v_{cm} + \omega R = 2v_{cm}$.
Given $v_{top} = 8 \text{ m/s}$, we find $v_{cm} = 4 \text{ m/s}$.
At the level of the center, the velocity is the vector sum of $v_{cm}$ (horizontal) and $v_{tan} = \omega R$ (vertical), which are perpendicular.
Thus, the speed at this point is $v = \sqrt{v_{cm}^2 + (\omega R)^2} = \sqrt{v_{cm}^2 + v_{cm}^2} = \sqrt{2} v_{cm}$.
Substituting $v_{cm} = 4 \text{ m/s}$, we get $v = 4\sqrt{2} \text{ m/s}$.

$\;V_B=\sqrt{V_A^{2-}2 g l (1-\;{{\sqrt{3}}}/{2}) }$ $\;=\sqrt{60+20 {\sqrt{3}}}$ $\;V_C=\sqrt{V_A^{2-}2 g (\;\frac{3 l}{2}) }$ $\;={\sqrt{100-60}}$ $\;={\sqrt{40}}$ $\;\;\frac{K_B}{K_C}= (\;\frac{V_B}{V_C}) ^2=\;{3+{\sqrt{3}}}/{2}$

$\;F=m \omega ^2 r_{ {\text{cm}}}$ $\;r_{ {\text{cm}}}=\;\frac{1}{2} {\text{m}}$ $\;\omega =\sqrt{\;\frac{2 F}{m}}$ $\;m=2 M$ $\;\omega =\sqrt{\;\frac{F}{M}}$ $\;\alpha =1$

$\;\omega =\;\frac{3}{\pi } \times 2 \pi =6 rad / s$ $\;R=L sin \;\theta$ $\;\;\text{ and }\; T=M {\sqrt{g^{2+}\omega ^4 R^2}}$ $\;\;\text{ Also, }\; T sin \;\theta =M \omega ^2 \cdot L sin \;\theta$ $\;\Rightarrow T=M(36) L$ $\;\Rightarrow T=36 ML$

$v=n {\sqrt{g R}}$ $\;v_0={\sqrt{v^{2+}2 g(2 R)}}$ $\;v_0={\sqrt{n^2 g R+4 g R}}$ $\; \therefore \;\; \;{k_{\;\text{bottom }\;}}/{k_{\;\text{top }\;}}=\;\frac{v_0^2}{v^2}=\;\frac{n^{2+}4}{n^2}$

$\;v_B=\sqrt{v_0^{2-}2 g (\;\frac{I}{2}) }$ $\;v_B={\sqrt{4 g l}}$ $\;v_C={\sqrt{v_0^{2-}3 g l}}={\sqrt{2 g l}}$ $\;\;\frac{K_B}{K_C}=\;\frac{v_B^2}{v_C^2}=2$

Given maximum possible speed $=v_0$For this So, $N=m g \cos \theta +\;\frac{m v_0^2}{r} sin \;\theta$ $f_r=\mu m g \cos \theta +\;\frac{\mu m v_0^2}{r} sin \;\theta$And $\;\frac{m v_0^2}{r} \cos \theta =m g sin \;\theta +f_r$ $\;\Rightarrow \;\frac{m v_0^2}{r} \cos \theta -m g sin \;\theta =\mu (m g \cos \theta +\;\frac{m v_0^2}{r} sin \;\theta )$ $\;\Rightarrow (v_0^{2-}g r \tan \theta ) =\mu (v_0^2 \tan \theta +g r)$ $\Rightarrow \mu =\;\frac{v_0^{2-}g r \tan \theta }{g r+v_0^2 \tan \theta }$

$\;\;\text{ Distance }\;=3 \times \pi r=3 \pi r$ $\;\;\text{ Displacement }\;=2 r$

The time period $T$ of a charged particle moving in a uniform magnetic field $B$ is given by $T = \frac{2\pi m}{qB}$.
Given: $q = 1.6 \, \mu\text{C} = 1.6 \times 10^{-6} \, \text{C}$, $m = 16 \, \mu\text{g} = 16 \times 10^{-9} \, \text{kg}$, $B = 6.28 \, \text{T}$, and $\pi = 3.14$.
Substituting these values: $T = \frac{2 \times 3.14 \times (16 \times 10^{-9})}{(1.6 \times 10^{-6}) \times 6.28}$.
Since $2 \times 3.14 = 6.28$, the expression simplifies to $T = \frac{6.28 \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28}$.
Dividing the terms: $T = \frac{16 \times 10^{-9}}{1.6 \times 10^{-6}} = 10 \times 10^{-3} = 0.01 \, \text{s}$.
The time required for the particle to return to its original location is $0.01 \, \text{s}$.

[CORRECT_OPTION: N/A]

$\;{| \vec{a}_c |}={| \vec{a}_t |}$ $\;\;\frac{v^2}{r}=\;\frac{d v}{d t}$ $\;\Rightarrow ∫_{4}^{v} \;\frac{d v}{v^2}=∫_{0}^{t} \;\frac{d t}{r}$ $\;\Rightarrow [\;\frac{-1}{v}] _4^v=\;\frac{t}{r}$ $\;\Rightarrow \;\frac{-1}{v}+\;\frac{1}{4}=2 t$ $\;\Rightarrow {\text{v}}=\;{4}/{1-8 {\text{t}}}=\;{ {ds}}/{ {\text{dt}}}$ $\;4 ∫_{0}^ { {\text{t}}} \;{ {\text{dt}}}/{1-8 {\text{t}}}=∫_{0}^{ {\ s}} {ds}$ $\;( {\text{r}}=0.5 {\text{m}}$ $\; {\ s}=2 \pi {\text{r}}=\pi )$ $\;4 \times \;{[\ln (1-8 {\text{t}})]_0^{ {\text{t}}}}/{-8}=\pi$ $\;\ln (1-8 {\text{t}})=-2 \pi$ $\;1-8 {\text{t}}= {\text{e}}^{-2 \pi }$ $\; {\text{t}}= (1- {\text{e}}^{-2 \pi }) \;\frac{1}{8} {\ s}$So, $\alpha =8$

$\therefore$ Displacement $= {\text{R}} {\sqrt{2}}=2 {\sqrt{2}}={\sqrt{8}} {\text{km}}$

Given : ${\text{R}}=9 {\text{m}}$,120 revolution in $3 {\min}$ $\omega =\;{120 {\text{Rev}} .}/{3 {\min} .}=\;{120 \times 2 \pi {\text{rad}}}/{3 \times 60 {\ sec}}=\;\frac{4 \pi }{3} {\text{rad}} / {\ s}$ ${\text{a}}_{\;\text{centripetal }\;}=\omega ^2 {\text{R}}= (\;\frac{4 \pi }{3}) ^2 \times 9=16 \pi ^2 {\text{m}} / {\ s}^2$

$\; {\text{N}}= {\text{mg}}$ $\; {\text{f}}= {\text{m}} \omega ^2 {\text{r}}$ $\; {\text{f}}=\mu {\text{N}}$ $\;\mu {\text{mg}}= {\text{mr}} \omega ^2$ $\;\omega =\sqrt{\;\frac{\mu g}{r}}$

Given that $\; {\text{m}}=900 {\text{gm}}=\;\frac{900}{1000} {\text{kg}}=\;\frac{9}{10} {\text{kg}}$ $\; {\text{r}}=1 {\text{m}}$ $\;\omega =\;{2 \pi {\text{N}}}/{60}=\;\frac{2 \pi (10)}{60}=\;\frac{\pi }{3} {\text{rad}} / {\ sec}$ $\; {\text{T}}- {\text{mg}}= {\text{mr}} \omega ^2$ $\; {\text{T}}= {\text{mg}}+ {\text{mr}} \omega ^2$ $\;=\;\frac{9}{10} \times 9.8+\;\frac{9}{10} \times 1 (\;\frac{\pi }{3}) ^2$ $\;=8.82+\;\frac{9}{10} \times \;\frac{\pi ^2}{9}$ $\;=8.82+0.98$ $\;=9.80 {\text{N}}$

$\; {\text{T}}= {\text{m}} \omega ^2 ℓ$ $\;400=0.5 \omega ^2 \times 0.5$ $\;\omega =40 {\text{rad}} / {\ s} .$

$\;{2 \pi {\text{R}}}/{ {\text{T}}}= {\text{V}}$Maximum height ${\text{H}}=\;{ {\text{v}}^2 sin \;^2 \theta }/{2 {\text{g}}}$ $\;4 {\text{R}}=\;{4 \pi ^2 {\text{R}}^2}/{ {\text{T}}^2 2 {\text{g}}} sin \;^2 \theta$ $\; sin \;\theta =\sqrt{\;{2 {\text{gT}}^2}/{\pi ^2 {\text{R}}}}$ $\;\theta = sin \;^{-1} (\;{2 {\text{gT}}^2}/{\pi ^2 {\text{R}}}) ^{\;\frac{1}{2}}$

$\; {\text{x}}_{\min }=\pi \times {\text{r}}_{\;\text{min }\;}$ $\;=\pi \times \;\frac{60}{100} {\text{m}} \;\text{. }\;$ $\; {\text{x}}_{\;\text{second }\;}=30 \times 2 \pi \times {\text{r}}_{\;\text{second }\;}$ $\;=30 \times 2 \pi \times \;\frac{75}{100}$ $\; {\text{x}}= {\text{x}}_{\;\text{second }\;}- {\text{x}}_{\min }$ $\;=139.4 {\text{m}}$ $\;$