Circular Motion Practice Questions

The bead is at rest with respect to the rotating ring. In the frame of the ring, the bead is in equilibrium under the action of:

1. Weight ($mg$): Acting vertically downwards.
2. Normal Reaction ($N$): Acting towards the center of the ring.
3. Centrifugal Force ($mr\omega^2$): Acting horizontally outwards.

1. Vertical Equilibrium:

The vertical component of the normal force balances the weight:
$N \sin 60^\circ = mg \quad \dots \text{(i)}$

2. Horizontal Equilibrium:

The horizontal component of the normal force provides the required centripetal acceleration (or balances the centrifugal force):
$N \cos 60^\circ = m \left( \frac{r}{2} \right) \omega^2 \quad \dots \text{(ii)}$
(Note: The radius of the bead's circular path is $r \cos 60^\circ = r/2$)

3. Solving for $\omega^2$:

Divide equation (i) by equation (ii):
$\frac{N \sin 60^\circ}{N \cos 60^\circ} = \frac{mg}{m \left( \frac{r}{2} \right) \omega^2}$
$\tan 60^\circ = \frac{2g}{r\omega^2}$

Since $\tan 60^\circ = \sqrt{3}$:
$\sqrt{3} = \frac{2g}{r\omega^2}$
$\omega^2 = \frac{2g}{r\sqrt{3}}$

Correct Option: (A)

$|\Delta v^{-}|$ = $\sqrt{v_1^{2+}v_2^{2+}2v_1v_1cos(\pi -\theta )}$ = 2v sin $\theta /2$ since $[|v_1^{-}|=|v_2^{-}|]$ = (2 × 10) × sin (30°) = 10 m/s

$v_x$ = 10 cos 60° = 5 m/s $v_y$ = 10 cos 30° = $5\sqrt3$ m/s velocity after t = 1 sec. $x_x$ = 5 m/s $v_y$ = $|(5\sqrt3-10)|$ m/s = 10 - $5\sqrt3$ $a_n$ = $v^2/R$ \Rightarrow R = ${v_x^{2+}v_y^2}/a_n$ = $\frac{25+100+75-100\sqrt3}{10cos\theta }$ tan θ = ${10-5\sqrt3}/5$ = 2 - $\sqrt3$ \Rightarrow θ = 15° R = $\frac{100(2-\sqrt3)}{10cos15}$ = 2.8 m

The centripetal force required for the coin to remain at rest on the disc is provided by the static friction force, so $m \omega^2 r \le \mu mg$, which simplifies to $\mu \ge \frac{\omega^2 r}{g}$.
Given frequency $f = 3.5 \text{ rev/s}$, the angular velocity is $\omega = 2 \pi f = 2 \pi \times 3.5 = 7\pi \text{ rad/s}$.
The radius is $r = 1.25 \text{ cm} = 0.0125 \text{ m}$ and acceleration due to gravity $g = 10 \text{ m/s}^2$.
Substituting these values: $\mu = \frac{(7\pi)^2 \times 0.0125}{10} = \frac{49 \times \pi^2 \times 0.0125}{10}$.
Using the approximation $\pi^2 \approx 9.86$ (or $\approx 10$ for standard calculations), we get $\mu \approx \frac{49 \times 10 \times 0.0125}{10} = 49 \times 0.0125 \approx 0.6125$.
Rounding to the given options, we find $\mu = 0.6$.

For a conical pendulum, the forces acting on the bob are the tension $T$ and the weight $mg$. Resolving these forces gives the equations:

1. Horizontal (centripetal) component: $T \sin\theta = \frac{mv^2}{r}$
2. Vertical equilibrium: $T \cos\theta = mg$

Dividing the first equation by the second eliminates $T$:
$\tan\theta = \frac{v^2}{rg} \implies v = \sqrt{rg \tan\theta}$
Substituting the given values $r = 0.4 \text{ m}$, $g = 10 \text{ m/s}^2$, and $\theta = 45^\circ$:
$v = \sqrt{0.4 \times 10 \times \tan(45^\circ)} = \sqrt{4 \times 1} = 2 \text{ m/s}$

We know, $a=r w^2=r \times (\;\frac{2 \pi }{T}) ^2$Given, $T_1=T_2=T$ $\;a_1=r_1 \times (\;\frac{2 \pi }{T}) ^2$ $\;a_2=r_2 \times (\;\frac{2 \pi }{T}) ^2$ $\; \therefore \;\frac{a_1}{a_2}=\;\frac{r_1}{r_2}$

For a particle in uniform circular motion,$a_c=\;\frac{v^2}{R}$ towards the center of the circleFrom figure, $\vec{a}=a_c \cos \theta (-{i}^{∧})+a_c sin \;\theta (-{j}^{∧})$ $=\;\frac{-v^2}{R} \cos \theta {i}^{∧}-\;\frac{v^2}{R} sin \;\theta {j}^{∧}$

Given $s=t^{3+}5$ $\Rightarrow$ Speed, $v=\;\frac{d s}{d t}=3 t^2$ Tangential acceleration $a_t=\;\frac{d v}{d t}=6 t$Radial acceleration $a_c=\;\frac{v^2}{R}=\;\frac{9 t^4}{R}$ At $t=2 {\ s}, a_t=6 \times 2=12 {\text{m}} / {\ s}^2$ $a_c=\;\frac{9 \times 16}{20}=7.2 {\text{m}} / {\ s}^2$ $\therefore$ Net acceleration$\;={\sqrt{a_t^{2+}a_c^2}}$ $\;={\sqrt{(12)^{2+}(7.2)^2}}$ $\;={\sqrt{144+51.84}}$ $\;={\sqrt{195.84}}$ $\;=14 m / s^2$

Only option $(b)$ is false since acceleration vector acts along the radius of the circle or towards center of the circle for uniform circular motion and velocity vector always acts along the tangent of the circle.

For no skidding along curved track, The maximum velocity possible $v_{\max }=\sqrt{\mu r g}$ Here $\mu =0.6, r=150 m, g=9.8$ $\therefore v_{\max }=\sqrt{0.6 \times 150 \times 9.8} ≃ 30 \frac{\text{m}}{\text{s}}$