Circular Motion

Let the point where the tension becomes zero be P .

At any point in vertical circular motion, the net radial force provides the necessary centripetal force. The forces acting on the particle at point P are :

• 1. Tension (T) is acting towards the centre.

• 2. Weight (mg) is acting vertically downwards.

The component of weight acting away along the radius is $\mathrm{mg} \sin \left(30^{\circ}\right)$. Thus, the equation of motion is :

$ \mathrm{T}+\mathrm{mg} \sin \left(30^{\circ}\right)=\frac{\mathrm{m} \mathrm{v}_{\mathrm{P}}^2}{\mathrm{r}} $

The problem states that at this point, the tension $\mathrm{T}=0$. Substituting this :

$ 0+m g \sin \left(30^{\circ}\right)=\frac{m v_p^2}{r} $

$\Rightarrow $ $\mathrm{mg}\left(\frac{1}{2}\right)=\frac{\mathrm{mv}_{\mathrm{p}}^2}{\mathrm{r}}$

$\Rightarrow $ $ \mathrm{v}_{\mathrm{p}}^2=\frac{\mathrm{gr}}{2} \ldots (i)$

We need to find the vertical height of point P relative to the bottom point A .

From bottom A to the centre, the height is r . And from the centre to point P , the vertical displacement is $\mathrm{r} \sin \left(30^{\circ}\right)$.

Total height h from A to $\mathrm{P}=\mathrm{h}=\mathrm{r}+\mathrm{r} \sin \left(30^{\circ}\right)=\mathrm{r}+\frac{\mathrm{r}}{2}=\frac{3 \mathrm{r}}{2}$

Applying law of conservation of energy between points A and P ,

Total Mechanical Energy at $\mathrm{A}=$ Total Mechanical Energy at P

Let $\mathrm{v}_{\mathrm{A}}$ be the velocity at the bottom point. Taking point A as the reference for zero potential energy :

$ \frac{1}{2} m v_{\mathrm{A}}^2=\frac{1}{2} m v_{\mathrm{P}}^2+m g h $

$\Rightarrow $ $\frac{v_A^2}{2}=\frac{1}{2}\left(\frac{g r}{2}\right)+g\left(\frac{3 r}{2}\right)$

$\Rightarrow $ $\frac{\mathrm{v}_{\mathrm{A}}^2}{2}=\frac{\mathrm{gr}}{4}+\frac{3 \mathrm{gr}}{2}$

$\Rightarrow $ $\mathrm{v}_{\mathrm{A}}^2=\frac{\mathrm{gr}}{2}+3 \mathrm{gr}=\frac{7}{2} \mathrm{gr}$

$\Rightarrow $ $ \mathrm{v}_{\mathrm{A}}=\sqrt{\frac{7}{2} \mathrm{gr}} $

The velocity at the bottom point A is $\sqrt{\frac{7}{2} \mathrm{gr}}$.

Hence, the correct option is (A).

When the drum rotates, the mass M experiences forces :

• 1. Weight $(\mathrm{Mg})$ acts vertically downwards.

• 2. Normal reaction ( N ) acts horizontally toward the centre of the drum. This provides the necessary centripetal force.

• 3. Frictional force (f) acts vertically upwards, opposing the tendency of the mass to slide down due to gravity.

For the body to remain rest with respect to the cylindrical surface, the vertical forces must be balanced :

$ \mathrm{f}=\mathrm{Mg} $

The horizontal normal force provides the centripetal acceleration ( $\mathrm{a}_{\mathrm{c}}=\mathrm{R} \omega^2$ ) :

$ N=M R \omega^2 $

The frictional force (f) is a self-adjusting force up to a maximum limit (limiting friction). For the body to stay at rest relative to the wall :

$ \mathrm{f} \leq \mathrm{f}_{\max } $

$\Rightarrow $ $\mathrm{f} \leq \mu \mathrm{N}$

Substituting the values of f and N into this inequality :

$ M g \leq \mu\left(M R \omega^2\right) $

$\Rightarrow $ $g \leq \mu R \omega^2$

$\Rightarrow $ $\omega^2 \geq \frac{g}{\mu R}$

$\Rightarrow $ $ \omega \geq \sqrt{\frac{g}{\mu R}} $

Therefore, the minimum angular speed is $\omega_{\min }=\sqrt{\frac{\mathrm{g}}{\mu \mathrm{R}}}$.

Hence, the correct option is (D).

Speed of car $A, v_A=72 \times \frac{5}{18}=20 \mathrm{~m} / \mathrm{s}$

Speed of car B, $\mathrm{v}_{\mathrm{B}}=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{s}$

Since both cars are moving in the same direction, the relative velocity of car $A$ with respect to $\operatorname{car} B\left(\vec{v}_{A B}\right)$ is :

$ \overrightarrow{\mathrm{v}}_{\mathrm{AB}}=\overrightarrow{\mathrm{v}}_{\mathrm{A}}-\overrightarrow{\mathrm{v}}_{\mathrm{B}} $

$\Rightarrow $ $\overrightarrow{\mathrm{v}}_{\mathrm{AB}}=20-10=10 \mathrm{~m} / \mathrm{s}$

Angular momentum of a point mass relative to an observer is defined as:

$ \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}} $

Where $\overrightarrow{\mathrm{r}}$ is the position vector and $\overrightarrow{\mathrm{p}}$ is the linear momentum ( $\mathrm{m} \overrightarrow{\mathrm{v}}$ ).

The perpendicular distance $\left(\mathrm{r}_{\perp}=\mathrm{d}\right)$ between the two parallel tracks:

$ \mathrm{L}=\mathrm{m} \cdot \mathrm{v}_{\text {relative }} \cdot \mathrm{r}_{\perp} $

Given values are,

• Mass $\mathrm{m}=10^3 \mathrm{~kg}$

• Relative speed $\mathrm{v}_{\mathrm{AB}}=10 \mathrm{~m} / \mathrm{s}$

• Perpendicular distance $\mathrm{r}_{\perp}=10 \mathrm{~m}$

Substituting the values into the formula :

$ L=\left(10^3\right) \times(10) \times(10) $

$\Rightarrow $ $ \mathrm{L}=10^5 \mathrm{~J} . \mathrm{s} $

Therefore, the angular momentum of the car A with respect to B is $10^5 \mathrm{~J}$. s.

Hence, the correct option is (B).

The speed of the car is given in $\mathrm{km} / \mathrm{h}$. We must convert it to $\mathrm{m} / \mathrm{s}$ to match the units of gravity and radius.

$ \mathrm{v}=54 \times \frac{5}{18}=3 \times 5=15 \mathrm{~m} / \mathrm{s} $

When the car takes a turn, from the perspective of an observer inside the car, the pendulum bob experiences gravity ( mg ) which is acting vertically downwards and centrifugal force ( $\mathrm{F}_{\mathrm{c}}$ ) a pseudo-force acting horizontally outwards due to the circular motion.

$ \mathrm{F}_{\mathrm{c}}=\frac{\mathrm{mv}^2}{\mathrm{r}} $

Where m is mass, v is velocity, and r is radius.

Let the string make an angle $\theta$ with the vertical. The tension T in the string can be resolved into two components:

The vertical component is $\mathrm{T} \cos \theta$

The horizontal component is $\mathrm{T} \sin \theta$

In equilibrium (relative to the car) :

$\mathrm{T} \cos \theta=\mathrm{mg} \tag{i}$ .....(i)

$\mathrm{T} \sin \theta=\frac{\mathrm{mv}^2}{\mathrm{r}} \ldots (ii)$

Dividing equation (ii) by the equation (i) :

$ \frac{\mathrm{T} \sin \theta}{\mathrm{~T} \cos \theta}=\frac{\mathrm{mv}^2 / \mathrm{r}}{\mathrm{mg}} $

$\Rightarrow $ $ \tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}} $

Substituting the given values, where speed of car is $\mathrm{v}=15 \mathrm{~m} / \mathrm{s}$, radius of turn is $\mathrm{r}=20 \mathrm{~m}$ and the acceleration due to gravity is $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$

$ \tan \theta=\frac{15^2}{20 \times 10} $

$\Rightarrow $ $\tan \theta=\frac{9}{8}=1.125$

$\Rightarrow $ $ \theta=\tan ^{-1}(1.125) $

Therefore, the angle made by the string with the vertical is $\tan ^{-1}(1.125)$.

Hence, the correct option is (C).

The potential energy at height $h$ converts into kinetic energy as the object slides down.

To move in a circle of radius $R$, the net force toward the center must be $\frac{m v^2}{R}$.

Let the bottom of the loop be the reference level (Potential Energy $=0$ ).

At release point, the gravitational potential energy is $\mathrm{PE}_1=\mathrm{mgh}$, and the kinetic energy is $\mathrm{KE}_1=0$.

So, total energy of the body at releasing point is $\mathrm{E}_1=\mathrm{PE}_1+\mathrm{KE}_1=\mathrm{mgh}+0=\mathrm{mgh}$

At the highest point of circle the height is 2 R (the diameter).

So, the gravitational potential energy is $\mathrm{PE}_2=\mathrm{mg}(2 \mathrm{R})$.

Let velocity at the top be $\mathrm{v}_2$. So, the kinetic energy of body at top is $\mathrm{KE}_2=\frac{1}{2} \mathrm{mv}_2^2$.

So, the total energy of the body at top of the circle is $\mathrm{E}_2=2 \mathrm{mgR}+\frac{1}{2} \mathrm{mv}_2^2$.

Using law of conservation of energy :

$ \mathrm{mgh}=2 \mathrm{mgR}+\frac{1}{2} \mathrm{mv}_2^2 $

$ g h=2 g R+\frac{v_2^2}{2} \Rightarrow v_2^2=2 g(h-2 R) \ldots (i)$

At the top of the loop, normal force ( N ) acts on the body from the track.

It is given that the body exerts a force of 3 mg on the plane, so by Newton's 3rd Law, the track exerts $\mathrm{N}=3 \mathrm{mg}$ on the body.

So, total centripetal force is $\mathrm{N}+\mathrm{mg}=\frac{\mathrm{mv}_2^2}{\mathrm{R}}$

Substituting N = 3mg:

$ 3 \mathrm{mg}+\mathrm{mg}=\frac{\mathrm{mv}_2^2}{\mathrm{R}} \Rightarrow 4 \mathrm{mg}=\frac{\mathrm{mv}_2^2}{\mathrm{R}} $

$ \mathrm{v}_2^2=4 \mathrm{gR} \ldots \text { (ii) } $

From both the equations,

$ 2 \mathrm{~g}(\mathrm{~h}-2 \mathrm{R})=4 \mathrm{gR} $

$\Rightarrow $ $h-2 R=2 R$

$\Rightarrow $ $ h=4 R=\alpha R \Rightarrow \alpha=4 $

Therefore, the value of $\alpha$ is 4 . Thus, the correct option is (B).

For the mass to remain stuck to the inner wall without falling, the upward frictional force must balance its weight.

Step 1: Forces acting on the mass

• Weight downward:

  $mg$

• Normal reaction by the wall toward the centre:

  $N$

• Friction upward:

  $f$

At the minimum rotational speed, the friction is just enough to prevent slipping, so

$f_{\max} = \mu N = mg$

Step 2: Normal reaction provides centripetal force

Since the drum is rotating, the normal force provides the centripetal force:

$N = m\omega^2 r$

Step 3: Use the condition for minimum speed

So,

$\mu N = mg$

Substitute $N = m\omega^2 r$:

$\mu (m\omega^2 r) = mg$

Cancel $m$:

$\mu \omega^2 r = g$

Hence,

$\mu = \frac{g}{\omega^2 r}$

Step 4: Substitute values

Given:

• $g = 10\ \mathrm{m/s^2}$

• $\omega = 5\ \mathrm{rad/s}$

• $r = 4\ \mathrm{m}$

$\mu = \frac{10}{(5)^2 \cdot 4} = \frac{10}{25 \cdot 4} = \frac{10}{100} = 0.1$

Answer:

$\boxed{0.1}$

So, the correct option is A.

If $V_B=2 V$

Point A is instantaneous center of rotation

Given $V_B=8 \mathrm{~m} / \mathrm{s}$

$\begin{aligned} & \mathrm{V}=4 \mathrm{~m} / \mathrm{s} \\ & \mathrm{~V}_{\mathrm{P}}=\sqrt{2} \mathrm{v} \Rightarrow V_{\mathrm{p}}=4 \sqrt{2} \mathrm{~m} / \mathrm{s} \\ & \operatorname{correct}(1) \end{aligned}$

To determine the distance traveled and displacement of a sportsman running around a circular track, we begin by analyzing the path described as $ A \rightarrow B \rightarrow A \rightarrow B $.

Displacement:

The displacement is the straight line distance from the initial point to the final point. In this case, the sportsman completes one full circle (from $ A $ back to $ A $) and then continues to point $ B $ (which is opposite point $ A $ on the circle). Therefore, the displacement is the diameter of the circle:

$ \text{Displacement} = 2r $

Distance:

The distance is the total path length traveled by the sportsman. Starting at $ A $, he completes one full circle back to $ A $, which has a circumference of:

$ 2 \pi r $

He then runs half the circle from $ A $ to $ B $, which is another:

$ \frac{1}{2} \times 2 \pi r = \pi r $

Thus, the total distance is the circumference of the full circle plus half of another circle:

$ \text{Distance} = 2 \pi r + \pi r = 3 \pi r $

In summary, the distance traveled by the sportsman is $ 3\pi r $ and the displacement is $ 2r $.

$\begin{aligned} & \mathrm{V}_{\text {Top }}=\sqrt{\mathrm{n}^2 \mathrm{gR}} \\ & \mathrm{~V}_{\text {Botom }}=\sqrt{\mathrm{n}^2 \mathrm{gR}+4 \mathrm{gR}} \\ & \text { Ratio }=\frac{\mathrm{n}^2+4}{\mathrm{n}^2} \end{aligned}$

$ N \cos\theta - \mu N \sin\theta = mg $

$ N \sin\theta + \mu N \cos\theta = \frac{mv_0^2}{r} $

Dividing the second equation by the first gives

$ \frac{v_0^2}{r} = \frac{g\left(\sin\theta + \mu \cos\theta\right)}{\cos\theta - \mu \sin\theta}. $

Multiplying both sides by $\cos\theta - \mu\sin\theta$ leads to

$ \frac{v_0^2}{r}(\cos\theta - \mu \sin\theta) = g(\sin\theta + \mu \cos\theta). $

Expanding and grouping the terms with $\mu$:

$ \frac{v_0^2}{r}\cos\theta - g\sin\theta = \mu\left(\frac{v_0^2}{r}\sin\theta + g\cos\theta\right). $

Thus, solving for $\mu$:

$ \mu = \frac{\frac{v_0^2}{r}\cos\theta - g\sin\theta}{\frac{v_0^2}{r}\sin\theta + g\cos\theta}. $

Multiplying the numerator and the denominator by $r$ gives

$ \mu = \frac{v_0^2\cos\theta - rg\sin\theta}{v_0^2\sin\theta + rg\cos\theta}. $

Dividing both the numerator and denominator by $\cos\theta$ (assuming $\cos\theta \neq 0$):

$ \mu = \frac{v_0^2 - rg\tan\theta}{v_0^2\tan\theta + rg}. $

This result matches the expression

$ \mu=\frac{v_0^2 - rg\,\tan\theta}{rg + v_0^2\,\tan\theta}. $

Thus, the correct answer is:

Option C

Step 1: Use Energy Conservation at Points A and B

At point A, the kinetic energy is from velocity and the potential energy is set at reference (usually zero). At point B, both kinetic and potential energy are present.

The conservation of energy formula between A and B is: $ \frac{1}{2} m \times 100 + 0 = \frac{1}{2} m V_B^2 + mg \left(R - \frac{R \sqrt{3}}{2}\right) $

Step 2: Use Given Values

Given: $ R = 2 $ m, $ g = 10\,\mathrm{m/s^2} $, and initial velocity at A is $ 10\,\mathrm{m/s} $, so $ (10)^2 = 100 $.

Substitute these into the equation: $ 100 = V_B^2 + 2gR\left(1 - \frac{\sqrt{3}}{2}\right) $ Here, $ m $ cancels out because it is on both sides.

Step 3: Simplify for $ V_B^2 $

Calculate $ 2gR = 2 \times 10 \times 2 = 40 $. So, $ 100 = V_B^2 + 40 \left(1 - \frac{\sqrt{3}}{2}\right) $

$ 40(1 - \frac{\sqrt{3}}{2}) = 40 - 20\sqrt{3} $

Rearrange to solve for $ V_B^2 $: $ V_B^2 = 100 - 40 + 20\sqrt{3} $ $ V_B^2 = 60 + 20\sqrt{3} $

So, kinetic energy at point B: $ K.E._B = \frac{1}{2} m V_B^2 = \frac{m}{2}(60 + 20\sqrt{3}) $

Step 4: Use Energy Conservation at Points A and C

At point C, use conservation of energy again: $ \frac{1}{2} m (10)^2 = \frac{1}{2} m V_C^2 + mg \left(\frac{3R}{2}\right) $

The potential energy term is $ mg \times \frac{3R}{2} $. Substitute values: $ 100 = V_C^2 + 2gR \times \frac{3}{2} $ $ 2gR = 40 $, so $ 40 \times \frac{3}{2} = 60 $

So: $ 100 = V_C^2 + 60 $ $ V_C^2 = 100 - 60 = 40 $

So, kinetic energy at point C: $ K.E._C = \frac{1}{2} m V_C^2 = \frac{1}{2} m (40) $

Step 5: Find the Ratio

The ratio of kinetic energies at B and C is: $ \frac{K.E._B}{K.E._C} = \frac{\frac{1}{2} m (60 + 20\sqrt{3})}{\frac{1}{2} m (40)} = \frac{60 + 20\sqrt{3}}{40} = \frac{3}{2} + \frac{\sqrt{3}}{2} = \frac{3 + \sqrt{3}}{2} $

As the string becomes slack when the bob reaches at the point D so tension is zero at this point. And the bob just completes the circle.

We know, to complete a vertical loop of radius R, the minimum velocity required at the bottom of loop is $\sqrt{5gR}$.

So, ${V_0} = \sqrt {5gl} = {V_A}$ (as R = $l$)

So, using energy conservation,

${1 \over 2}mv_A^2 = {1 \over 2}mv_B^2 + mgh$

$ \Rightarrow {1 \over 2}m(5gl) = {1 \over 2}mv_B^2 + mg\left( {{l \over 2}} \right)$

$ \Rightarrow K{E_B} = {{mgl} \over 2}(5 - 1) = 2mgl$

For point C,

${1 \over 2}mv_A^2 = {1 \over 2}mv_c^2 + mg\left( {l + {l \over 2}} \right)$

$ \Rightarrow {5 \over 2}mgl = K{E_C} + {{3mgl} \over 2}$

$K{E_C} = {{mgl} \over 2}(5 - 3) = mgl$

So, ${{K{E_B}} \over {K{E_C}}} = {{2mgl} \over {mgl}} = 2$

To determine the distance traveled by the tips of the second hand and the minute hand, we need to calculate the circumference of the circles they make. Let's start by calculating the distances traveled by both hands over a period of 30 minutes.

First, the circumference formula is given by:

$C = 2\pi r$

For the second hand:

The length of the second hand is $75 \space \text{cm}$. The second hand completes one full revolution every 60 seconds, so in 1 minute, the second hand travels:

$ \text{Distance per minute} = 2\pi \times 75 \text{ cm} $

In 30 minutes, the second hand will travel:

$ \text{Total distance traveled by second hand} = 30 \times 2\pi \times 75 \text{ cm} $

Now, let's calculate it with $\pi = 3.14$:

$ \text{Total distance traveled by second hand} = 30 \times 2 \times 3.14 \times 75 \text{ cm} $

$ \text{Total distance traveled by second hand} = 30 \times 471 \text{ cm} = 14130 \text{ cm} $

For the minute hand:

The length of the minute hand is $60 \text{ cm}$. The minute hand completes one full revolution every 60 minutes, so in 30 minutes, the minute hand travels:

$ \text{Total distance traveled by minute hand} = 0.5 \times 2\pi \times 60 \text{ cm} $

Again, let's calculate it with $\pi = 3.14$:

$ \text{Total distance traveled by minute hand} = 0.5 \times 2 \times 3.14 \times 60 \text{ cm} $

$ \text{Total distance traveled by minute hand} = 0.5 \times 376.8 \text{ cm} = 188.4 \text{ cm} $

The difference in distance $x$ is:

$ x = 14130 \text{ cm} - 1884 \text{ cm} $

$ x = 13941.6 \text{ cm} $

Convert this distance into meters:

$ x = 13941.6 \text{ cm} \times \frac{1 \text{ meter}}{100 \text{ cm}} $

$ x \approx 139.4 \text{ meters} $

Thus, the value of $x$ in meters is nearly 139.4 meters, making the correct option:

Option C

When a car takes a turn on a banked road, the forces involved are the gravitational force, the normal force from the surface, and frictional force (if any). The net force provides the necessary centripetal force for the circular motion. The angle of banking and static friction contribute to the maximum speed the car can achieve without slipping.

The forces acting on the car are as follows:

1. The normal force ($N$) acts perpendicular to the surface of the road.

2. Gravitational force ($mg$) acts downward.

3. Frictional force ($f$), which can provide additional centripetal force if needed. It acts parallel to the surface of the road, towards the center of the circle.

The normal force and the gravitational force components can be resolved into two directions: perpendicular and parallel to the road surface. The maximum speed is achieved when all available forces (normal, frictional) are utilized to provide the necessary centripetal force ($F_c$) without slipping.

The centripetal force required for circular motion is given by:

$F_c = \frac{mv^2}{r}$

Where:

• $m$ = mass of the car = $800 kg$
• $v$ = speed of the car
• $r$ = radius of the turn = $300 m$

On a banked curve, the maximum velocity can be calculated using the formula:

$v = \sqrt{rg(\tan\theta + \mu)\Big/\big(1-\mu\tan\theta)}$

Where:

• $\theta$ = angle of banking = $30^{\circ}$
• $\mu$ = coefficient of static friction = $0.2$
• $g$ = acceleration due to gravity = $10 m/s^2$

First, calculate the tangent of the angle:

$\tan30^{\circ} = \frac{1}{\sqrt{3}} = \frac{1}{1.73}$

Substitute all the values into the formula:

$v = \sqrt{300 \times 10 \times \left(\frac{1}{1.73} + 0.2\right)\Big/\big(1 - 0.2 \times \frac{1}{1.73}\big)}$

$v = \sqrt{3000 \times \left(\frac{1}{1.73} + 0.2\right)\Big/\big(1 - \frac{0.2}{1.73}\big)}$

$v = \sqrt{3000 \times \frac{1 + 1.73 \times 0.2}{1.73 - 0.2}}$

$v = \sqrt{3000 \times \frac{1 + 0.346}{1.73 - 0.2}}$

$v = \sqrt{3000 \times \frac{1.346}{1.53}}$

$v = \sqrt{3000 \times 0.8797}$

$v = \sqrt{2639.1} \approx 51.37 \, m/s$

Hence, the closest option to the calculated maximum speed without slipping is:

Option A: 51.4 m/s

First, let's calculate the centripetal acceleration experienced by the monkey while the man does cycling smoothly on a circular track. The formula for centripetal acceleration ($a_c$) is given by:

$a_c = \frac{v^2}{r}$

where

• $v$ is the velocity of the monkey and the man cycling,


• $r$ is the radius of the circular track.

To find $v$, we first need to find the circumference of the circle, which is given by:

$C = 2\pi r$

Given the radius $r = 9 \, \mathrm{m}$, the circumference $C$ is:

$C = 2\pi \times 9 = 18\pi \, \mathrm{m}$

Then, we determine the total distance travelled by calculating how many times they complete the circle in the given time. With 120 revolutions in 3 minutes (180 seconds), the total distance $D$ travelled is:

$D = 120 \times C = 120 \times 18\pi = 2160\pi \, \mathrm{m}$

To find the speed $v$, which is distance over time, we divide the total distance by the total time in seconds:

$v = \frac{D}{t} = \frac{2160\pi}{180} = 12\pi \, \mathrm{m/s}$

Now, using the formula for centripetal acceleration $a_c = \frac{v^2}{r}$, we can plug in the values:

$a_c = \frac{(12\pi)^2}{9} = \frac{144\pi^2}{9} = 16\pi^2 \mathrm{~m/s}^2$

Therefore, the magnitude of the centripetal acceleration of the monkey is $16\pi^2 \, \mathrm{m/s}^2$, which corresponds to Option B.

$\begin{aligned} \text { Displacement } & =\sqrt{2} R \\ & =\sqrt{8} \mathrm{~km} \end{aligned}$

To find the maximum possible angular velocity (ω) of the ball, we need to consider the maximum tension the string can bear without breaking. This tension provides the centripetal force needed to keep the ball in a circular path.

The centripetal force (F_c) required for circular motion is given by the formula:

$ F_{c} = m r \omega^2 $

Where:

m = mass of the ball (0.5 kg)

r = radius of the circle (0.5 m, since 50 cm = 0.5 m)

ω = angular velocity in rad/s

The maximum tension the string can bear is also the maximum centripetal force (F_c,max) that can be provided by the string, which is 400 N.

Now we can set up the equation with the given values:

$ 400 = 0.5 \times 0.5 \times \omega^2 $

$ \Rightarrow $ $ 400 = 0.25 \times \omega^2 $

$ \Rightarrow $ $ \omega^2 = \frac{400}{0.25} $

$ \Rightarrow $ $ \omega^2 = 1600 $

$ \Rightarrow $ $ \omega = \sqrt{1600} $

$ \Rightarrow $ $ \omega = 40 \, \mathrm{rad/s} $

Therefore, the maximum possible angular velocity of the ball is 40 rad/s, which corresponds to Option C.

To solve for the angle of projection $\theta$, we will first establish the relationship between the variables given and then derive the formula using kinematics.

The time $T$ for one revolution at speed $v$ in a circle of radius $R$ is related to the circumference of the circle by the formula:

$ v = \frac{2\pi R}{T} $

When the particle is projected with the same speed $v$ at an angle $\theta$ to the horizontal, its vertical component of velocity is given by $v_y=v\sin\theta$.

The maximum height $H$ reached by the projectile can be found from the kinematic equation:

$ H = \frac{v_y^2}{2g} = \frac{(v\sin\theta)^2}{2g} $

We are given that the maximum height attained $H$ is equal to $4R$, so:

$ 4R = \frac{(v\sin\theta)^2}{2g} $

Substitute $v$ from the first equation into the second one:

$ 4R = \frac{((\frac{2\pi R}{T})\sin\theta)^2}{2g} $

$ 4R = \frac{(2\pi R\sin\theta)^2}{2gT^2} $

$ 4R = \frac{4\pi^2 R^2 \sin^2\theta}{2gT^2} $

$ 2gT^2 = \pi^2 R \sin^2\theta $

Now solve for $\sin\theta$:

$ \sin\theta =\left[\frac{2gT^2}{\pi^2 R}\right]^{1/2} $

Finally, to solve for $\theta$, take the inverse sine of both sides:

$ \theta = \sin^{-1}\left(\left[\frac{2gT^2}{\pi^2 R}\right]^{1/2}\right) $

Therefore, the correct answer is:

Option A

$ \sin^{-1}\left[\frac{2gT^2}{\pi^2 R}\right]^{1/2} $

When the coin is on the disc and the disc starts rotating, centrifugal force acts on the coin, trying to push it away from the center. The friction between the coin and the disc opposes this motion. For the coin to not slip, the frictional force must be equal to the centrifugal force acting on the coin.

The normal force (N) acting on the coin is equal to the weight of the coin, which can be represented as:

$N = mg$

where $m$ is the mass of the coin, and $g$ is the acceleration due to gravity.

The centrifugal force ($f$) that acts on the coin due to the rotation of the disc is given by:

$f = m\omega^2r$

where $\omega$ is the angular velocity, and $r$ is the distance of the coin from the center of the disc.

Friction force (f) is also given by the formula:

$f = \mu N$

where $\mu$ is the coefficient of static friction between the coin and the disc.

For the coin to not slip, the centrifugal force must be equal to the frictional force, hence:

$\mu mg = m\omega^2r$

Dividing both sides by $mr$, we get:

$\omega = \sqrt{\frac{\mu g}{r}}$

This equation shows that the maximum angular velocity ($\omega$) that can be given to the disc to prevent the coin from slipping off depends on the coefficient of friction ($\mu$), the acceleration due to gravity ($g$), and the distance of the coin from the center ($r$).

Given that

$\begin{aligned} & \mathrm{m}=900 \mathrm{~gm}=\frac{900}{1000} \mathrm{~kg}=\frac{9}{10} \mathrm{~kg} \\ & \mathrm{r}=1 \mathrm{~m} \\ & \omega=\frac{2 \pi \mathrm{N}}{60}=\frac{2 \pi(10)}{60}=\frac{\pi}{3} \mathrm{rad} / \mathrm{sec} \\ & \mathrm{T}-\mathrm{mg}=\mathrm{mr} \omega^2 \\ & \mathrm{~T}=\mathrm{mg}+\mathrm{mr} \omega^2 \\ & =\frac{9}{10} \times 9.8+\frac{9}{10} \times 1\left(\frac{\pi}{3}\right)^2 \\ & =8.82+\frac{9}{10} \times \frac{\pi^2}{9} \\ & =8.82+0.98 \\ & =9.80 \mathrm{~N} \end{aligned}$

Given $\mathrm{m}_1=\mathrm{m}_2$

$\text { and } \frac{r_1}{r_2}=\frac{3}{4}$

As centripetal force $\mathrm{F}=\frac{\mathrm{mv}^2}{\mathrm{r}}$

In order to have constant (same in this question) centripetal force

$\begin{aligned} & \mathrm{F}_1=\mathrm{F}_2 \\ & \frac{\mathrm{m}_1 \mathrm{v}_1^2}{\mathrm{r}_1}=\frac{\mathrm{m}_2 \mathrm{v}_2^2}{\mathrm{r}_2} \\ & \Rightarrow \frac{\mathrm{v}_1}{\mathrm{v}_2}=\sqrt{\frac{\mathrm{r}_1}{\mathrm{r}_2}}=\frac{\sqrt{3}}{2} \end{aligned}$

$\tan \theta=\frac{v^2}{R g}=\frac{12 \times 12}{10 \times 400}$

$\begin{aligned} & \tan \theta=\frac{\mathrm{h}}{1.5} \\\\ & \Rightarrow \frac{\mathrm{h}}{1.5}=\frac{144}{4000} \\\\ & \mathrm{~h}=5.4 \mathrm{~cm} \end{aligned}$

To calculate the centrifugal force acting on the child, we need to find the angular velocity of the merry-go-round and then apply the formula for centrifugal force.

The merry-go-round makes 1 rotation in 3.14 seconds, so its angular velocity ($\omega$) can be calculated as:

$\omega = \frac{2\pi}{T}$, where $T$ is the time period for one rotation.

$\omega = \frac{2\pi}{3.14} = 2 \, \text{radians/s}$

Now, the formula for centrifugal force ($F$) is:

$F = m \cdot r \cdot \omega^2$, where $m$ is the mass of the child, $r$ is the radius of the merry-go-round, and $\omega$ is the angular velocity.

$F = 5\,\text{kg} \cdot 2\,\text{m} \cdot (2\,\text{radians/s})^2 = 5\,\text{kg} \cdot 2\,\text{m} \cdot 4\,(\text{radians/s})^2$

$F = 40\,\text{N}$

Therefore, the centrifugal force on the child is $40\,\text{N}$.

In this problem, the spring is stretched due to the circular motion of the block, so the effective radius of the circular motion becomes the natural length of the spring plus the extension in the spring $(r=0.2+x)$.

The spring force, which is also the centripetal force, is given by $F_c=Kx=m\omega^2 r$, where $K$ is the spring constant, $x$ is the extension of the spring, $m$ is the mass of the block, $\omega$ is the angular velocity, and $r$ is the radius of the circular path.

Plugging in the values and solving for $x$ and $Kx$ gives the extension in the spring as $x = 0.1$ m and the tension in the spring (which is the spring force) as $Kx=7.5 \times 0.1 = 0.75$ N.

So, the correct answer is $0.75$ N.

$\overrightarrow v = 4\widehat j$ (m/s)

$a = {{{v^2}} \over R} = {{16} \over 2} = 8$ m/s$^2$

$\overrightarrow a = 8\left( {m/{s^2}} \right)\left( {\widehat i} \right)$

In car’s frame, FBD of bob

where $a_{P}=$ Pseudoforce or centrifugal force

$\theta=\tan ^{-1}\left(\frac{a_{P}}{g}\right)=\tan ^{-1}\left(\frac{v^{2}}{R g}\right)=\tan ^{-1}\left(\frac{400}{40 \times 10}\right)$ $=45^{\circ}$

Block is moving in the circular path, that means there should be a force which is keeping the block in contact with the wall here that force is normal force, which is towards the center of the circular path. And we know this force is equal to ${{m{v^2}} \over r}$.

$ \therefore $ $N = {{m{v^2}} \over r}$

$ \Rightarrow $ N $ \propto $ v²

$\Rightarrow$ The graph given in option A suits the best for the above relation.

Distance travelled = 60 m

$\Rightarrow$ Angle covered = 135$^\circ$

Displacement $ = 2R\sin \left( {{{135^\circ } \over 2}} \right)$

$ = 2\left( {{{60} \over {135}} \times {{180} \over \pi }} \right){\left[ {{{1 - \cos (135^\circ )} \over 2}} \right]^{1/2}}$

$ = 2\left( {{{80} \over \pi }} \right){(0.85)^{1/2}}$

$ \approx 47$ m

${a_r} = {k^2}r{t^2} = {{{v^2}} \over r}$

$ \Rightarrow {v^2} = {k^2}{r^2}{t^2}$ or $v = krt$

and ${{d|v|} \over {dt}} = kr$

$ \Rightarrow {a_t} = kr$

$ \Rightarrow |\overline F \,.\,\overline v | = (mkr)(krt)$

$ = m{k^2}{r^2}t = $ power delivered

$\overrightarrow v = \sqrt {{u^2} - 2gL} \widehat j$

$\overrightarrow u = u\widehat i$

$\therefore$ $\left| {\overrightarrow v - \overrightarrow u } \right| = \sqrt {({u^2} - 2gL) + {u^2}} $

$ = \sqrt {2u - 2gL} $

$\therefore$ $x = 2$

At Q ;

The force that prevents the beaker from sliding is the static friction force. For an object in uniform circular motion, the net force acting on the object (which is the friction force in this case) is equal to the centripetal force.

The static friction force is given by the normal force (which is equal to the weight of the object) multiplied by the coefficient of static friction, which is :

$F_{\text{friction}} = \mu mg.$

The centripetal force needed to keep an object moving in a circle of radius R at angular velocity ω is given by :

$F_{\text{centripetal}} = mR\omega^2.$

For the beaker not to slide off, the static friction force must be at least as large as the centripetal force. Therefore, we have :

$\mu mg \geq mR\omega^2.$

After canceling the mass m from both sides, we get :

$R \leq \frac{\mu g}{\omega^2}.$

So, the correct answer is Option B :

$R \leq \frac{\mu g}{\omega^2}.$

As the particle in uniform circular motion experiences only centripetal acceleration of magnitude $\omega$²R or ${{{v^2}} \over R}$ directed towards centre so from diagram,

$\overrightarrow a = {{{v^2}} \over R}\cos \theta ( - \widehat i) + {{{v^2}} \over R}\sin ( - \widehat j)$

At any $\theta :T - mg\cos \theta = {{m{v^2}} \over R}$

$ \Rightarrow T = mg\cos \theta + {{m{v^2}} \over R}$

Since v is constant,

$\Rightarrow$ T will be minimum when cos$\theta$ is minimum.

$\Rightarrow$ $\theta$ = 180$^\circ$ corresponds to T_minimum.

${\theta _1} = {1 \over 2}\alpha (2 \times 1 - 1) = 5$ rad

$\Rightarrow$ $\alpha$ = 10 rad/sec²

So ${\theta _2} = {1 \over 2} \times \alpha (2 \times 2 - 1) = 15$ rad