Circular Motion

In car’s frame, FBD of bob

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

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

Given,

Ratio of radius of two circular paths $=1: 2$

Ratio of masses $=1: 1$

Centripetal force is constant.

Centripetal force is given by

$ \begin{array}{lr} \begin{array}{lr} F_c=\frac{m v^2}{r} & \\ F_{c_1}=\frac{m_1 v_1^2}{r_1} & \text { [For 1 }{ }^{\text {st }} \text { particle ] } \\ F_{c_2}=\frac{m_2 v_2^2}{r_2} & \text { [For 2 }{ }^{\text {nd }} \text { particle ] } \\ F_{c_1}=F_{c_2} & \text { [given] } \\ \frac{m_1 v_1^2}{r_1}=\frac{m_2 v_2^2}{r_2} & \\ \frac{v_1^2}{v_2^2}=\frac{m_2 r_1}{m_1 r_2}\left[m_1=m_2, \frac{r_1}{r_2}=\frac{1}{2} \text { (given) }\right] & \\ \frac{v_1}{v_2}=\frac{1}{\sqrt{2}} & \end{array} \end{array} $

Hence, the ratio of speed is $1: \sqrt{2}$.

Velocity $=v$,

radius $=r$

Radial acceleration, $a_r=\frac{v^2}{r}$

Tangential acceleration, $a_l=a$

Total acceleration $=\sqrt{a_r^2 \times a_x^2}$

$ \left(\frac{v^2}{r}\right)^2=\sqrt{\frac{v^4}{r^2}+a^2} $

To understand how the centripetal acceleration changes when the angular speed of a particle moving in a circular path is doubled, let's start by defining the initial conditions:

Assume the initial angular speed of the particle is $\omega$.

The formula for centripetal acceleration is given by:

$ a_c = \omega^2 r $

Now, if the angular speed is doubled, the new angular speed becomes:

$ \omega^{\prime} = 2\omega $

Using the new angular speed, the revised centripetal acceleration is:

$ a_c^{\prime} = (\omega^{\prime})^2 r = (2\omega)^2 r = 4\omega^2 r $

Thus, the new centripetal acceleration is:

$ a_c^{\prime} = 4a_c $

This shows that when the angular speed is doubled, the centripetal acceleration becomes four times the initial centripetal acceleration. Therefore, the correct description is that the centripetal acceleration is 4 times the initial centripetal acceleration.

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

$T = m{\omega ^2}r$

$ \Rightarrow 80 = 0.1 \times {\left( {2\pi \times {K \over \pi } \times {1 \over {60}}} \right)^2} \times 2$

$ \Rightarrow {{800} \over 2} = {{{K^2}} \over {900}}$

$ \Rightarrow K = 30 \times 20 = 600$

For rotating bob, tension provides necessary centripetal force.

$ \Rightarrow \quad T=m r \omega^2 $

Here,

$ \begin{aligned} & T=72 \mathrm{~N}, m=250 \mathrm{~g}=0.25 \mathrm{~kg} \\ & r=50 \mathrm{~cm}=0.5 \mathrm{~m} \end{aligned} $

So, $72=0.25 \times 0.5 \times \omega^2$

$ \begin{array}{ll} \Rightarrow & m r \omega^2=69.5 \\ \Rightarrow & \omega^2=\frac{72}{0.25 \times 0.5}=576 \\ \Rightarrow & \omega=\sqrt{576} \approx 24 \mathrm{rad} / \mathrm{s} \end{array} $

Cyclist has a centripetal acceleration $a_c$ and a tangential acceleration $a_t$.

So, net acceleration of cyclist is

$ a=\sqrt{a_c^2+a_t^2} $

Here, $a_c=\frac{v^2}{r}$ and $a_t=\frac{d v}{d t}$

From given data we have,

$ a_c=\left(36 \times \frac{5}{18}\right)^2 / 50=2 \mathrm{~ms}^{-2} $

and $\quad a_t=\frac{0.5}{1}=\frac{1}{2} \mathrm{~ms}^{-2}$

So, $a=\sqrt{2^2+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{17}}{2} \mathrm{~ms}^{-2}$ Also, angle $\theta$ is such that,

$ \begin{aligned} & \tan \theta =\frac{a_c}{a_t}=\frac{2}{1 / 2}=4 \\ \Rightarrow & \theta =\tan ^{-1}(4) \end{aligned} $

The given situation is shown below.

Radius of vertical circle, $r=\frac{A B}{2}=2 \mathrm{~m}$

$ \Rightarrow A B=\text { diameter }=d=2 \times 2=4 \mathrm{~m} $

Using the concept of conservation of energy,

Total energy at point $X=$ Total energy at point $Y$.

$ \begin{aligned} \Rightarrow \quad m g h+0 & =\frac{1}{2} m v^2+0 \\ h & =\frac{v^2}{2 g} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Since, body just completes motion in vertical circle, hence

$ v=\sqrt{5 g r} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii), we get

$ \begin{aligned} h & =\frac{(\sqrt{5 g r})^2}{2 g} \\ & =\frac{5 g r}{2 g}=\frac{5 r}{2} \\ & =\frac{5 \times 2}{2}=5 \mathrm{~m} \end{aligned} $

Frequency of merry-go-round,

$ \begin{aligned} f & =9 \text { rotation per } 18 \text { second } \\ & =9 / 18 \text { rotation per second } \\ & =1 / 2 \text { rotation per second } \end{aligned} $

∴ Angular velocity,

$ \begin{aligned} \omega & =2 \pi f \\ & =2 \pi \times \frac{1}{2}=\pi \mathrm{rad} / \mathrm{s} \end{aligned} $

Given,

Mass of body, $m=10 \mathrm{~kg}$

Length of wire, $l=0.3 \mathrm{~m}$

Breaking stress, $\sigma=4.8 \times 10^7 \mathrm{~N} / \mathrm{m}^2$

Area of cross-section, $A=10^{-6} \mathrm{~m}^2$

Let angular velocity be $\omega$

$\begin{aligned} & \because \text { Stress, } \sigma=\frac{\text { Force }(F)}{A} \\ & \Rightarrow \quad F=\sigma A=m \omega^2 l \Rightarrow \omega=\sqrt{\frac{\sigma A}{m l}} \\ & \Rightarrow \quad \omega=\sqrt{\frac{4.8 \times 10^7 \times 10^{-6}}{10 \times 0.3}}=\sqrt{16}=4 \mathrm{~rad} \mathrm{~s}^{-1} \end{aligned}$

Given, mass of car, $m=500 \mathrm{~kg}$

Radius of track, $r=50 \mathrm{~m}$

Velocity of car, $v=36 \mathrm{~km} / \mathrm{h}=\frac{36 \times 10^3}{60 \times 60}=10 \mathrm{~ms}^{-1}$

Let centripetal force be $F$.

As we know that,

$\begin{aligned} & F=\frac{m v^2}{r} \\ & \therefore \quad F=\frac{500 \times 10^2}{50}=10 \times 10^2=1000 \mathrm{~N} \\ & \end{aligned}$

Given, radius of wall, R = 8 m

Minimum speed, $v=5\sqrt5$ ms$^{-1}$

Acceleration due to gravity, g = 10 ms$^{-1}$

Let coefficient of friction be $\mu$.

According to diagram,

$f=mg$ ..... (i)

$\therefore \quad f=\mu N$ ..... (ii)

where, $f$ is friction force and N is normal reaction = centripetal force

$\Rightarrow N=\frac{mv^2}{R}$

Substituting in Eq. (ii), we get

$f=\mu \frac{mv^2}{R}$ .... (iii)

Now, on equating Eqs. (i) and (iii), we get

$\begin{aligned} f & =m g=\mu \frac{m v^2}{R} \\ \Rightarrow \quad \mu & =\frac{R g}{v^2}=\frac{8 \times 10}{(5 \sqrt{5})^2}=\frac{80}{125}=0.64\end{aligned}$

As we know that,

Earth rotates from West to East and, in assertion two trains are moving in opposite direction.

$\therefore$ The train going towards East will have high speed with respect to earth from the train going towards West.

$\because$ Normal reaction is proportional to (speed)$^2$ and speeds of different trains are different.

Hence, magnitude of normal reaction is different.

$\therefore$ Assertions will be false and reason will be true because their relative speeds are different. But actual speeds of both the trains are same.

Given, $v_1=14 \mathrm{~m} / \mathrm{s}$ and $v_2=28 \mathrm{~m} / \mathrm{s}$

As we know that, $\tan \theta=\frac{v^2}{r g}$

$ \Rightarrow \quad r=\frac{v^2}{g \tan \theta} $

In the first case, $r_1=\frac{v_1^2}{g \tan \theta}=\frac{(14)^2}{g \tan \theta}$

In the second case,

$ r_2=\frac{(28)^2}{g \tan \theta} $

[for similar ramp remains same]

$ \begin{aligned} r_2 & =\frac{(14 \times 2)^2}{g \tan \theta}=4 \frac{(14)^2}{g \tan \theta} \\ \Rightarrow \quad r_2 & =4 r_1 \quad[\text { using Eq. (i) }] \end{aligned} $

Hence, radius should be increased by factor 4.

According to the question, the situation is as shown below

Balancing the forces,

$ \begin{align*} & N \sin 30^{\circ}=m g \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \text { and } \quad & N \cos 30^{\circ}=\frac{m v^2}{R} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

Dividing Eq. (i) by Eq. (ii), we get

$ \begin{aligned} \tan 30^{\circ} & =\frac{g R}{v^2}=\frac{g R}{\tan 30^{\circ}} \\ \Rightarrow \quad & 10 \times 20 \sqrt{3} \times \sqrt{3} \\ & \quad\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right] \\ & =600=\sqrt{600} \\ & =10 \sqrt{6} \mathrm{~m} / \mathrm{s} \end{aligned} $

The given situation is shown in the following figure

Acceleration, $\mathbf{a}=6 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}$

Clearly, acceleration a is perpendicular to velocity $\mathbf{v}$, hence

$ \mathbf{v} \cdot \mathbf{a}=0 $

Again, position vector $\mathbf{r}$ is parallel to acceleration $\mathbf{a}$, hence

$ \mathbf{r} \times \mathbf{a}=0 $

$ \begin{aligned} &\text { Acceleration of a body moving over a circular path is }\\ &a=\frac{v^2}{r}=\frac{100}{r} \quad(\because v=10 \mathrm{~m} / \mathrm{s}) \end{aligned} $

$ \Rightarrow \quad a \propto \frac{1}{r} $

We have

$m{\omega ^2}r = \mu mg$ ...... (1)

Given : rate of rotation = 3.5 rev/s

$\Rightarrow$ 1 revolution = 2$\pi$ rad

That is, 3.5 revolutions = 3.5 $\times$ 2$\pi$ rad

Therefore, $\omega$ = 3.5 $\times$ 2$\pi$ rad/s

r = 1.25 cm = 1.25 $\times$ 10^$-$2 m

Thus, from Eq. (1), we have

$m{\omega ^2}r = \mu mg \Rightarrow {\omega ^2}r = \mu g$

$ \Rightarrow \mu = {{{\omega ^2}r} \over g} = {{{{(3.5 \times 2\pi )}^2}(1.25 \times {{10}^{ - 2}})} \over {10}}$

$ \Rightarrow \mu = 0.60$