Circular Motion

To find the fastest speed at which the car can go without slipping, use this formula:

$ v_{\max }=\sqrt{R g \frac{\tan \theta+\mu_s}{1-\mu_s \tan \theta}} $

Here:

• $R = 75$ m (radius of the path)
• $g = 9.8$ m/s$^2$ (acceleration due to gravity)
• $\tan \theta = 0.2$ (angle of banking)
• $\mu_s = 0.1$ (friction coefficient)

Now, put all the values into the formula: $ v_{\max }=\sqrt{75 \times 9.8 \times \frac{0.2+0.1}{1-0.1 \times 0.2}} $

First, add $0.2 + 0.1 = 0.3$ in the numerator.

Multiply $0.1 \times 0.2 = 0.02$, so the denominator is $1 - 0.02 = 0.98$.

So, $ v_{\max} = \sqrt{75 \times 9.8 \times \frac{0.3}{0.98}} $

This gives $ v_{\max} = \sqrt{225} = 15~\mathrm{m/s} $

The maximum speed the car can go without slipping is about $15~\mathrm{m/s}$.

We know that force ( $\mathbf{F}$ ) is directly proportional to acceleration (a) by Newton's second law ( $\mathbf{F}=m \mathbf{a}$ ). Therefore, if the force is always perpendicular to the velocity, then the acceleration is also always perpendicular to the velocity. The work done by a force is given by $W=\int \mathbf{F} \cdot d \mathbf{s}$, where $d \mathbf{s}$ is the displacement.

Since, $\mathbf{F}$ is perpendicular to $\mathbf{v}$ and $d \mathbf{s}$ is in the direction of $\mathbf{v}$, the dot product $\mathbf{F} \cdot d \mathbf{s}=0$. This means the work done by the force is zero.

According to the work-energy theorem, the net work done on a particle equals the change in its kinetic energy ( $\Delta \mathrm{KE}$ ). Since the work done is zero, the change in kinetic energy is zero, meaning the kinetic energy remains constant.

Given,

Radius, $ r=9 \mathrm{~km}=9 \times 10^{3} \mathrm{~m} $

Speed, $ u=540 \mathrm{~km} / \mathrm{h}=150 \mathrm{~m} / \mathrm{s} $

Angle, $ \theta= $ ?

$ g=10 \mathrm{~m} / \mathrm{s}^{2} $

Here, $ F=\frac{m u^{2}}{r} $

$ m= $ mass of plane

$ m g=N \cos \theta $

and due to centripetal force

$ N \sin \theta=\frac{m u^{2}}{r} $

From Eqs. (i) and (ii)

$ \begin{aligned} \tan \theta & =\frac{u^{2}}{r g} \\\\ & =\frac{(150)^{2}}{9 \times 10^{3} \times 10}=\frac{225}{900} \\\\ \tan \theta & =\frac{1}{4} \\\\ \cot \theta & =4 \\\\ \theta & =\cot ^{-1}(4) \end{aligned} $

Given, $m=20 \mathrm{~T}=20 \times 10^3 \mathrm{~kg}$

$ r=240 \mathrm{~m} $

distance between wheels, $d=1.5 \mathrm{~m}$

Centre of gravity above ground, $h=2 \mathrm{~m}$

Equating torque by gravity and torque by centrifugal force about outer wheel, we get

$ \begin{aligned} \frac{m v_{\max }^2}{r} \times h & =m g \frac{d}{2} \\\\ \Rightarrow \quad v_{\max } & =\sqrt{\frac{g r d}{2 h}} \\\\ & =\sqrt{\frac{10 \times 240 \times 1.5}{2 \times 2}} \\\\ & =\sqrt{900}=30 \mathrm{~m} / \mathrm{s} \end{aligned} $

As we know,

$ v_{\max }=\sqrt{\frac{r g(\tan \theta+\mu)}{1-\mu \tan \theta}} \quad \ldots \text { (i) } $

For optimum speed,

$ v_{\mathrm{opt}}=\sqrt{\mathrm{rg} \tan \theta} $

where $\theta=45^{\circ}$

$ \begin{aligned} & v_{\mathrm{opt}}=\sqrt{\mathrm{rg} \tan 45^{\circ}} \\\\ & v_{\mathrm{opt}}=\sqrt{r g} \quad \ldots \text { (ii) } \end{aligned} $

From Eq. (i),

$ \begin{aligned} \left(v_{\mathrm{max}}\right)^2 & =\frac{\left(v_{\mathrm{opt}}\right)^2(\tan \theta+\mu)}{1-\mu \tan \theta} \\\\ \left(2 v_{\mathrm{opt}}\right)^2 & =\frac{\left(v_{\mathrm{opt}}\right)^2(\tan \theta+\mu)}{1-\mu \tan \theta} \\\\ \Rightarrow 4(1-\mu) & =1+\mu \\\\ 4-4 \mu & =1+\mu \Rightarrow 3=5 \mu \\\\ \mu & =0.6 \end{aligned} $

The coefficient of static friction between the wheels of the car and the road is 0.6 .

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.

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, $v_1=14 \mathrm{~m} / \mathrm{s}$ 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} $