Circular Motion

To solve this problem, note that when a charged particle with mass $m$ and charge $q$ is fired perpendicular to a magnetic field of strength $B$, it undergoes uniform circular motion. The period (time to complete one full circle) is given by:

$ T = \frac{2\pi m}{qB} $

Here's how to calculate it step by step:

Convert the given quantities to SI units:

Charge: $q = 1.6\,\mu\text{C} = 1.6 \times 10^{-6}\,\text{C}$

Mass: $m = 16\,\mu\text{g} = 16 \times 10^{-9}\,\text{kg}$

Magnetic field: $B = 6.28\,\text{T}$

Write down the period formula:

$ T = \frac{2\pi m}{qB} $

Substitute the values:

$ T = \frac{2\pi \times (16 \times 10^{-9}\,\text{kg})}{(1.6 \times 10^{-6}\,\text{C})(6.28\,\text{T})} $

Notice that $2\pi \approx 6.28$, which cancels with the given magnetic field value, simplifying the expression:

$ T = \frac{6.28 \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28} = \frac{16 \times 10^{-9}}{1.6 \times 10^{-6}} $

Simplify the fraction:

$ T = \frac{16}{1.6} \times \frac{10^{-9}}{10^{-6}} = 10 \times 10^{-3} = 0.01\,\text{s} $

Thus, the time required for the particle to return to its original location is:

$ \boxed{0.01\,\text{s}} $

$\begin{aligned} & \mathrm{T} \cos \theta=\mathrm{mg} \\ & \mathrm{~T} \sin \theta=\mathrm{M} \omega^2 \mathrm{R} \\ & \text { Using equation (2) } \\ & \mathrm{T} \sin \theta=\mathrm{M} \omega^2(\mathrm{~L} \sin \theta) \\ & \mathrm{T}=\mathrm{M} \omega^2 \mathrm{~L}=\mathrm{M}\left(\frac{3}{\pi} \times 2 \pi\right)^2 \mathrm{~L} \\ & \mathrm{~T}=36 \mathrm{ML} \end{aligned}$

$\begin{aligned} & \mathrm{F}=2 \mathrm{M} \omega^2 \frac{\ell}{2}=\mathrm{Mw}^2 \ell \\ & \omega=\sqrt{\frac{\mathrm{F}}{\mathrm{M} \ell}} \end{aligned}$

$\begin{aligned} & \left|\vec{a}_c\right|=\left|\vec{a}_t\right| \\ & \frac{v^2}{r}=\frac{d v}{d t} \\ & \Rightarrow \int_\limits4^v \frac{d v}{v^2}=\int_\limits0^t \frac{d t}{r} \\ & \Rightarrow\left[\frac{-1}{v}\right]_4^v=\frac{t}{r} \\ & \Rightarrow \frac{-1}{v}+\frac{1}{4}=2 t \end{aligned}$

$\begin{aligned} & \Rightarrow \mathrm{v}=\frac{4}{1-8 \mathrm{t}}=\frac{\mathrm{ds}}{\mathrm{dt}} \\ & 4 \int_0^{\mathrm{t}} \frac{\mathrm{dt}}{1-8 \mathrm{t}}=\int_0^{\mathrm{s}} \mathrm{ds} \\ & (\mathrm{r}=0.5 \mathrm{~m} \\ & \mathrm{s}=2 \pi \mathrm{r}=\pi) \\ & 4 \times \frac{[\ell \mathrm{n}(1-8 \mathrm{t})]_0^{\mathrm{t}}}{-8}=\pi \\ & \ell \mathrm{n}(1-8 \mathrm{t})=-2 \pi \\ & 1-8 \mathrm{t}=\mathrm{e}^{-2 \pi} \\ & \mathrm{t}=\left(1-\mathrm{e}^{-2 \pi}\right) \frac{1}{8} \mathrm{~s} \end{aligned}$

So, $\alpha=8$

Acceleration of stone $a=\frac{v^{2}}{r}=\omega^{2} R$

$ \begin{aligned} & a=\left(\frac{28 \times 2}{60} \times \frac{22}{7}\right)^{2} \times 1.8 \\\\ & =\frac{1936}{125} \end{aligned} $

So, $x=125$

${{dv} \over {dt}} = {{{v^2}} \over R} \Rightarrow {{{v^2}} \over R} = v{{dv} \over {ds}}$

$ \Rightarrow {{dv} \over v} = {{ds} \over R} \Rightarrow \left. {\ln v} \right|_{15}^v = {s \over R}$

$ \Rightarrow v = 15{e^{\Delta /R}} = {{ds} \over {dt}} \Rightarrow dt = {1 \over {15}}{e^{ - \Delta /R}}ds$

$\Delta t = {R \over {15}}[1 - {e^{ - \Delta /R}}]$

$ = 40[1 - {e^{ - \pi /2}}]$ seconds

$ \Rightarrow t = 40$

$ \begin{aligned} & V_{a v}=\frac{S_{\text {total }}}{T_{\text {total }}}=\frac{200+200+\frac{2 \pi \times 200}{4}}{238} \\\\ &=\frac{400+100 \pi}{238}=\frac{714}{238}=3 \mathrm{~m} / \mathrm{s} \end{aligned} $

Let minimum velocity at the lowest point for the bob to compleate the circular path = v'

And we know, v' = $\sqrt{5 r g}$

From the conservation of momentum, we have

$ 75 \times 10^{-3} \times v=50 \times 10^{-3} \times v^{\prime}+75 \times 10^{-3} \times \frac{v}{3} $

$75 \times 10^{-3} v=50 \times 10^{-3} \sqrt{5 r g} \times\left(75 \times 10^{-3} \times \frac{v}{3}\right)$

According to question, the bob completes a vertical circle of $2 \mathrm{~m}$ radius, therefore

$ \begin{aligned} & r=2 \mathrm{~m}, g=10 \mathrm{~ms}^{-2} \\\\ & 75 \times 10^{-3} v=50 \times 10^{-3} \times \sqrt{5 \times 2 \times 10}+75 \times 10^{-3} \times \frac{v}{3} \\\\ & 75 \times 10^{-3}\left(\frac{2 v}{3}\right)=50 \times 10^{-3} \times 10 \\\\ & 150 \times 10^{-3} \times v=(150) \times 10^{-2} \\\\ & v=10 \mathrm{~m} / \mathrm{s} \end{aligned} $

Note :



At the point $3$, both the tension $T_{3}$ and the weight $m g$ of the body act towards the centre of the circle. So $T_{3}+m g$ provides the centripetal force necessary for the rotation of the body.

$ \therefore T_{3}+m g=\frac{m v_{3}^{2}}{r} $

At the point $1$, the tension $T_{1}$ acts vertically upwards i.e., towards the centre of the circle and the weight $m g$ of the body acts vertically downwards i.e., in an opposite direction. So $T_{1}-m g$ provides the necessary centripetal force here.

$ \therefore T_{1}-m g=\frac{m v_{1}{ }^{2}}{r} $

If the tension in the string just vanishes at $3$ i.e., if $T_{3}=0$, then

$ m g=\frac{m v_{3}^{2}}{r} \text { or, } v_{3}=\sqrt{g r} $

If the velocity of the body at the highest point $3$ be less than $\sqrt{g r}$, the string will slack and the body will drop down instead of rotating in the circular path. So this minimum velocity of the body at the highest point is called the critical velocity.

Minimum velocity at the lowest point for maintaining the critical velocity :

Now, as the body goes from $3$ to $1$, its height increases by $2 r$. So its potential energy increases by $m g \times 2 r$. From the principle of conservation of mechanical energy, we have

The K.E. of the body at $3$ - Its K.E. at $1=$ Increase in P.E.

or, $1 / 2 m v_{3}^{2}-1 / 2 m v_{1}^{2}=2 m g r$

or, $v_{3}^{2}=v_{1}^{2}+4 g r$

When $v_{3}=\sqrt{g r}, v_{1}$ is minimurn

$\therefore\left(v_{1}\right)_{\min }=\sqrt{g r+4 g r}=\sqrt{5 g r}$

Minimum tension :

When the body moves with the critical velocity at the highest point, the tension in the string becomes zero; then $m g=\frac{m v_{3}^{2}}{r}$ When this condition is satisfied, the tension in the string at the lowest point $1$ becomes minimum.

So, $ \left(T_{1}\right)_{\min }=\frac{m\left(v_{1}\right)^{2} \min }{r}+\frac{m\left(v_{3}\right)^{2} \min }{r}=\frac{m}{r}(5 g r+g r)=6 \mathrm{mg} $

For a car to move safely around a curved road without skidding, the centripetal force needed is provided by the frictional force between the tires and the road. The relationship governing this scenario can be expressed as:

$ f_{\text{friction}} = f_{\text{centripetal}} $

The frictional force is given by:

$ f_{\text{friction}} = \mu \cdot m \cdot g $

And the centripetal force is given by:

$ f_{\text{centripetal}} = \frac{m \cdot v^2}{r} $

Where:

$\mu$ is the coefficient of friction,

$m$ is the mass of the car,

$g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$),

$v$ is the speed of the car,

$r$ is the radius of the curve.

Since the frictional force equals the centripetal force, we have:

$ \mu \cdot m \cdot g = \frac{m \cdot v^2}{r} $

Cancelling $m$ from both sides (assuming $m \neq 0$) gives:

$ \mu \cdot g = \frac{v^2}{r} $

Rewriting for $v$ gives the speed:

$ v = \sqrt{\mu \cdot g \cdot r} $

For the first scenario with $r = 75 \, \text{m}$ and $v = 30 \, \text{m/s}$:

$ v_1^2 = \mu \cdot g \cdot r_1 $

Thus,

$ \mu \cdot g = \frac{v_1^2}{r_1} = \frac{30^2}{75} = \frac{900}{75} = 12 $

For the second scenario with $r_2 = 48 \, \text{m}$, the new speed $v_2$ is:

$ v_2 = \sqrt{\mu \cdot g \cdot r_2} = \sqrt{12 \times 48} $

Calculating inside the square root:

$ 12 \times 48 = 576 $

So,

$ v_2 = \sqrt{576} = 24 $

Hence, the maximum allowed speed for a curve with a radius of $48 \, \text{m}$ is 24 m/s.

Let the speed of bob at lowest position be v₁ and at the highest position be v₂.

Maximum tension is at lowest position and minimum tension is at the highest position. Now, using, conservation of mechanical energy,

${1 \over 2}mv_1^2 = {1 \over 2}mv_2^2 + mg2l$

$ \Rightarrow {v_1}^2 = {v_2}^2 + 4gl$ ..........(1)

Now, ${T_{\max }} - mg = {{mv_1^2} \over l}$

$ \Rightarrow {T_{\max }} = mg + {{mv_1^2} \over l}$

& ${T_{\min }} + mg = {{mv_2^2} \over l}$

$ \Rightarrow {T_{\min }} = {{mv_2^2} \over l} - mg$

${{{T_{\max }}} \over {{T_{\min }}}} = {5 \over 1}$

$ \Rightarrow {{mg + {{mv_1^2} \over l}} \over {{{mv_2^2} \over l} - mg}} = {5 \over 1}$

$ \Rightarrow mg + {{mv_1^2} \over l} = \left[ {{{mv_2^2} \over l} - mg} \right]5$

$ \Rightarrow mg + {m \over l}\left[ {v_2^2 + 4gl} \right] = {{5mv_2^2} \over l} - 5mg$

$ \Rightarrow mg + {{mv_2^2} \over l} + 4mg = {{5mv_2^2} \over l} - 5mg$

$ \Rightarrow 10mg = {{4mv_2^2} \over l}$

${v_2}^2 = {{10 \times 10 \times 1} \over 4}$

$ \Rightarrow {v_2}^2 = 25 \Rightarrow {v_2} = 5$ m/s

Thus, velocity of bob at highest position 5 m/s.