Calculation of Motion of a body in a vertical Circle: Let us consider, A body that has mass $m$ moving in a vertical circle of radius $r$. If at any instant the body is at position $P$ with angular displacement $\theta$ from the lower position $L$ of the circle. As shown in the figure below.
The various forces acting on the body are:
The weight $mg$ of the body, acting vertically downwards
Tension $T$in the string acting along the $PO$. Now $mg$ can be resolved into two components:
1.) The horizontal component $mg cos\theta$ opposite to $T$
2.) The vertical component $mg sin\theta$ act along tangent to the circle at $P$
So the net force on the body at position $P$ provides the necessary centripetal force required by the body
$T-mg cos\theta = \frac{m v^{2}}{r}$
$T= \frac{m v^{2}}{r} + mg \: cos\theta \qquad(1)$
Tension and velocity at the highest position $H$ of the verticle circle: At the highest position $H$ the tension $T_{H}$ of the will be minimum and the velocity $v_{H}$ will also be minimum.
If $cos\theta = -1$ i.e. $\theta=180^{\circ}$
Then Tension $T$ and velocity $v$ will be minimum i.e $T_{H}$ and $v_{H}$
so from equation $(1)$
$T_{H}= \frac{m v_{H}^{2}}{r} - mg $
So from the above equation, we can conclude that The body will move along the vertical circle only when $T_{H} \geq 0$
$\left( \frac{m v_{H}^{2}}{r} - mg \right) \geq 0$
$ \frac{m v_{H}^{2}}{r} \geq mg $
$ \frac{ v_{H}^{2}}{r} \geq g $
$ v_{H} \geq \sqrt {rg} $
Thus the minimum value of the velocity at the highest position is $\sqrt{rg}$
Tension and velocity at lowest position $L$ of the verticle circle: At the lowest position $L$ the tension and velocity will be maximum. Now applying the principle of energy conservation for the position $H$ and $L$ of the body.
Total energy at $L$ =Total energy at $H$
$\frac{1}{2}m v_{L}^{2}=\frac{1}{2}m v_{H}^{2} +mg(2r)$
Now substitute the value of $v_{H} \geq \sqrt{gr}$ in above equation then we get
$\frac{1}{2}m v_{L}^{2}=\frac{1}{2}m (gr) +mg(2r)$
$\frac{1}{2}m v_{L}^{2}=\frac{5mgr}{2}$
$v_{L} \geq \sqrt{5gr}$
For the maximum value of tension $T_{L}$ at the lowest position $L$ of the vertical circle. Now put $cos \theta =1$ i.e $\theta=0^{\circ}$ in the equation $(1)$ then we get
$T_{L}= \frac{m v_{L}^{2}}{r} + mg $
Now substitute the value of $v_{L} \geq \sqrt{5gr}$ in the above equation then
$T_{L} \geq \frac{m 5gr}{r} + mg $
$T_{L} \geq 6gr$
Tension and velocity at horizontal position $M$ of the verticle circle:
Now apply the principle of conservation of energy for position $M$ and position $L$
Total energy at $M$ =Total energy at $L$
$ \frac{1}{2}m v_{M}^{2} + mgr = \frac{1}{2} m v_{L}^{2}$
$ \frac{1}{2}m v_{M}^{2} = \frac{1}{2} m v_{L}^{2} - mgr$
Now substitute the value of $v_{L} \geq \sqrt{5gr}$ in the above equation then
$ \frac{1}{2}m v_{M}^{2} \geq \frac{1}{2} m 5gr - mgr$
$ \frac{1}{2}m v_{M}^{2} \geq \frac{3 mgr}{2}$
$v_{M}^{2} \geq 3 gr$
$v_{M} \geq \sqrt{3 gr}$
To find the tension $T_{M}$ at position $M$ substitute the $cos \theta =0$ i.e $\theta =90^{\circ}$ in equation $(1)$. Then we get
$T_{M} = \frac{m v_{M}^{2}}{r} + 0$
$T_{M} = \frac{m v_{M}^{2}}{r}$
Now substitute the value of $v_{M} \geq \sqrt{3gr}$ in the above equation. Then we get
$T_{M} \geq 3mg$
Practical Application of motion in a vertical Circle:
1.) When a bucket containing water is rotated in a vertical circle with a velocity at the lowest point $v_{L} \geq \sqrt{5gr}$, water shall not spill even at the highest point, when the bucket is upside down. If the bucket is whirled slowly, so that $ \left(mg \gt \frac{mv_{H}^{2}}{r} \right) $, then a part of the weight shall provide the necessary centripetal force $\left( \frac{mv_{H}^{2}}{r} \right)$; and the rest of the weight of water $\left( mg - \frac{mv_{H}^{2}}{r} \right) $ causes some water to accelerate downwards and spill. Only this much water shall leave the bucket.
2.) A pilot of an aircraft can successfully loop a vertical loop without falling at the top of the loop (being without a belt) when its velocity at the bottom of the loop is $\geq \sqrt{5gr}$
3.) In a circus, a motorcyclist is able to perform the feat of driving the motorcycle along a vertical circle in a cage. The motorcyclist does not fall even at the highest point, when his velocity at the bottom of the cage is $\geq \sqrt{5gr}$.
To acquire this velocity at the lowest point $L$ of the vertical circle of radius $r$, he has to roll down a vertical height $h$, As shown in the figure below:
From the equation of motion $v^{2}=u^{2}+2as$
Let $u=0$, $a=+g$ and $s=h$
we get, $v^{2}=0 +2gh$
$v=\sqrt{2gh}$
To move in a vertical circle, the velocity $v$ acquired at $L$ must at least be equal to $\sqrt{5gr}$. i.e.
$\sqrt{2gh}= \sqrt{5gr}$
$h=\frac{5r}{2}$
Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater than the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of the incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less than the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of lig
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