(D) By conservation of energy,the velocity $v$ of the particle at an angle $\theta$ is given by: $mgH = mgR(1 + \cos \theta) + \frac{1}{2}mv^2$,which

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(D) By conservation of energy,the velocity $v$ of the particle at an angle $\theta$ is given by: $mgH = mgR(1 + \cos \theta) + \frac{1}{2}mv^2$,which simplifies to $v^2 = 2g(H - R - R \cos \theta)$.Applying Newton's second law in the radial direction: $N - mg \cos \theta = \frac{mv^2}{R}$.Substituting $v^2$: $N = mg \cos \theta + \frac{m}{R}[2g(H - R - R \cos \theta)] = mg \cos \theta + \frac{2mgH}{R} - 2mg - 2mg \cos \theta = \frac{2mgH}{R} - 2mg - mg \cos \theta$.At $\theta = 0$,$N = \frac{2mgH}{R} - 3mg$,which is the maximum value.At $\theta = \pi$,$N = \frac{2mgH}{R} - mg$,which is the minimum value (if $H > 5R/2$,$N > 0$ at the top).Thus,all statements are correct.

Class 11 Physics Work, Energy, Power and Collision Vertical Circular Motion

Medium

$A$ particle of mass $m$ is released from a height $H$ on a smooth curved surface which ends into a vertical loop of radius $R$,as shown. If $\theta$ is the instantaneous angle which the line joining the particle and the centre of the loop makes with the vertical,then identify the correct statement$(s)$ related to the normal reaction $N$ between the block and the surface.

A
The maximum value of $N$ occurs at $\theta = 0$.
B
The minimum value of $N$ occurs at $\theta = \pi$ for $H > 5R/2$.
C
The value of $N$ becomes zero only when $\theta \geq \pi / 2$.
D
All of the above.

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Work, Energy, Power and Collision

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Vertical Circular Motion

$A$ mass $m$ is attached to a string and revolves in a vertical circle. What is the tension in the string when the mass is at the lowest position?

EasyAIIMS 2001

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$A$ stone of mass $m$ tied to the end of a string revolves in a vertical circle of radius $R$. The net forces at the lowest and highest points of the circle directed vertically downwards are:
Lowest Point Highest Point
$(a) \ mg - T_1$ $mg + T_2$
$(b) \ mg + T_1$ $mg - T_2$
$(c) \ mg + T_1 - \frac{mv_1^2}{R}$ $mg - T_2 + \frac{mv_2^2}{R}$
$(d) \ mg - T_1 - \frac{mv_1^2}{R}$ $mg + T_2 + \frac{mv_2^2}{R}$

$T_1$ and $v_1$ denote the tension and speed at the lowest point. $T_2$ and $v_2$ denote corresponding values at the highest point.

Medium

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In a circus,a stuntman rides a motorbike on a vertical circular track of radius $r$. Find the minimum speed he must maintain at the highest point of the track.

MediumAP EAMCET 2020

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$A$ body of mass $m \ kg$ slides from rest along the curve of a vertical circle from point $A$ to $B$ on a frictionless path. The velocity of the body at $B$ is: (Given: $R = 14 \ m$,$g = 10 \ m/s^2$,and $\sqrt{2} = 1.4$) (in $m/s$)

DifficultJEE MAIN 2024

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$A$ bob attached to a string is held horizontal and released. The tension $T$ and vertical distance $h$ from the point of suspension can be represented by which of the following graphs?

Medium

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Lowest Point	Highest Point
$(a) \ mg - T_1$	$mg + T_2$
$(b) \ mg + T_1$	$mg - T_2$
$(c) \ mg + T_1 - \frac{mv_1^2}{R}$	$mg - T_2 + \frac{mv_2^2}{R}$
$(d) \ mg - T_1 - \frac{mv_1^2}{R}$	$mg + T_2 + \frac{mv_2^2}{R}$

Similar Questions

$A$ mass $m$ is attached to a string and revolves in a vertical circle. What is the tension in the string when the mass is at the lowest position?

In a circus,a stuntman rides a motorbike on a vertical circular track of radius $r$. Find the minimum speed he must maintain at the highest point of the track.

$A$ body of mass $m \ kg$ slides from rest along the curve of a vertical circle from point $A$ to $B$ on a frictionless path. The velocity of the body at $B$ is: (Given: $R = 14 \ m$,$g = 10 \ m/s^2$,and $\sqrt{2} = 1.4$) (in $m/s$)

$A$ bob attached to a string is held horizontal and released. The tension $T$ and vertical distance $h$ from the point of suspension can be represented by which of the following graphs?

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