(A) The maximum tension the string can bear is $T_{max} = 3.7 \, kg \cdot wt = 3.7 \times 10 \, N = 37 \, N$.In a vertical circular motion,the tension

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(A) The maximum tension the string can bear is $T_{max} = 3.7 \, kg \cdot wt = 3.7 \times 10 \, N = 37 \, N$.In a vertical circular motion,the tension is maximum at the lowest point of the path.The formula for tension at the lowest point is $T = mg + m\omega^2r$.Given mass $m = 500 \, g = 0.5 \, kg$,radius $r = 4 \, m$,and $g = 10 \, m/s^2$.Substituting the values: $37 = (0.5 \times 10) + (0.5 \times \omega^2 \times 4)$.$37 = 5 + 2\omega^2$.$32 = 2\omega^2$.$\omega^2 = 16$.$\omega = 4 \, rad/s$.

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

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$A$ weightless thread can bear a tension of up to $3.7 \, kg \, wt$. $A$ stone of mass $500 \, g$ is tied to it and revolved in a circular path of radius $4 \, m$ in a vertical plane. If $g = 10 \, m/s^2$,then the maximum angular velocity of the stone will be ........ $rad/s$.

A
$4$
B
$16$
C
$\sqrt{21}$
D
$2$

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Vertical Circular Motion

The mass of the $bob$ of a simple pendulum of length $L$ is $m$. If the $bob$ is released from its horizontal position,then the speed of the $bob$ and the tension in the thread at the lowest position of the $bob$ will be respectively:

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$A$ body of mass $m$ hangs at one end of a string of length $l$,the other end of which is fixed. It is given a horizontal velocity such that the string just reaches a position where it makes an angle of $60^{\circ}$ with the vertical. The tension in the string at the mean position is:

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$A$ body of mass $100 \ g$ is moving in a circular path of radius $2 \ m$ on a vertical plane as shown in the figure. The velocity of the body at point $A$ is $10 \ m/s$. The ratio of its kinetic energies at point $B$ and $C$ is: (Take acceleration due to gravity as $10 \ m/s^2$)

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$A$ stone of mass $1 \ kg$ is tied to one end of a light string and is whirled in a vertical circle as shown in the figure. The tension in the string when the string becomes horizontal is ......... $N$.

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$A$ body of mass $1\,kg$ is rotating in a vertical circle of radius $1\,m$. What will be the difference in its kinetic energy at the top and bottom of the circle? $(g = 10\,m/s^2)$

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$A$ weightless thread can bear a tension of up to $3.7 \, kg \, wt$. $A$ stone of mass $500 \, g$ is tied to it and revolved in a circular path of radius $4 \, m$ in a vertical plane. If $g = 10 \, m/s^2$,then the maximum angular velocity of the stone will be ........ $rad/s$.

Similar Questions

The mass of the $bob$ of a simple pendulum of length $L$ is $m$. If the $bob$ is released from its horizontal position,then the speed of the $bob$ and the tension in the thread at the lowest position of the $bob$ will be respectively:

$A$ body of mass $m$ hangs at one end of a string of length $l$,the other end of which is fixed. It is given a horizontal velocity such that the string just reaches a position where it makes an angle of $60^{\circ}$ with the vertical. The tension in the string at the mean position is:

$A$ body of mass $100 \ g$ is moving in a circular path of radius $2 \ m$ on a vertical plane as shown in the figure. The velocity of the body at point $A$ is $10 \ m/s$. The ratio of its kinetic energies at point $B$ and $C$ is: (Take acceleration due to gravity as $10 \ m/s^2$)

$A$ stone of mass $1 \ kg$ is tied to one end of a light string and is whirled in a vertical circle as shown in the figure. The tension in the string when the string becomes horizontal is ......... $N$.

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