A stone of mass m tied to end of a string revolves in a vertical circle of radius R . net forces at

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5.Work, Energy, Power and Collision

medium

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:
Choose the correct alternative

Lowest Point Highest Point

$(a)$ ${mg - {T_1}}$ ${mg + {T_2}}$

$(b)$ ${{m_g} + {T_1}}$ ${{m_g} - {T_2}}$

$(c)$ ${mg + {T_1} - \frac{{mv_1^2}}{R}}$ ${mg - {T_2} + \frac{{mv_1^2}}{R}}$

$(d)$ ${mg - {T_1} - \frac{{mv_1^2}}{P}}$ ${mg + {T_2} + \frac{{mv_1^2}}{p}}$

$T_{1}$ and $v_{1}$ denote the tension and speed at the lowest point. $T_{2}$ and $v_{2}$ denote corresponding values at the highest point.

Option A

Option B

Option C

Option D

Solution

The free body diagram of the stone at the lowest point is shown in the following figure.
According to Newton’s second law of motion, the net force acting on the stone at this point is equal to the centripetal force, i.e.,
$F_{net}=T-m g=\frac{m v_{1}^{2}}{R}$
Where, $v_{1}=$ Velocity at the lowest point
The free body diagram of the stone at the highest point is shown in the following figure.
Using Newton's second law of motion, we have:
$T+m g=\frac{m v_{2}^{2}}{R}$
Where, $v_{2}=$ Velocity at the highest point It is clear from above equations that the net force acting at the lowest and the highest points are respectively $(T-m g )$ and $(T+m g )$

Standard 11

Physics

A particle of mass $m$ , attached with a string of length $l$ is moving in a vertical circle. If the particle is just looping the loop without slackening of the string and $v_A, v_B, v_D$ are speeds at positions $A, B, D$ shown in figure, then

hard

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medium

A stone of mass $m$ is tied to a string and is moved in a vertical circle of radius $r$ making $n$ revolutions per minute. The total tension in the string when the stone is at its lowest point is

medium

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medium

A bucket tied at the end of a $1.6\, m$ long string is whirled in a vertical circle with constant speed. What should be the minimum speed so that the water from the bucket does not spill, when the bucket is at the highest position …….. $m/sec$ (Take $g = 10\,m/{\sec ^2}$)

easy

(AIIMS-1987)

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

Solution

Similar Questions

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