(D) To complete a full vertical circle,the tension at the highest point must be at least zero $(T \ge 0)$. At the highest point,the forces acting are

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(D) To complete a full vertical circle,the tension at the highest point must be at least zero $(T \ge 0)$. At the highest point,the forces acting are gravity $(mg)$ and tension $(T)$,providing the centripetal force: $T + mg = \frac{mv_h^2}{l}$. Setting $T = 0$ for the minimum velocity at the top,we get $mg = \frac{mv_h^2}{l}$,which implies $v_h = \sqrt{gl}$. Using the law of conservation of energy between the lowest point (velocity $v$) and the highest point (velocity $v_h$): $\frac{1}{2}mv^2 = \frac{1}{2}mv_h^2 + mg(2l)$. Substituting $v_h^2 = gl$: $\frac{1}{2}mv^2 = \frac{1}{2}m(gl) + 2mgl = \frac{5}{2}mgl$. Solving for $v$,we get $v = \sqrt{5gl}$.

Class 11 Physics Work, Energy and Power Vertical Circular Motion

Difficult

$A$ body of mass $m$ is suspended from a string of length $l$. What is the minimum horizontal velocity that should be given to the body in its lowest position so that it may complete one full revolution in a vertical plane with the point of suspension as the center of the circle?

A
$v = \sqrt{2gl}$
B
$v = \sqrt{3gl}$
C
$v = \sqrt{4gl}$
D
$v = \sqrt{5gl}$

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Work, Energy and Power

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Vertical Circular Motion

$A$ bob of mass $m$ is tied to a string of length $l$. What is the minimum kinetic energy that must be provided to the bob at the lowest point so that it completes a full vertical circle (in $;mgl$)?

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$A$ steel wire can withstand a load up to $2940 \ N$. $A$ load of $150 \ kg$ is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position,so that the wire does not break when the load passes through the position of equilibrium,is (in $^{\circ}$)

MediumAP EAMCET 2008

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$A$ bullet of mass $m$ moving with velocity $v$ passes through a pendulum bob of mass $M$ and emerges with a velocity $v/2$. What is the minimum value of $v$ so that the pendulum bob completes a full vertical circle of length $\ell$?

Medium

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In vertical circular motion of a bob,match the entries of List-$I$ with entries of List-$II$. Here,$v_0$ is the velocity of the bob at the lowest point.

List-$I$ (Speed at lowest point) List-$II$ (Possible situation)

$(P) v_0 = \sqrt{5 g \ell}$ $(1)$ Tension at lowest point $= 6 mg$

$(Q) v_0 = \sqrt{g \ell}$ $(2)$ String will slack for a finite time

$(R) v_0 = 2 \sqrt{g \ell}$ $(3)$ Bob will oscillate

$(S) v_0 = 3 \sqrt{g \ell}$ $(4)$ Tension at highest point $= 4 mg$

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$A$ stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is

DifficultAIPMT 2004

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List-$I$ (Speed at lowest point)	List-$II$ (Possible situation)
$(P) v_0 = \sqrt{5 g \ell}$	$(1)$ Tension at lowest point $= 6 mg$
$(Q) v_0 = \sqrt{g \ell}$	$(2)$ String will slack for a finite time
$(R) v_0 = 2 \sqrt{g \ell}$	$(3)$ Bob will oscillate
$(S) v_0 = 3 \sqrt{g \ell}$	$(4)$ Tension at highest point $= 4 mg$

$A$ body of mass $m$ is suspended from a string of length $l$. What is the minimum horizontal velocity that should be given to the body in its lowest position so that it may complete one full revolution in a vertical plane with the point of suspension as the center of the circle?

Similar Questions

$A$ bob of mass $m$ is tied to a string of length $l$. What is the minimum kinetic energy that must be provided to the bob at the lowest point so that it completes a full vertical circle (in $;mgl$)?

$A$ bullet of mass $m$ moving with velocity $v$ passes through a pendulum bob of mass $M$ and emerges with a velocity $v/2$. What is the minimum value of $v$ so that the pendulum bob completes a full vertical circle of length $\ell$?

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