(D) Let the lowest point be $A$ and the horizontal position be $B$. At $A$, velocity is $\vec{u} = u \hat{i}$.Using the law of conservation of energy

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(D) Let the lowest point be $A$ and the horizontal position be $B$. At $A$, velocity is $\vec{u} = u \hat{i}$.Using the law of conservation of energy between $A$ and $B$:$\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgL$$v^2 = u^2 - 2gL$$v = \sqrt{u^2 - 2gL}$.At position $B$, the velocity is directed vertically upward, so $\vec{v} = v \hat{j} = \sqrt{u^2 - 2gL} \hat{j}$.The change in velocity is $\Delta \vec{v} = \vec{v} - \vec{u} = v \hat{j} - u \hat{i}$.The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{v^2 + u^2} = \sqrt{(u^2 - 2gL) + u^2} = \sqrt{2u^2 - 2gL} = \sqrt{2(u^2 - gL)}$.

Class 11 Physics Work, Energy and Power Vertical Circular Motion

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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

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A
$\sqrt{u^2 - 2gL}$
B
$\sqrt{2gL}$
C
$\sqrt{u^2 - gL}$
D
$\sqrt{2(u^2 - gL)}$

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Work, Energy and Power

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Vertical Circular Motion

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$A$ bob of mass $m$ is suspended at a point $O$ by a light string of length $l$ and is set to perform vertical circular motion as shown in the figure. Initially,by applying a horizontal velocity $v_0$ at point $A$,the string becomes slack when the bob reaches point $D$. The ratio of the kinetic energy of the bob at points $B$ and $C$ is:

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$A$ small sphere of mass $m$ is suspended by a light and inextensible string of length $l$ from a point $O$ fixed on a smooth inclined plane of inclination $\theta$ with the horizontal. The sphere is moving in a circle on the incline as shown. If the sphere has a velocity $u$ at the top most position $A$, then:

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$A$ 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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The string of a pendulum of length $L$ is displaced through $90^{\circ}$ from the vertical and released. The maximum tension the string must withstand as the pendulum passes through the mean position is ($m =$ mass of the pendulum,$g =$ acceleration due to gravity).

EasyMHT CET 2024

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$A$ weightless thread can support a tension of up to $30 \, N$. $A$ stone of mass $0.5 \, kg$ is tied to it and is revolved in a circular path of radius $2 \, m$ in a vertical plane. If $g = 10 \, m/s^2$,then the maximum angular velocity of the stone will be ........ $rad/s$.

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$A$ 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 $;mgl$)?

The string of a pendulum of length $L$ is displaced through $90^{\circ}$ from the vertical and released. The maximum tension the string must withstand as the pendulum passes through the mean position is ($m =$ mass of the pendulum,$g =$ acceleration due to gravity).

$A$ weightless thread can support a tension of up to $30 \, N$. $A$ stone of mass $0.5 \, kg$ is tied to it and is revolved in a circular path of radius $2 \, m$ in a vertical plane. If $g = 10 \, m/s^2$,then the maximum angular velocity of the stone will be ........ $rad/s$.

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