(D) By the law of conservation of energy,the velocity $v$ at the base of the loop is given by $\frac{1}{2}mv^2 = mgH$. Given $H = 2R$,we have $v = \sq

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(D) By the law of conservation of energy,the velocity $v$ at the base of the loop is given by $\frac{1}{2}mv^2 = mgH$. Given $H = 2R$,we have $v = \sqrt{2gH} = \sqrt{4gR}$.For a particle to complete a vertical loop,the minimum velocity required at the bottom is $\sqrt{5gR}$. Since $\sqrt{4gR} When the velocity at the bottom is between $\sqrt{2gR}$ and $\sqrt{5gR}$,the particle leaves the circular path at a height $h$ above the center,where $R Thus,both statements $(B)$ and $(C)$ are correct.

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

Difficult

$A$ 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. Choose the correct alternative$(s)$ if $H = 2R$.

A
The particle reaches the top of the loop with zero velocity.
B
The particle cannot reach the top of the loop.
C
The particle breaks off at a height $R < h < 2R$.
D
Both $(B)$ and $(C)$.

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Work, Energy, Power and Collision

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Vertical Circular Motion

$A$ bob of mass $m$ suspended by a light string of length $L$ is whirled in a vertical circle as shown in the figure. What will be the trajectory of the particle if the string is cut at $(a)$ point $B$,$(b)$ point $C$,and $(c)$ point $X$?

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$A$ particle of mass $m$ is tied to a string of length $l$ and whirled in a vertical circle. The difference in tension and kinetic energy at the highest and lowest positions of the circular path will be:

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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 $\ell$?

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$A$ 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. Choose the correct alternative$(s)$ if $H = 2R$.

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$A$ bob of mass $m$ suspended by a light string of length $L$ is whirled in a vertical circle as shown in the figure. What will be the trajectory of the particle if the string is cut at $(a)$ point $B$,$(b)$ point $C$,and $(c)$ point $X$?

$A$ pendulum consisting of a small sphere of mass $m$ suspended by an inextensible and massless string of length $l$ is made to swing in a vertical plane. If the breaking strength of the string is $2mg$,then the maximum angular amplitude of the displacement from the vertical can be ....... $^o$

$A$ particle of mass $m$ is tied to a string of length $l$ and whirled in a vertical circle. The difference in tension and kinetic energy at the highest and lowest positions of the circular path will be:

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