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Law of Conservation of Mechanical Energy

A particle is...

Question

A particle is suspended from a string of length $R$ and it is given a velocity $u = 3 \sqrt{g R}$ at the bottom keeping the other end fixed. Identify the correct statement from the options given below:

v_{B} < v_{C}

T_{B} > T_{C}

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v_{B} > v_{C}

T_{B} > T_{C}

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v_{B} > v_{C}

T_{B} < T_{C}

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v_{B} < v_{C}

T_{B} < T_{C}

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Solution

The correct option is B $v_{B} > v_{C}$ , $T_{B} > T_{C}$
Given : $v_{A} = u = 3 \sqrt{g R} m / s$
Applying energy conservation between A and B:
$\frac{1}{2} m v_{A}^{2} = \frac{1}{2} m v_{B}^{2} + m g R$
$v_{B}^{2} = v_{A}^{2} - 2 g R = (3 \sqrt{g R})^{2} - 2 g R = 7 g R$
$\Rightarrow v_{B} = \sqrt{7 g R} m/s$
Simillarly applying energy conservation between A and C:
$\frac{1}{2} m v_{A}^{2} = \frac{1}{2} m v_{C}^{2} + m g (2 R)$
$v_{C}^{2} = v_{A}^{2} - 4 g R = 9 g R - 4 g R = 5 g R$

$\Rightarrow v_{C} = \sqrt{5 g R}$
Applying Newton's $2^{n d}$ law,

At B:
$T_{B} = \frac{m v_{B}^{2}}{R} = 7 m g$

Simillarly, at C:
$m g + T_{C} = \frac{m v_{C}^{2}}{R}$
$\Rightarrow T_{C} = \frac{m v_{C}^{2}}{R} - m g$
$∴ T_{c} = 4 m g$

$∴ v_{B} > v_{C}$ and $T_{B} > T_{C}$

Hence option B is the correct answer

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A particle is suspended from a string of length R and it is given a velocity u=3√gR at the bottom keeping the other end fixed. Identify the correct statement from the options given below:

A particle is suspended from a string of length $R$ and it is given a velocity $u = 3 \sqrt{g R}$ at the bottom keeping the other end fixed. Identify the correct statement from the options given below: