A particle tied to a string describes a vertical circular motion of radius R continuously- If it has velocity -3 gR at the highest point- then the ratio of respective tensions in the string at the highest and lowest point isA- 4- 3B- 3- 2C- 5- 4D- 1- 4

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Radial & Tangential Acceleration for Non Uniform Circular Motion

A particle ti...

Question

A particle tied to a string describes a vertical circular motion of radius $R$ continuously. If it has velocity $\sqrt{3 g R}$ at the highest point, then the ratio of respective tensions in the string at the highest and lowest point is

4 : 3

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5 : 4

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1 : 4

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3 : 2

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Solution

The correct option is C $1 : 4$
Solving from the frame of reference of the particle, there will be a centrifugal force acting radially outward.
At top most point ( $A$ ):

$T_{T o p} + m g = \frac{m v^{2}}{R}$
$\Rightarrow T_{T o p} + m g = \frac{m}{R} (3 g R)$
(given $v = \sqrt{3 g R}$ at the top)
$\Rightarrow T_{T o p} = 3 m g - m g = 2 m g$
$\Rightarrow T_{T o p} = 2 m g . . . . . . . (i)$

Step - 2
Applying law of conservation of energy
$U_{A} + K E_{A} = U_{B} + K E_{B}$
$\Rightarrow m g (2 R) + \frac{1}{2} m (3 g R) = \frac{1}{2} m v_{B}^{2}$
$(∵ U_{B} = 0)$
$\Rightarrow v_{B} = \sqrt{7 g R}$

Step - 3
Now at bottom most point

$T_{B o t t o m} - m g = \frac{m v_{B}^{2}}{R}$
$T_{B o t t o m} = m g + \frac{m}{R} \times 7 g R$
$T_{B o t t o m} = 8 m g . . . . . . . (i i)$

From ( $i$ ) & $(i i)$
$\frac{T_{T o p}}{T_{B o t t o m}} = \frac{2 m g}{8 m g} = \frac{1}{4}$

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A particle tied to a string describes a vertical circular motion of radius R continuously. If it has velocity √3gR at the highest point, then the ratio of respective tensions in the string at the highest and lowest point is

A particle tied to a string describes a vertical circular motion of radius $R$ continuously. If it has velocity $\sqrt{3 g R}$ at the highest point, then the ratio of respective tensions in the string at the highest and lowest point is