A simple pendulum of mass m is performing vertical circular motion with its velocity at the highest point being V -2 g 1- where l is the length of the string- Find the tension in the string when it reaches the lowest point- A- 2 mgB- 6 mgC- 3 mgD- 7 mg

The correct option is D 7 mg From WET, k_i+ u_1+ W_NC= k_f+ u_f (W_NC=0) 1/2 m (2gl)+mg (2l)= 1/2 mu^2+0 ⇒ u^2=6gl Tension at the lowest point is T = ...

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Standard XII

Physics

Damped Oscillations

A simple pend...

Question

A simple pendulum of mass m is performing vertical circular motion with its velocity at the highest point being $V = \sqrt{2 g l}$ , where l is the length of the string. Find the tension in the string when it reaches the lowest point.

2 mg

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6 mg

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7 mg

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3 mg

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Solution

The correct option is C 7 mg

From WET,
$k_{i} + u_{1} + W_{N C} = k_{f} + u_{f}$ $(W_{N C} = 0)$
$\frac{1}{2} m (2 g l) + m g (2 l) = \frac{1}{2} m u^{2} + 0$
$\Rightarrow u^{2} = 6 g l$
Tension at the lowest point is
$T = m g + \frac{m u^{2}}{l} (\frac{m u^{2}}{l} : c e n t r i f u g a l f o r c e)$
$= m g + \frac{m}{l} (6 g l)$
$= 7 m g$

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A simple pendulum of mass m is performing vertical circular motion with its velocity at the highest point being V=√2gl , where l is the length of the string. Find the tension in the string when it reaches the lowest point.

The correct option is C 7 mg From WET, ki+u1+WNC=kf+uf (WNC=0) 12m(2gl)+mg(2l)=12mu2+0 ⇒u2=6gl Tension at the lowest point is T=mg+mu2l(mu2l:centrifugal force) =mg+ml(6gl) =7mg

A simple pendulum of mass m is performing vertical circular motion with its velocity at the highest point being $V = \sqrt{2 g l}$ , where l is the length of the string. Find the tension in the string when it reaches the lowest point.