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

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

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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.

2mg

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

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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(2I)=12mu2+0 ⇒u2=6gl Tension at the lowest point is T=mg+mu2I(mu2I:centrifugal force) =mg+mI(6gl) =7mg option (c)

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.