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Dynamics of uniform and non uniform circular motion

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Question

A particle of mass 1 kg is moving on a circular path of radius 1 m. Its kinetic energy is K=bt⁴, where b= 1 J/s. The force acting on the particle at t = 1s is (2x)^½N. The value of x is____________.

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Solution

$Here, \frac{1}{2} m v^{2} = b t^{4} or m v^{2} = 2 b t^{4}$
The centripetal force on the body is $F_{c} = \frac{m v^{2}}{R} = \frac{2 b t^{4}}{R}$
$\Rightarrow v = (\sqrt{\frac{2 b}{m}}) t^{2}$
$and a_{t} = \frac{d v}{d t} = \sqrt{\frac{2 b}{m}} \frac{d (t)^{2}}{d t} = \sqrt{\frac{8 b}{m}} t$
$∴ Tangential force is F_{t} = m a_{t} = m \sqrt{\frac{8 b}{m}} t$
The force on the particle is
$\begin{array}{l} F = \sqrt{{(\frac{2 b t^{4}}{R})}^{2} + 8 m b t^{2}} = \sqrt{4 + 8} = \sqrt{12} = 2 \sqrt{3} N \\ ∴ 2 \sqrt{3} = (2 x)^{1 / 2} \Rightarrow 4 \times 3 = 2 x \\ \Rightarrow x = 6 \end{array}$

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