Question

any physics student here,whi can solve it

100 points will be awarded

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Answer 1

Hey there!

It's given to us that the cabin rotates with the uniform angular velocity ω in a circular path of radius R.

Length of Groove = L.

And, groove makes the angle $\theta$ with the radius of OA.

The particle A will experience a force mω²R and the particle will start to move along the groove.

And when particle starts moving along the groove the force that act on it will be cos component of mω²R. i.e mω²Rcos$\theta$.

$F = m\omega^2R \cos \theta\\or, ma = m\omega^2\cos\theta\\\boxed{a = \omega^2R\cos\theta}$

Using Second Equation of motion,(for u=0)(s=L);

$h=ut + \dfrac{1}{2}at^2\\\\L = 0 + \dfrac{1}{2}\omega^2R\cos\theta t^2\\\\t^2 = \dfrac{2L}{\omega^2R\cos\theta}\\\\\boxed{t = \sqrt{\dfrac{2L}{\omega^2R\cos\theta}}}$

Hence, Time taken by particle to reach point B $= \sqrt{\dfrac{2L}{\omega^2R\cos\theta}}}$

Answer 2

Explanation:

t = root 2L /W square R cos theta , thanks