While reading HC Verma chapter 7 circular motion I came across a derivation which I couldnt understand. I have marked my doubt with red. I don't understand from where +dw/dt [- i sine +j cos0] came from.
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$\begingroup$ $\left[ -\vec{i} \sin\theta + \vec{j} \cos\theta \right]$ is a unit vector (check its size), so ask yourself "Which way does it point?" $\endgroup$– dmckee --- ex-moderator kittenCommented May 24, 2016 at 17:14
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$\begingroup$ derivative for multiplication, from equation (ii): d(xy)/dt=x dy/dt+y dx/dt $\endgroup$– user83548Commented May 24, 2016 at 17:16
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It's because of the product rule of derivatives, that states d(fg(t))/dt=f(t)(dg/dt)+g(t)(df/dt). In this case, let's call this part: (-isinθ+jcosθ)=g(t), and solve it: d(rωg)/dt=r(dωg/dt)=r(ω*(dg/dt)+g(dω/dt)). If you substitute g for (-i*sinθ+jcosθ), you get the equation for acceleration your book presents.
