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Work, Energy, and Power — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsWork, Energy, and Power5 Marks
Question
A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle $\theta$ it slides.
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Answer



Let the sphere move towards left with an acceleration ‘a'
Let m = mass of the particle
The particle ‘m’ will also experience the inertia due to acceleration ‘a’ as it is on the sphere. It will also experience the tangential inertia force $\Big(\text{m}\Big(\frac{\text{dv}}{\text{dt}}\Big)\Big)$ and centrifugal force $\Big(\frac{\text{mv}^2}{\text{R}}\Big).$
$\text{m}\frac{\text{dv}}{\text{dt}}=\text{ma}\cos\theta+\text{mg}\sin\theta$
$\Rightarrow\text{mv}\frac{\text{dv}}{\text{dt}}=\text{ma}\cos\theta\Big(\text{R}\frac{\text{d}\theta}{\text{dt}}\Big)+\text{mg}\sin\theta\Big(\text{R}\frac{\text{d}\theta}{\text{dt}}\Big)$
Because, $\text{v}=\text{R}\frac{\text{d}\theta}{\text{dt}}$
$\Rightarrow\text{vdv}=\text{aR}\cos\theta\text{ d}\theta+\text{gR}\sin\theta\text{ d}\theta$
Integrating both sides we get,
$\frac{\text{v}^2}{2}=\text{aR}\sin\theta-\text{gR}\sin\theta+\text{C}$
Given that, at $\theta=0,\text{v}=0$
So, $\text{C}=\text{gR}$
So, $\frac{\text{v}^2}{2}=\text{aR}\sin\theta-\text{gR}\sin\theta+\text{gR}$
$\therefore\ \text{v}^2=2\text{R}(\text{a}\sin\theta+\text{g}-\text{g}\cos\theta)$
$\Rightarrow\text{v}=[2\text{R}(\text{a}\sin\theta+\text{g}-\text{g}\cos\theta)]^\frac{1}{2}$

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