Calculating Particle Speed on a Moving Sphere: Integration Method

troy611
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A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a particle is released from the top of sphere as soon as the sphere starts accelerating (i.e. at time = 0). Find the speed of the particle with respect to the sphere as a function of angle it makes with the upward vertical (take it as \theta)

hint: integration is to be used

i am just missing one term in the answer, i have taken, the horizontal pseudo force, Normal reaction, and weight of particle amd then used COE.

ps: answer is \sqrt{2R(a*sin theta + g - g*cos theta)}
 
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Hi troy611! :wink:

Show us your full calculations, and then we'll see what's missing, and we'll know how to help! :smile:
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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