A particle is given an initial speed is inside a smooth spherical shell of radius R =1 it that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is:
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Step 1
Given initial velocity at the lowest point, \(v = \sqrt{5gR}\). Using energy conservation, we have \(v^2 = u^2 - 2gh\). Substitute \(u = \sqrt{5gR}\) and \(h = R\), we get \(v^2 = 3gR\). Show more…
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