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A particle moves clockwise in a circle of radius `1 m` with centre at `(x, y) = (1m, 0)`. It startsx at rest at the origin at time `t = 0`. Its speed increase at the constant rate of `((pi)/(2)) bm//s^(2)`. If the net acceleration at `t = 2 sec` is `(pi)/(2) sqrt((1 + Npi^(2))` then what is the value of `N` ? |
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Answer» Correct Answer - 4 `R = 1m,7` `a_(t) = (dv)/(dt) = (pi)/(2) m//s^(2)` at `t = 0, u = 0, omega_(0) = 0` `alpha = (a_(1))/(R) = (pi)/(2) rad//s^(2)` `v = u + a_(t)t = 0 + (pi)/(2) xx 2 = pi m//s` `a_(t) = (pi)/(2)m//s^(2) , a_(C) = (v^(2))/(r) = pi^(2) m//s^(2)` `a = sqrt(a_(1)^(2) + a_(c)^(2)) = sqrt((pi^(2))/(4) + pi^(4)) = (pi)/(2)sqrt(1 + 4pi^(2)) m//s^(2)` Hene `N = 4` |
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