InterviewSolution
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InterviewSolution
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The tangential acceleration of a particle moving in a circular path of radius `5 cm is 2 m s^-2`. The angular velocity of the particle increases from `10 rad s^-1 "to" 20 rad s^-1` during some time. Find (a) this duration od time and (b) the number of revolutions completed during this time. |
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Answer» `r = 5 cm = 0.05 m, a_t = 2 ms^-2, omega_1 = 10 rad s^-1` `omega_2 = 20 rad s^-1, prop = a_t//r = 2//0.05 = 40 rad s^-2` (a) `omega_2 = omega_1 + omega_t rArr 20 = 10 + 40 t rArr t = 0.25 s` (b) `theta = omega_1 t + (1)/(2) prop t^2 = 10 xx (1)/(4) + (1)/(2) xx 40 xx ((1)/(4))^2 = (15)/(4) rad` Number of revolutions = `(theta)/( 2pi) = (15)/( 4 xx 2 pi) = 0.6`. |
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