JEE MAIN - Physics (2022 - 28th June Evening Shift - No. 2)
A ball is spun with angular acceleration $$\alpha$$ = 6t2 $$-$$ 2t where t is in second and $$\alpha$$ is in rads$$-$$2. At t = 0, the ball has angular velocity of 10 rads$$-$$1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is :
$${3 \over 2}{t^4} - {t^2} + 10t$$
$${{{t^4}} \over 2} - {{{t^3}} \over 3} + 10t + 4$$
$${{2{t^4}} \over 3} - {{{t^3}} \over 6} + 10t + 12$$
$$2{t^4} - {{{t^3}} \over 2} + 5t + 4$$
Explanation
$$\alpha = {{d\omega } \over {dt}} = 6{t^2} - 2t$$
$$\int_0^\omega {d\omega = \int_0^t {(6{t^2} - 2t)dt} } $$
so $$\omega = 2{t^3} - {t^2} + 10$$
and $${{d\theta } \over {dt}} = 2{t^3} - {t^2} + 10$$
so $$\int_4^\theta {d\theta = \int_0^t {(2{t^3} - {t^2} + 10)dt} } $$
$$\theta = {{{t^4}} \over 2} - {{{t^3}} \over 3} + 10t + 4$$
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