A ball is spun with angular acceleration $\alpha$ = 6t2 $-$ 2t where t is in second and $\alpha$ is in rads$-$2. At t = 0, the ball has angular velocity of 10 rads$-$1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is :
Solution
<p>$\alpha = {{d\omega } \over {dt}} = 6{t^2} - 2t$</p>
<p>$\int_0^\omega {d\omega = \int_0^t {(6{t^2} - 2t)dt} }$</p>
<p>so $\omega = 2{t^3} - {t^2} + 10$</p>
<p>and ${{d\theta } \over {dt}} = 2{t^3} - {t^2} + 10$</p>
<p>so $\int_4^\theta {d\theta = \int_0^t {(2{t^3} - {t^2} + 10)dt} }$</p>
<p>$\theta = {{{t^4}} \over 2} - {{{t^3}} \over 3} + 10t + 4$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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