Question 151: Centre of Mass and Rotational Motion

Question

A ball is spun with angular acceleration $\alpha$ = 6t
2
$-$ 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 :

Accepted Answer

$\alpha  = {{d\omega } \over {dt}} = 6{t^2} - 2t$
$\int_0^\omega  {d\omega  = \int_0^t {(6{t^2} - 2t)dt} } $
so $\omega  = 2{t^3} - {t^2} + 10$
and ${{d	heta } \over {dt}} = 2{t^3} - {t^2} + 10$
so $\int_4^	heta  {d	heta  = \int_0^t {(2{t^3} - {t^2} + 10)dt} } $
$	heta  = {{{t^4}} \over 2} - {{{t^3}} \over 3} + 10t + 4$

Answer

${3 \over 2}{t^4} - {t^2} + 10t$

Answer

${{{t^4}} \over 2} - {{{t^3}} \over 3} + 10t + 4$

Answer

${{2{t^4}} \over 3} - {{{t^3}} \over 6} + 10t + 12$

Answer

$2{t^4} - {{{t^3}} \over 2} + 5t + 4$

Centre of Mass and Rotational Motion

A ball is spun with angular acceleration \alpha = 6t 2 - 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 :

Choose the correct answer: