Question 156: Centre of Mass and Rotational Motion

Question

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 :

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 = 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 :

Choose the correct answer: