#### Answer

(a) $A = 1.50~rad/s^2$
$B = 1.10~rad/s^4$
(b) (i) $L = 58.9~kg~m^2/s~$
(ii) $\tau = 56.1~N~m$

#### Work Step by Step

(a) $\theta(t) = (At^2+Bt^4)~rad$
The units of $At^2$ are $radians$.
Thus $A$ has units of $rad/s^2$
$A = 1.50~rad/s^2$
The units of $Bt^4$ are $radians$.
Thus $B$ has units of $rad/s^4$
$B = 1.10~rad/s^4$
(b) (i) We can find the moment of inertia of the sphere.
$I = \frac{2}{3}mR^2$
$I = \frac{2}{3}(12.0~kg)(0.240~m)^2$
$I = 0.4608~kg~m^2$
We can find the angular velocity at time $t$.
$\omega(t) = \frac{d\theta}{dt}$
$\omega(t) = (2At+4Bt^3)~rad/s$
We can find the angular velocity at $t = 3.00~s$:
$\omega = [(2)(1.50~rad/s^2)(3.00~s)+(4)(1.10~rad/s^4)(3.00~s)^3)]$
$\omega = 127.8~rad/s$
We can find the angular momentum at $t = 3.00~s$:
$L = I\omega$
$L = (0.4608~kg~m^2)(127.8~rad/s)$
$L = 58.9~kg~m^2/s~$
(ii) We can find the angular acceleration at time $t$.
$\alpha(t) = \frac{d\omega}{dt}$
$\alpha(t) = (2A+12Bt^2)~rad/s^2$
We can find the angular acceleration at $t = 3.00~s$.
$\alpha = (2)(1.50~rad/s^2)+(12)(1.10~rad/s^4)(3.00~s)^2$
$\alpha = 121.8~rad/s^2$
We can find the torque at $t = 3.00~s$:
$\tau = I \alpha$
$\tau = (0.4608~kg~m^2)(121.8~rad/s^2)$
$\tau = 56.1~N~m$