## University Physics with Modern Physics (14th Edition)

(a) $A = 1.50~rad/s^2$ $B = 1.10~rad/s^4$ (b) (i) $L = 58.9~kg~m^2/s~$ (ii) $\tau = 56.1~N~m$
(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 $t = 3.00~s$: $\tau = I \alpha$ $\tau = (0.4608~kg~m^2)(121.8~rad/s^2)$ $\tau = 56.1~N~m$