A hollow, thin-walled sphere of mass $\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{\text{kg}}$and diameter $\mathbf{48}\mathbf{.}\mathbf{0}\mathbf{\text{cm}}$is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by $\mathit{\theta }\left(t\right)\mathbf{=}\mathit{A}{\mathit{t}}^{\mathbf{2}}\mathbf{+}\mathit{B}{\mathit{t}}^{\mathbf{4}}$, where $\mathit{A}$has numerical value $\mathbf{1}\mathbf{.}\mathbf{50}$ and has numerical value $\mathbf{1}\mathbf{.}\mathbf{10}$. (a) What are the units of the constants $\mathit{A}$and $\mathit{B}$? (b) At the time $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\text{s}}$, find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.

Angular velocity is measured in angles per unit time or in radians per second. The rate of change of angular velocity is angular acceleration.

Consider the known data as below.

Mass of a hollow, thin-walled sphere, $M=12\text{kg}$

Diameter of a hollow, thin walled sphere,$D=48\text{cm}$

Radius of a hollow, thin walled sphere is,

$\begin{array}{rcl}R& =& \frac{D}{2}\\ & =& \frac{48\text{cm}}{2}\\ & =& 24\text{cm}\\ & =& 0.24\text{m}\end{array}$

Equation of angle, $\theta \left(t\right)=A{t}^2+B{t}^4$

Here, the constants $A$ and $B$ are $1.50$and $1.10$respectively.

Determine the unit of constant $A$as below.

role="math" localid="1662313812718" $\text{Unit of}\left(A{t}^2\right)=\text{Unit of}\theta \left(t\right)$

role="math" localid="1662313801781" $\begin{array}{rcl}\text{Unit of}A& =& \frac{\text{Unit of}\theta \left(t\right)}{\text{Unit of}{t}^2}\\ & =& \frac{\text{rad}}{{\text{s}}^{\text{2}}}\end{array}$

Hence, the unit of a constant $A$is role="math" localid="1662313835173" $\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^2$}\right.$.

Determine the unit of constant $B$as below.

$\text{Unit of}\left(B{t}^4\right)=\text{Unit of}\theta \left(t\right)$

$\begin{array}{rcl}\text{Unit of}B& =& \frac{\text{Unit of}\theta \left(t\right)}{\text{Unit of}{t}^4}\\ & =& \frac{\text{rad}}{{\text{s}}^4}\end{array}$

Hence, the unit of a constant $B$is$\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^4$}\right.$.

Now, calculate the angular acceleration by using the following equation.

${\omega }^2={\omega }_0^2+2\alpha \theta$

Substitute $\left(\frac{v}{r}\right)$for $\omega$and $0$for ${\omega }_0$in the above equation.

${\left(\frac{v}{r}\right)}^2=0+2\alpha \theta$

$\alpha =\frac{v^2}{2\theta r^2}$ ..... (1)

Convert the angular displacement into radian,

localid="1662314280616" $\begin{array}{rcl}\theta & =& \left(20.0\text{rev}\right)\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\\ & =& 20\times 2\pi \text{rad}\\ & =& 40\pi \text{rad}\end{array}$

Substitute known values in the above equation.

localid="1662314579614" $\begin{array}{rcl}\alpha & =& \frac{{\left(15.336{\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}\right)}^2}{2\left(40\pi \text{rad}\right){\left(0.260\text{m}\right)}^2}\\ & =& \frac{235.193{\left({\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}\right)}^2}{\left(5.408\pi \text{rad}·{\text{m}}^2\right)}\\ & =& 13.8\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^2$}\right.\end{array}$

Hence, the required angular acceleration is $\begin{array}{rcl}& & 13.8\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^2$}\right.\end{array}$.