Exercises 10.39: 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 $\mathit{B}$ 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 of time or in radians per second. The rate of change of angular velocity is angular acceleration.

Consider the known data below.

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

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

The 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.

$\text{Unit of}\left(A{t}^2\right)=\text{Unit of}\theta \left(t\right)$

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

Hence, the unit of a constant $A$is role="math" localid="1661883966008" $\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{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,

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

Substitute $40\pi \text{rad}$ for $\theta$, role="math" localid="1661883746304" $15.336\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$for $v$, and $0.260\text{m}$for $r$into equation (1).$40\pi \text{rad}$

role="math" localid="1661883893997" $\begin{array}{rcl}\alpha & =& \frac{{\left(15.336{\displaystyle \raisebox{1ex}{$\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}$}\right.}\right)}^2}{5.408\pi \text{rad}·{\text{m}}^2}\\ & =& 13.8\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\end{array}$

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