Q39E - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

The angular acceleration is $13.8\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^2$}\right.$.

Definition:

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

Given data:

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.

(i) Define the unit of the constants:

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 $\raisebox{1ex}{$\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$}\right.$.

Define angular acceleration:

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 known values in the above equation.

$\begin{array}{rcl}\alpha & =& \frac{{\left(15.336{\displaystyle \raisebox{1ex}{$\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$}\right.\end{array}$.