A disk of radius $\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{\text{cm}}$is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (Fig. P9.59). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation $\mathit{a}\left(t\right)\mathbf{=}\mathit{A}\mathit{t}$, where$\mathit{t}$is in seconds and $\mathit{A}$is a constant. The cylinder starts from rest, and at the end of the third second, the ball’s acceleration is $\mathbf{1}\mathbf{.}\mathbf{80}\raisebox{1ex}{${\displaystyle \mathbf{\text{m}}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\mathbf{\text{s}}}^{\mathbf{2}}}$}\right.$. (a) Find $\mathit{A}$. (b) Express the angular acceleration of the disk as a function of time. (c) How much time after the disk has begun to turn does it reach an angular speed of $\mathbf{15}\mathbf{.}\mathbf{0}\raisebox{1ex}{${\displaystyle \mathbf{\text{rad}}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \mathbf{\text{s}}}$}\right.$? (d) Through what angle has the disk turned just as it reaches $\mathbf{15}\mathbf{.}\mathbf{0}\raisebox{1ex}{${\displaystyle \mathbf{\text{rad}}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \mathbf{\text{s}}}$}\right.$? (Hint: See Section 2.6.)

Angular motion is defined as, the movement of a body about a fixed point or a fixed axis. It is equal to the angle through which a line drawn to a body passes at a point or axis.

Consider the given data as below.

The radius of a disk, $R=25\text{cm}=0.25\text{m}$

Time, $t=3\text{s}$

Acceleration, $a=1.8\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^2$}\right.$

The equation of an acceleration of the ball is,

$a\left(t\right)=At$ ….. (1)

Substitute the known values in the above equation, and you get

role="math" localid="1662299252029" $$\begin{gathered} 1.8\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^2$}\right.=A\left(3\text{s}\right) \\ A=\frac{1.8{\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^2$}\right.}}{3\text{s}} \\ A=0.60\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^3$}\right. \end{gathered}$$

Hence, the value$A$ of is $0.60\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^3$}\right.$.

Calculate the angular acceleration of the disc by using the following formula.

$\begin{array}{rcl}\alpha \left(t\right)& =& \frac{a\left(t\right)}{R}\\ & =& \frac{At}{R}\end{array}$

Substitute known values in the above equation.

$\begin{array}{rcl}\alpha \left(t\right)& =& \frac{0.60{\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^3$}\right.}\times \left(t\right)}{0.25\text{m}}\\ & =& 2.4\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^3$}\right.\times \left(t\right)\end{array}$

Hence, the expression of the angular acceleration of the disk as a function of time is $\begin{array}{rcl}\alpha \left(t\right)& =& 2.4\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^3$}\right.\times \left(t\right)\end{array}$.

Angular speed is defined by,

$\omega \left(t\right)=\int \alpha \left(t\right)\text{dt}$

localid="1662299977421" $\begin{array}{rcl}\omega \left(t\right)& =& \int 2.4\times \left(t\right)\text{dt}\\ & =& \left(2.4\right)\int t\text{dt}\end{array}$

$\omega \left(t\right)=\left(2.4\right)\frac{t^2}{2}+{\text{C}}_{\text{1}}$ ..... (1)

Here, $C_1$is an integration constant.

At time $t=0$, the angular speed is,

$\omega \left(t\right)=0$

Apply this condition to equation (1), and you get

$$\begin{gathered} 0=0+{\text{C}}_{\text{1}} \\ {\text{C}}_{\text{1}}=0 \end{gathered}$$

Substitute the above value into equation (1).

$\omega \left(t\right)=\left(2.4\right)\frac{t^2}{2}$

$\omega \left(t\right)=1.2t^2$ ..... (2)

The given angular speed is,

$\omega \left(t\right)=15\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$

Substitute this value into equation (2).

localid="1662300532585" $\begin{array}{rcl}15\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{s}}$}\right.& =& 1.2t^2\end{array}$

$\begin{array}{rcl}{t}^2& =& \frac{{\displaystyle 15\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{s}}$}\right.}}{{\displaystyle 1.2\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^3}$}\right.}}\\ & =& 12.5{\text{s}}^2\end{array}$

$\begin{array}{rcl}t& =& \sqrt{12.5{\text{s}}^2}\\ & =& 3.535\text{s}\end{array}$

Hence, the required time is $3.535\text{s}$after the disk has begun to turn does it reach an angular speed of$15.0\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{s}}$}\right.$.

The angular displacement is defined by,

$\theta \left(t\right)=\int \omega \left(t\right)dt$

$\begin{array}{rcl}\theta \left(t\right)& =& \int \left(1.2\right)t^{2}dt\\ & =& \left(1.2\right)\int t^{2}dt\\ & =& \left(1.2\right)\frac{t^3}{3}+{\text{C}}_2\end{array}$

Here, role="math" localid="1662301275164" $C_2$is the integration constant.

At time $t=0$, the angular displacement is $\theta \left(t\right)=0$, and so$C_2=0$

Substitute the above value into the equation of angular displacement.

$\theta \left(t\right)=\left(1.2\right)\frac{t^3}{3}$

$\theta \left(t\right)=0.4t^3$ .... (3)

The time taken to reach on an angular speed is,

$t=3.535\text{s}$

Substitute this value into equation (3) as below.

$\begin{array}{rcl}\theta \left(t\right)& =& 0.4\times {\left(3.535\text{s}\right)}^3\\ & =& 0.4\times 44.2\\ & =& 17.67\text{rad}\end{array}$

Hence, the required angle is$17.67\text{rad}$.