91Ó°ÊÓ

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.

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

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

Acceleration, role="math" localid="1660324583139" $a=1.8\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.$

Equation of an acceleration of the ball is,

$a\left(t\right)=At$

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

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

Hence, the value of $A$is role="math" localid="1660324884629" $0.60\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^3$}\right.$.

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

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

Substitute known values in the above equation.

role="math" localid="1660325011334" $\begin{array}{rcl}\mathrm{Î\pm }\left(t\right)& =& \frac{0.60{\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^3$}\right.}×\left(t\right)}{0.25\text{m}}\\ & =& 2.4\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^3$}\right.×\left(t\right)\end{array}$

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

Angular speed is defined by,

$\mathrm{Ó\neg }\left(t\right)=∫\mathrm{Î\pm }\left(t\right)dt$

$\begin{array}{rcl}\mathrm{Ó\neg }\left(t\right)& =& ∫2.4×\left(t\right)dt\\ & =& \left(2.4\right)∫tdt\end{array}$

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

Here, ${\text{C}}_{\text{1}}$is integration constant.

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

$\mathrm{Ó\neg }\left(t\right)=0$

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).

$\mathrm{Ó\neg }\left(t\right)=\left(2.4\right)\frac{t^2}{2}$

$\mathrm{Ó\neg }\left(t\right)=1.2t^2$ ..... (2)

The given angular speed is,

role="math" localid="1660325446226" $\mathrm{Ó\neg }\left(t\right)=15\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$

Substitute this value into equation (2).

role="math" localid="1660325600646" $15\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\text{=1.2}{t}^2$

role="math" localid="1660325663526" $\begin{array}{rcl}{t}^2& =& \frac{15{\displaystyle \raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}}{1.2{\displaystyle \raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^3$}\right.}}\\ & =& 12.5{\text{s}}^2\end{array}$

role="math" localid="1660325675953" $\begin{array}{rcl}t& =& \sqrt{12.5{\text{s}}^2}\\ & =& 3.535\text{s}\end{array}$

Hence, the required time is $\begin{array}{rcl}& & 3.535\text{s}\end{array}$ 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,

$\mathrm{θ}\left(t\right)=∫\mathrm{Ó\neg }\left(t\right)dt$

role="math" localid="1660325941976" $\begin{array}{rcl}\mathrm{θ}\left(t\right)& =& ∫\left(1.2\right)t^{2}dt\\ & =& \left(1.2\right)∫t^{2}dt\\ & =& \left(1.2\right)\frac{t^3}{3}+{\text{C}}_2\end{array}$

Here, ${\text{C}}_{\text{2}}$is the integration constant.

At time $t=0$, the angular displacement is $\mathrm{θ}\left(t\right)=0$, and so $\begin{array}{rcl}& & {\text{C}}_2\text{=0}\end{array}$.

Substitute the above value into equation of angular displacement.

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

$\mathrm{θ}\left(t\right)=0.4t^3$ ..... (3)

The time taken to reach an angular speed is,

$t=3.535\text{s}$

Substitute this value into equation (3) as below.

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

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