Starting from rest, a DVD steadily accelerates to $500\text{rpm}$in $1.0\text{s}$, rotates at this angular speed for $3.0\text{s}$, then steadily decelerates to a halt in $2.0\text{s}$. How many revolutions does it make?

The DVD steady accelerate to $500\text{rpm}$in $1.0\text{s}$and rotates for$3.0\text{s}$at this angular velocity, then steadily decelerate to a halt in$2.0\text{s}$.

Formula to calculate the total angular displacement of the DVD is,

$\theta ={\theta }_1+{\theta }_2+{\theta }_3\dots \mathrm{..}\left(\text{l}\right)$

Here,$\theta$ is the total angular displacement of the DVD, ${\theta }_1$is the angular displacement of the DVD in initial steady acceleration period, ${\theta }_2$is the angular displacement of the DVD in the period of constant angular velocity, and${\theta }_3$ is the angular displacement of the DVD in the decelerate period.

The angular displacement of the DVD during the steady acceleration is,

${\theta }_1={\omega }_o{\left(\Delta t\right)}_1+\frac{1}{2}\left(\frac{{\omega }_1-{\omega }_o}{{\left(\Delta t\right)}_1}\right){\left(\Delta t\right)}_1^2$

Here, ${\omega }_0$is the initial angular velocity of the DVD, ${\left(\Delta t\right)}_1$is the time period of steady acceleration of the DVD, and ${\omega }_1$is the angular velocity of the DVD at the end of $t=1\text{s}$.

The angular displacement of the DVD during the constant angular velocity is, ${\theta }_2={\omega }_1{\left(\Delta t\right)}_2$

Here, ${\left(\Delta t\right)}_2$is the time period of constant angular velocity of the DVD.

The angular displacement of the DVD during the constant deceleration is,

${\theta }_3={\omega }_1{\left(\Delta t\right)}_3+\frac{1}{2}\left(\frac{{\omega }_3-{\omega }_1}{{\left(\Delta t\right)}_3}\right){\left(\Delta t\right)}_3^2$${\left(\Delta t\right)}_3$

Here, ${\left(\Delta t\right)}_3$is the time period of steady deceleration of the DVD, and ${\omega }_3$is the angular velocity of the DVD at the end.

Substitute

${\omega }_o{\left(\Delta t\right)}_1+\frac{1}{2}\left(\frac{{\omega }_1-{\omega }_o}{{\left(\Delta t\right)}_1}\right){\left(\Delta t\right)}_1^2$for ${\theta }_1,{\omega }_1{\left(\Delta t\right)}_2$ for ${\theta }_2$and

${\omega }_1{\left(\Delta t\right)}_3+\frac{1}{2}\left(\frac{{\omega }_3-{\omega }_1}{{\left(\Delta t\right)}_3}\right){{\left(\Delta t\right)}_3}^2$for ${\theta }_3$in the equation (I) to find $\theta$.

$\theta =\left[\begin{array}{l}{\omega }_o{\left(\Delta t\right)}_1+\frac{1}{2}\left(\frac{{\omega }_1-{\omega }_o}{{\left(\Delta t\right)}_1}\right){\left(\Delta t\right)}_1^2+{\omega }_1{\left(\Delta t\right)}_2+{\omega }_1{\left(\Delta t\right)}_3+\frac{1}{2}\left(\frac{{\omega }_3-{\omega }_1}{{\left(\Delta t\right)}_3}\right){\left(\Delta t\right)}_3^2\end{array}\right]\dots \dots (\text{(II)}$

$=\left[\begin{array}{l}\left(0\text{rpm}\right)(1.0\text{s})+\frac{1}{2}\left(\frac{500\text{rpm}\times \frac{1\text{min}}{60\text{s}}-0\text{rpm}}{(1.0\text{s})}\right)+\left(500\text{rpm}\times \frac{1\text{min}}{60\text{s}}\right)(3.0\text{s})\\ +\left(500\text{rpm}\times \frac{1\text{min}}{60\text{s}}\right)(2.0\text{s})+\frac{1}{2}\left(\frac{0\text{rpm}-500\text{rpm}}{(2.0\text{s})}\right)\end{array}\right]$

$\begin{array}{l}=4.17\text{rev}+25\text{rev}+8.33\text{rev}\\ =38\text{rev}\end{array}$

Therefore, the number of revolution made by the DVD during the motion is$38\text{rev}$.