A flywheel with a diameter of$\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{ }\mathbf{\text{m}}\mathbf{}$ is rotating at an angular speed of $\mathbf{200}\mathbf{ }\raisebox{1ex}{${\displaystyle \mathbf{\text{rev}}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \mathbf{\text{min}}}$}\right.$.

(a) What is the angular speed of the flywheel in radians per second?

(b) What is the linear speed of a point on the rim of the flywheel?

(c) What constant angular acceleration (in revolutions per minute-squared) will increase the wheel’s angular speed to $\mathbf{1000}\mathbf{ }\raisebox{1ex}{${\displaystyle \mathbf{\text{rev}}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \mathbf{\text{min}}}$}\right.$in$\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{s}}$ ?

(d) How many revolutions does the wheel make during that$\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{s}}$ ?

1. The initial angular speed of the flywheel is${\omega }_0=200\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.$ .
2. The diameter of flywheel is $d=1.20 \text{m}$.
3. The final angular speed after$t=60 \text{s}$is$1000\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.$.
4. The time for which the flywheel accelerates t is $60 \text{s}$.

The flywheel exhibits rotational motion about an axis passing through its axel. Hence, we can determine its angular speed, acceleration and linear speed of the points on its rim using rotational kinematic equations which are given as follows.

$$\begin{gathered} \omega =\frac{d\theta }{dt} \\ v=r\omega \\ \omega ={\omega }_0+\alpha t \\ \theta -{\theta }_0=\frac{1}{2}\left(\omega +{\omega }_0\right)t \end{gathered}$$

The initial angular velocity is given in the units of revolution/min. we will change the units to rad/s to gettherequired value.

$\begin{array}{rcl}\omega & =& 200\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.\\ & =& 200\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.\times \frac{2\pi \text{rad}}{1 \text{rev}}\times \frac{1 \text{min}}{60.0 \text{s}}\\ & =& 6.66\raisebox{1ex}{$\pi {\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\\ \omega & =& 20.9\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\\ & & \end{array}$

Hence the angular speed is $20.9 \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$, .

Step 3: (b) Calculation for thelinear speed of a point on the rim of the flywheel

We are given the diameter of the flywheel, so first we calculate the radius as

$\begin{array}{rcl}r& =& \frac{\text{d}}{2}\\ & =& \frac{1.20 \text{m}}{2}\\ r& =& 0.60 \text{m}\\ & & \end{array}$

Hence, we calculate linear velocity using the value of angular velocity as

$v=r\omega$

Substitute all the value in the above equation

$$\begin{gathered} v=0.60 \text{m×}20.9\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ v=12.5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Hence the linear speed is, $12.5 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

The flywheel accelerates from initial angular speed to final angular speed of$1000 \raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.$To determine the angular acceleration of the flywheel in this duration, we use the rotational kinematic equation.

Since the answer is expected in the units of $\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{min}}^{\text{2}}}$}\right.$ , we use ω₀ in the units of and time $t=60.0 \text{s}=1 \text{min}$.

$$\begin{gathered} \omega ={\omega }_0+\alpha t \\ \alpha =\frac{\omega -{\omega }_0}{t} \end{gathered}$$

Substitute all the value in the above equation

$$\begin{gathered} \alpha =\frac{{\displaystyle 1000\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min-200}\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.}$}\right.}}{{\displaystyle 1 \text{min}}} \\ \alpha =800\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{min}}^{\text{2}}}$}\right. \end{gathered}$$

Hence the angular acceleration is, .

The number of revolutions in this time interval can be calculated using the rotational kinematic equation.

$\theta -{\theta }_0=\frac{1}{2}\left(\omega +{\omega }_0\right)t$

We assume$n=\theta -{\theta }_0$from start of the interval

We also use the values of initial and final angular speeds in the units of rev/min

Substitute all the value in the above equation

$$\begin{gathered} n=\frac{1}{2}\left(1000\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min+}200\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.}$}\right.\right)\times 1 \text{min} \\ =600 \text{rev} \end{gathered}$$

Hence the number of revolution is, $600 \text{rev}$.