Question

As a result of friction, the angular speed of a wheel changes with time according to$\frac{d\theta }{dt}={\omega }_o{e}^{-\sigma T}$where ${\omega }_o$ and $\sigma$are constants. The angular speed changes from 3.50 rad/s at t = 0 to 2.00 rad/s at t= 9.30 s. (a) Use this information to determine$\sigma$and${\omega }_o$. Then determine (b) the magnitude of the angular acceleration at $t=3.00s$, (c) the number of revolutions the wheel makes in the first $2.50s$, and (d) the number of revolutions it makes before coming to rest.

Short answer

The solution is

a) The value of constants are${\omega }_o=3.5rad/s$ ,$\sigma =0.0602s^{-1}$

b) The magnitude of the angular acceleration is$\alpha =-0.176rad/s^2$

c) The number of revolutions the wheel makes in the first 2.50 s is $N=1.29$,

d) The number of revolutions wheel makes before coming to rest$N=9.258$

Step-by-step solution

Given Information

$\frac{d\theta }{dt}={\omega }_o{e}^{-\sigma t}$

Since$\frac{dv}{dt}=\omega$( angular velocity )

So that;$\omega ={\omega }_o{e}^{-\sigma t}\mathrm{...}\left(1\right)$

 Step 2: Determining the constants

(a)

At $t=0$, $\omega =3.5\text{\hspace{0.17em}rad/s}$

So,

$\begin{array}{l}3.5={\omega }_o{e}^{-\sigma \times 0}\\ 3.5={\omega }_o{e}^0\end{array}$$(since{e}^0=1)$

${\omega }_o=3.5\text{\hspace{0.17em}rad/s}$

At $t=9.3s$, $\omega =2rad/s$

So, from equation (1), we have,

$\begin{array}{l}2=3.5e^{-\sigma \times 9.3}\\ e^{9.3\sigma }=\frac{3.5}{2}\end{array}$

Take ‘ln’ both sides

$\begin{array}{c}\ln e^{9.3\sigma }=\ln \left(\frac{3.5}{2}\right)\\ 9.3\sigma =\ln \left(\frac{3.5}{2}\right)\end{array}$

Since, ln e =1, therefore

$\sigma =0.0602s^{-1}$

Calculating the angular acceleration

(b)

$\begin{array}{c}\alpha =\frac{d\omega }{dt}\\ =\frac{d({\omega }_o{e}^{-\sigma t})}{dt}\\ =({\omega }_o{e}^{-\sigma t}\left)\right(-\sigma \left)\right(1)\\ =-\sigma {\omega }_o{e}^{-\sigma t}\end{array}$

Angular acceleration at $t=3$s,

$\begin{array}{l}\alpha =-\left(0.0602\right)\left(3.5\right)e^{-0.0602\times 3}\\ \alpha =-0.176{\text{\hspace{0.17em}rad/s}}^2\end{array}$

Magnitude of the angular acceleration is$0.176rad/s^2$

the number of revolutions the wheel makes in the first 2.50 s.

(c)

We have,

$\begin{array}{l}\frac{dv}{dt}={\omega }_t{e}^{-\sigma t}\\ {\int }_0^{\theta }d\theta ={\int }_0^t{\omega }_0{e}^{-\sigma \dot{t}}dt\\ {{\left[0\right]}^{\theta }}_0={\omega }_0{\left[\frac{e^{-\sigma t}}{-\sigma }\right]}_0^t\end{array}$

Simplifying for the angle,

$\begin{array}{l}\theta =\frac{-{\omega }_o}{\sigma }(e^{{}^{-\sigma t}}-e^0)\\ \theta =\frac{{\omega }_o}{\sigma }(1-e^{-\sigma t})]\mathrm{...........}\left(2\right)\end{array}$

In first 2.50 s, the angle is

$\begin{array}{l}\theta =\frac{3.5}{0.0602}[1-e^{-0.0602\times 2.5}]\\ \theta =8\times 12326\text{radian}\end{array}$

So, number of revolution

$\begin{array}{l}N=\frac{\theta }{2\pi }=\frac{8.12326}{2\pi }\\ N=1.29\end{array}$

Hence, the number of revolution is 1.29.

The number of revolutions wheel makes before coming to rest

(d)

If wheel comes to rest then

$\omega =0$from equation (2)

We have,

$0={\omega }_o{e}^{-\sigma t}$

Only values satisfy is

$t=\infty$

From equation (2), we have

$\theta =\frac{{\omega }_o}{\sigma }(1-e^{-\infty })$

$\begin{array}{c}\theta =\frac{{\omega }_o}{\sigma }\\ =\frac{3.5}{0.0602}\\ =58.14\text{\hspace{0.17em}rad}\end{array}$

So, the number of revolution

$\begin{array}{c}N=\frac{Q}{2\pi }\\ =\frac{58.14}{2\pi }\\ =9.258\end{array}$

Hence, the number of revolution is 9.258.