Chapter 10, Problem 10.89CP

As a result of friction, the angular speed of a whorl changes with time according to

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

where ω₀ and σ 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 σ and ω_0. Then determine (b) the magnitude of the angular acceleration at t = 3.00 s, (c) the number of revolutions the wheel makes in the first 2.50 s, and (d) the number of revolutions it makes before coming to rest.

To determine

The value of $\sigma$ and ${\omega }_0$ from the given information.

Answer to Problem 10.89CP

The value of ${\omega }_0$ is $3.50\text{rad}/\text{s}$ and the value of $\sigma$ is $0.0602{\text{s}}^{-1}$.

Explanation of Solution

Given info: The initial angular speed of the wheel is $3.50\text{rad}/\text{s}$ at time $t=0$ and the final angular speed of the wheel is $2.00\text{rad}/\text{s}$ at time $9.30\text{s}$.

The given expression for the change in the angular in the angular speed of the wheel with respect to time is,

$\begin{array}{c}\frac{d\theta }{dt}={\omega }_0{e}^{-\sigma t}\\ \omega ={\omega }_0{e}^{-\sigma t}\end{array}$ (1)

Here,

$\sigma$ and ${\omega }_0$ are constant.

$\omega$ is the angular speed of the wheel.

Substitute $3.50\text{rad}/\text{s}$ for $\omega$ and $0$ for $t$ in the equation (1).

$\begin{array}{c}3.50\text{rad}/\text{s}={\omega }_0{e}^{-\sigma \left(0\right)}\\ {\omega }_0{e}^0=3.50\text{rad}/\text{s}\\ {\omega }_0\left(1\right)=3.50\text{rad}/\text{s}\\ {\omega }_0=3.50\text{rad}/\text{s}\end{array}$

Thus, the value of ${\omega }_0$ is $3.50\text{rad}/\text{s}$.

Substitute $3.50\text{rad}/\text{s}$ for ${\omega }_0$, $2.00\text{rad}/\text{s}$ for $\omega$ and $9.3\text{s}$ for $t$ in the equation (1).

$\begin{array}{c}2.00\text{rad}/\text{s}=\left(3.50\text{rad}/\text{s}\right)e^{-\sigma \left(9.3\text{s}\right)}\\ \frac{2.00\text{rad}/\text{s}}{3.50\text{rad}/\text{s}}=e^{-\sigma \left(9.3\text{s}\right)}\end{array}$

Take log on both side of the above equations.

$\begin{array}{c}\log \left(\frac{2.00\text{rad}/\text{s}}{3.50\text{rad}/\text{s}}\right)=\log \left(e^{-\sigma \left(9.3\text{s}\right)}\right)\\ -0.243=-\sigma \left(9.3\text{s}\right)\left(\log e\right)\\ \sigma =0.06017{\text{s}}^{-1}\\ \approx 0.0602{\text{s}}^{-1}\end{array}$

Conclusion:

Therefore, the value of ${\omega }_0$ is $3.50\text{rad}/\text{s}$ and the value of $\sigma$ is $0.0602{\text{s}}^{-1}$.

To determine

The magnitude of the angular acceleration at time $t=3.00\text{s}$.

Answer to Problem 10.89CP

The magnitude of the angular acceleration at time $t=3.00\text{s}$ is $-0.176{\text{rad}/\text{s}}^2$.

Explanation of Solution

Given info: The initial angular speed of the wheel is $3.50\text{rad}/\text{s}$ at time $t=0$ and the final angular speed of the wheel is $2.00\text{rad}/\text{s}$ at time $9.30\text{s}$.

The given expression for the change in the angular in the angular speed of the wheel with respect to time is,

$\begin{array}{c}\frac{d\theta }{dt}={\omega }_0{e}^{-\sigma t}\\ \omega ={\omega }_0{e}^{-\sigma t}\end{array}$

The rate of change of the angular velocity with respect to time is known as the angular acceleration.

Differentiate the equation (1) to find the angular acceleration.

$\begin{array}{c}\frac{d\omega }{dt}=\frac{d}{dt}\left({\omega }_0{e}^{-\sigma t}\right)\\ =\left(-\sigma t{\omega }_0\right)e^{-\sigma t}\end{array}$

Substitute $3.00\text{s}$ for $t$ and $0.0602{\text{s}}^{-1}$ for $\sigma$ in the above equation.

$\begin{array}{c}\frac{d\omega }{dt}=-\left(0.0602{\text{s}}^{-1}\right)\left(3.00\text{s}\right)\left(3.50\text{rad}/\text{s}\right)e^{-\left(0.0602{\text{s}}^{-1}\right)\left(3.00\text{s}\right)}\\ =-0.17579{\text{rad}/\text{s}}^2\\ \approx -0.176{\text{rad}/\text{s}}^2\end{array}$

Conclusion:

Therefore, the magnitude of the angular acceleration at time $t=3.00\text{s}$ is $-0.176{\text{rad}/\text{s}}^2$.

To determine

The number of revolution made by the wheel in first $2.50\text{s}$.

Answer to Problem 10.89CP

The number of revolution made by the wheel in first $2.50\text{s}$ is $1.29\text{rev}$.

Explanation of Solution

Given info: The initial angular speed of the wheel is $3.50\text{rad}/\text{s}$ at time $t=0$ and the final angular speed of the wheel is $2.00\text{rad}/\text{s}$ at time $9.30\text{s}$.

The given expression for the change in the angular in the angular speed of the wheel with respect to time is,

$\begin{array}{c}\frac{d\theta }{dt}={\omega }_0{e}^{-\sigma t}\\ d\theta =\left({\omega }_0{e}^{-\sigma t}\right)dt\end{array}$

Integrate both side of the above equation,

$\begin{array}{c}{\displaystyle \int d\theta }={\displaystyle \int {\omega }_0{e}^{-\sigma t}dt}\\ \theta =\left(\frac{{\omega }_0}{\sigma }\right)\left(1-e^{-\sigma t}\right)\end{array}$ (2)

From part (a) the value of ${\omega }_0$ is $3.50\text{rad}/\text{s}$ and the value of $\sigma$ is $0.0602{\text{s}}^{-1}$.

Substitute $3.50\text{rad}/\text{s}$ for ${\omega }_0$, $0.0602{\text{s}}^{-1}$ for $\sigma$ and $2.50\text{s}$ for $t$ in the equation (1).

$\begin{array}{c}\theta =\left(\frac{3.50\text{rad}/\text{s}}{0.0602{\text{s}}^{-1}}\right)\left(1-e^{-\left(0.0602{\text{s}}^{-1}\right)\left(2.5\text{s}\right)}\right)\\ =0.81234\text{rad}\end{array}$

The formula to calculate the number of revolution is,

$n=\frac{\theta }{2\pi }$ (3)

Here,

$n$ is the number of revolution made by the wheel.

Substitute $0.81234\text{rad}$ for $\theta$ in the above equation.

$\begin{array}{c}n=\frac{8.1234\text{rad}}{2\pi }\\ =1.2928\text{rev}\\ =1.29\text{rev}\end{array}$

Conclusion:

Therefore, the number of revolution made by the wheel in first $2.50\text{s}$ is $1.29\text{rev}$.

To determine

The number of revolution made by the wheel before its comes to rest.

Answer to Problem 10.89CP

The number of revolution made by the wheel before its comes to rest is $9.26\text{rev}$.

Explanation of Solution

Given info: The initial angular speed of the wheel is $3.50\text{rad}/\text{s}$ at time $t=0$ and the final angular speed of the wheel is $2.00\text{rad}/\text{s}$ at time $9.30\text{s}$.

The given expression for the change in the angular in the angular speed of the wheel with respect to time is,

$\frac{d\theta }{dt}={\omega }_0{e}^{-\sigma t}$

The time when the when comes to rest final velocity is $0$.

Substitute $0$ for $\frac{d\theta }{dt}$ in the above equation.

$\begin{array}{l}0={\omega }_0{e}^{-\sigma t}\\ e^{-\sigma t}=0\end{array}$

Substitute $3.50\text{rad}/\text{s}$ for ${\omega }_0$, $0$ for $e^{-\sigma t}$ and $0.0602{\text{s}}^{-1}$ for $\sigma$ in the equation (2).

$\begin{array}{c}\theta =\left(\frac{3.50\text{rad}/\text{s}}{0.0602{\text{s}}^{-1}}\right)\left(1-0\right)\\ =58.1395\text{rad}\\ \approx \text{58}\text{.14}\text{rad}\end{array}$

From part (a) the value of ${\omega }_0$ is $3.50\text{rad}/\text{s}$ and the value of $\sigma$ is $0.0602{\text{s}}^{-1}$.

Substitute $3.50\text{rad}/\text{s}$ for ${\omega }_0$, $0.0602{\text{s}}^{-1}$ for $\sigma$ and $2.50\text{s}$ for $t$ in the equation (1).

$\begin{array}{c}\theta =\left(\frac{3.50\text{rad}/\text{s}}{0.0602{\text{s}}^{-1}}\right)\left(1-e^{-\left(0.0602{\text{s}}^{-1}\right)\left(2.5\text{s}\right)}\right)\\ =0.81234\text{rad}\end{array}$

Substitute $\text{58}\text{.14}\text{rad}$ for $\theta$ in the above equation.

$\begin{array}{c}n=\frac{\text{58}\text{.14}\text{rad}}{2\pi }\\ =9.2533\text{rev}\\ =9.26\text{rev}\end{array}$

Conclusion:

Therefore, the number of revolution made by the wheel before its comes to rest is $9.26\text{rev}$.