Question

A fan blade rotates with angular velocity given by ${\mathbf{\omega }}_{\mathbf{z}}\mathbf{=}\mathbf{\gamma }\mathbf{-}{\mathbf{\beta t}}^{\mathbf{2}}$, where $\mathbf{\gamma }\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{rad}\mathbf{/}\mathbf{s}$and $\mathbf{\beta }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{800}\mathbf{rad}\mathbf{/}{\mathbf{s}}^{\mathbf{3}}$. (a) Calculate the angular acceleration as a function of time. (b) Calculate the instantaneous angular acceleration ${\mathbf{\alpha }}_{\mathbf{z}}\mathbf{}\mathbf{at}\mathbf{}\mathbf{t}\mathbf{=}\mathbf{3}\mathbf{s}$and the average angular acceleration for the time interval $\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{to}\mathbf{}\mathbf{t}\mathbf{=}\mathbf{3}\mathbf{}\mathbf{s}$. How do these two quantities compare? If they are different, why?

Short answer

1. The angular acceleration as a function of time is${\alpha }_z\left(t\right)=-1.6t$.
2. The instantaneous angular acceleration is ${\mathrm{\alpha }}_{\mathrm{z}}\left(\mathrm{t}\right)=-1.6\mathrm{t}-4.8\mathrm{rad}/{\mathrm{s}}^2$, average angular acceleration is $-2.4\mathrm{rad}/{\mathrm{s}}^2$, the instantaneous angular acceleration is twice of average angular acceleration, and they are different because the acceleration changes linearly with time.

Step-by-step solution

Identification of given data

The angular velocity of the flywheel obeys the following equation.

${\omega }_z\left(t\right)=A+B{t}^2$

Where, A = 2.75 and B = 1.50.

The first radius is$r_a=2.5m$ .

Concept/Significance of acceleration

$\mathbf{\alpha }\left(t\right)\mathbf{=}\frac{{\mathbf{d\omega }}_{\mathbf{z}}\left(t\right)}{\mathbf{dt}}$The derivative of the angular velocity is called angular acceleration.

Here,$\mathbf{\alpha }\left(t\right)$ is angular acceleration, and${\mathbf{\omega }}_{\mathbf{z}}\left(t\right)$ is angular velocity.

Determine the angular acceleration as a function of time(a)

The angular acceleration is given by,

${\alpha }_z\left(t\right)=\frac{d{\omega }_z\left(t\right)}{dt}$

Substitute ${\omega }_z\left(t\right)=\gamma -\beta t^2$in the above equation.

$$\begin{gathered} {\alpha }_z\left(t\right)=\frac{d}{dt}\left[\gamma -\beta t^2\right] \\ =-2\beta t \\ =-2\left(0.8\right)t \\ =-1.6t \end{gathered}$$ .......................(1)

Therefore, the angular acceleration as a function of time is ${\alpha }_z\left(t\right)=-1.6t$.

Determine the instantaneous angular acceleration at t = 3 s and the average angular acceleration αav-z for the time interval to t = 0 to t = 3 s , and compare the results.(b)

Substitute in equation .

$$\begin{gathered} {\mathrm{\alpha }}_{\mathrm{z}}\left(\mathrm{t}\right)=-1.6\left(3\mathrm{s}\right) \\ =-4.8\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Find the average angular acceleration as follows.

$$\begin{gathered} {\mathrm{\alpha }}_{\mathrm{av}-\mathrm{z}}=\frac{{\mathrm{\omega }}_{\mathrm{z}}\left({\mathrm{t}}_1\right)-{\mathrm{\omega }}_{\mathrm{z}}\left({\mathrm{t}}_0\right)}{{\mathrm{t}}_1-{\mathrm{t}}_0} \\ =\frac{\left(\mathrm{\gamma }-{\mathrm{\beta \beta t}}_1^2\right)-\left(\mathrm{\gamma }-{\mathrm{\beta \beta t}}_0^2\right)}{{\mathrm{t}}_1-{\mathrm{t}}_0} \\ =-\mathrm{\beta }\left(\frac{{\mathrm{t}}_1^2-{\mathrm{t}}_0^2}{{\mathrm{t}}_1-{\mathrm{t}}_0}\right) \end{gathered}$$

Substitute all the values in the above expression.

$$\begin{gathered} {\mathrm{\alpha }}_{\mathrm{av}-\mathrm{z}}=-\left(0.8\mathrm{rad}/{\mathrm{s}}^3\right)\left(\frac{3^2-0^2}{3-0}\right) \\ =-2.4\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

The average angular acceleration and the instantaneous angular acceleration are different, and the instantaneous angular acceleration is twice the size of average angular acceleration. They are different because the acceleration is not constant, and it is changing linearly with time.

Therefore, the instantaneous angular acceleration is $-4.8\mathrm{rad}/{\mathrm{s}}^2$, and average angular acceleration is $-2.4\mathrm{rad}/{\mathrm{s}}^2$.