Question

The angular position of a point on the rim of a rotating wheel is given by, $\mathbf{\theta }\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{t}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{t}}^{\mathbf{3}}$where $\mathbf{\theta }$ is in radians and $\mathbf{t}$ is in seconds. What are the angular velocities at

(a) $\mathbf{t}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}\mathbf{}$and

(b) $\mathbf{t}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$?

(c) What is the average angular acceleration for the time interval that begins at $\mathbf{t}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}\mathbf{}$and ends at $\mathbf{t}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}\mathbf{}$? What are the instantaneous angular accelerations at

(d) the beginning and

(e) the end of this time interval?

Short answer

1. The angular velocity at $\text{t}=2.0\text{s}$is ${\mathrm{\omega }}_2=4.0\frac{\text{rad}}{\text{s}}$
2. The angular velocity at $\text{t}=4.0\text{s}$is ${\mathrm{\omega }}_4=28\frac{\text{rad}}{\text{s}}$
3. The average angular acceleration for the time interval from $\mathrm{t}=2.0\text{s}$ to the $\mathrm{t}=4.0\text{s}$ is ${\mathrm{\alpha }}_{\mathrm{avg}}=12\frac{\text{rad}}{{\text{s}}^2}$.
4. The instantaneous angular acceleration at the beginning is ${\mathrm{\alpha }}_2=6.0\frac{\text{rad}}{{\text{s}}^2}$
5. The instantaneous angular acceleration at the end of the time interval ${\mathrm{\alpha }}_4=18\frac{\text{rad}}{{\text{s}}^2}$

Step-by-step solution

Write the given quantities

The angular position of a point on the rim of a rotating wheel is

$\mathrm{\theta }=4.0\mathrm{t}\left(\frac{\text{rad}}{\text{s}}\right)-3.0{\mathrm{t}}^2\left(\frac{\text{rad}}{{\text{s}}^{\text{2}}}\right)+{\mathrm{t}}^3\left(\frac{\text{rad}}{{\text{s}}^{\text{3}}}\right)$

Understanding the concept of angular velocity and displacement  

To find the angular velocity and angular acceleration, we can take the first and second derivative respectively. We can use the expression of average angular acceleration and find its value.

Formulae:

$\mathrm{\omega }=\frac{\text{d}\mathrm{\theta }}{\text{dt}}$

${\mathrm{\alpha }}_{\mathrm{avg}}=\frac{{\mathrm{\omega }}_{\mathrm{f}}-{\mathrm{\omega }}_{\mathrm{i}}}{\text{dt}}$

$\alpha =\frac{\text{d}\omega }{\text{dt}}$

Calculate the angular velocity

The angular position of a point on the rim of a rotating wheel is

$\mathrm{\theta }=4.0\mathrm{t}\left(\frac{\text{rad}}{\text{s}}\right)-3.0{\mathrm{t}}^2\left(\frac{\text{rad}}{{\text{s}}^{\text{2}}}\right)+{\mathrm{t}}^3\left(\frac{\text{rad}}{{\text{s}}^{\text{3}}}\right)$

The angular velocity of a point on the rim of a rotating wheel is

$\begin{array}{c}\mathrm{\omega }=\frac{\mathrm{d\theta }}{\mathrm{dt}}\\ =\frac{\mathrm{d}\left[4.0\mathrm{t}\left(\frac{\mathrm{rad}}{\mathrm{s}}\right)-3.0{\mathrm{t}}^2\left(\frac{\mathrm{rad}}{{\mathrm{s}}^2}\right)+{\mathrm{t}}^3\left(\frac{\mathrm{rad}}{{\mathrm{s}}^3}\right)\right]}{\text{dt}}\\ =4.0-6.0\mathrm{t}+3{\mathrm{t}}^2\end{array}$

(a) Calculate the angular velocity t=2.0 s 

The angular velocity of the point:

$\begin{array}{c}\mathrm{\omega }=4.0-6.0\mathrm{t}+3{\mathrm{t}}^2\\ {\mathrm{\omega }}_2=4.0-6.0\left(2.0\text{s}\right)+3{(2.0\text{s})}^2\\ =4.0\frac{\text{rad}}{\text{s}}\end{array}$

The angular velocity at $\text{t}=2.0\text{s}$ is ${\mathrm{\omega }}_2=4.0\frac{\text{rad}}{\text{s}}$

(b) Calculate the angular velocity t=4.0 s 

The angular velocity of the point is

$\begin{array}{l}\mathrm{\omega }=4.0-6.0\mathrm{t}+3{\mathrm{t}}^2\\ {\mathrm{\omega }}_4=4.0-6.0(4.0\text{s})+3{(4.0\text{s})}^2\\ {\mathrm{\omega }}_4=28\frac{\text{rad}}{\text{s}}\end{array}$

The angular velocity at $\text{t}=4.0\text{s}$ is ${\mathrm{\omega }}_4=28\frac{\text{rad}}{\text{s}}$

(c) Calculate the average angular acceleration for the time interval from  t=2.0 sto the t=4.0 s: 

$\begin{array}{c}{\mathrm{\alpha }}_{\mathrm{avg}}=\frac{{\mathrm{\omega }}_{\mathrm{f}}-{\mathrm{\omega }}_{\mathrm{i}}}{\text{dt}}\\ =\frac{{\mathrm{\omega }}_4-{\mathrm{\omega }}_2}{{\text{t}}_4-{\text{t}}_2}\end{array}$

Substitute the values and solve as:

$\begin{array}{c}{\mathrm{\alpha }}_{\mathrm{avg}}=\frac{28\frac{\text{rad}}{\text{s}}-4.0\frac{\text{rad}}{\text{s}}}{(4.0\text{s})-(2.0\text{s})}\\ =12\frac{\text{rad}}{{\text{s}}^2}\end{array}$

The average angular acceleration for the time interval from $\mathrm{t}=2.0\text{s}$ to the $\mathrm{t}=4.0\text{s}$ is ${\mathrm{\alpha }}_{\mathrm{avg}}=12\frac{\text{rad}}{{\text{s}}^2}$

Calculate the the instantaneous angular acceleration

The instantaneous angular acceleration of the point is

$\begin{array}{c}\mathrm{\alpha }=\frac{\text{d}\mathrm{\omega }}{\text{dt}}\\ =\frac{\text{d}[4.0-6.0\mathrm{t}+3{\mathrm{t}}^2]}{\text{dt}}\\ =-6.0+6.0\mathrm{t}\end{array}$

(d) Calculate the instantaneous angular acceleration at the beginning:

The instantaneous angular acceleration of the point is

$\begin{array}{c}\mathrm{\alpha }=-6.0+6.0\mathrm{t}\\ {\mathrm{\alpha }}_2=-6.0+6.0(2.0\text{s})\\ {\mathrm{\alpha }}_2=6.0\frac{\text{rad}}{{\text{s}}^2}\end{array}$

The instantaneous angular acceleration at the beginning is ${\mathrm{\alpha }}_2=6.0\frac{\text{rad}}{{\text{s}}^2}$

(e) Calculation of the instantaneous angular acceleration at the end of time interval:

The instantaneous angular acceleration at the end of the time interval:

$\begin{array}{l}\mathrm{\alpha }=-6.0+6.0\mathrm{t}\\ {\mathrm{\alpha }}_4=-6.0+6.0(4.0\text{s})\\ {\mathrm{\alpha }}_4=18\frac{\text{rad}}{{\text{s}}^2}\end{array}$

The instantaneous angular acceleration at the end of the time interval ${\mathrm{\alpha }}_4=18\frac{\text{rad}}{{\text{s}}^2}$