The angular position of a point on a rotating wheel is given by${}^{\mathbf{\theta }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}{\mathbf{t}}^{\mathbf{3}}}$, where${}^{\mathbf{\theta }}$is in radians and${}^{\mathbf{t}}$is in seconds. At ${}^{\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{s}}$, what are (a) the point’s angular position and (b) its angular velocity? (c) What is its angular velocity at time${}^{\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}}$? (d) Calculate its angular acceleration at.${}^{\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}}$. (e) Is its angular acceleration constant?

The angular position of a point on a rotating wheel,${}^{\theta =2.0+4.0t^2+2.0t^3}$

Angular velocity is the rate of change of displacement with time. Angular velocity is a vector quantity, with direction given by right hand rule. Angular acceleration is the time rate of change of angular velocity.

The expression for angular velocity is given as:

$\mathrm{\omega }=\frac{\mathrm{d\theta }}{\mathrm{dt}}$ … (i)

Here, $∆\theta$is the angular displacement and $∆t$is the time interval.

The expression for the angular acceleration is given as:

$\propto =\frac{d\omega }{dt}$ … (ii)

The angular position of a point on a rotating wheel is written as,

Thus, the angular position at${}^{t=0}$is${}^{2.0\mathrm{rad}}$

Using equation (i), the angular velocity of the point is calculated as:

$$\begin{gathered} \mathrm{\omega }\left(\mathrm{t}\right)=\frac{\mathrm{d\theta }}{\mathrm{dt}} \\ =\frac{\mathrm{d}(2.0+4.0{\mathrm{t}}^2+2.0{\mathrm{t}}^3}{\mathrm{dt}} \\ =8.0\mathrm{t}+6.0{\mathrm{t}}^2 \end{gathered}$$

Now, at $t=0s$the angular velocity is,

role="math" localid="1654663354225" $$\begin{gathered} \mathrm{\omega }(\mathrm{t}=0\mathrm{s})=8.0\left(0\mathrm{s}\right)+6.0{\left(0\mathrm{s}\right)}^2 \\ =0\mathrm{rad}/\mathrm{s} \end{gathered}$$

Thus, the angular velocity at is $t=0$is role="math" localid="1654663441148" $0\mathrm{rad}/\mathrm{s}$

$$\begin{gathered} \mathrm{The}\mathrm{angular}\mathrm{velocity}\mathrm{of}\mathrm{the}\mathrm{point}\mathrm{is}, \\ \mathrm{\omega }\left(\mathrm{t}\right)=8.0\mathrm{t}+6.0{\mathrm{t}}^2 \\ \mathrm{At}\mathrm{t}=4.0\mathrm{s},\mathrm{the}\mathrm{angular}\mathrm{velocity}\mathrm{is}, \\ \mathrm{\omega }(\mathrm{t}=4.0\mathrm{s})=8.0(4.0\mathrm{s})+6.0{(4.0\mathrm{s})}^2 \\ =32\mathrm{rad}/\mathrm{s}+96\mathrm{rad}/\mathrm{s} \\ =128\mathrm{rad}/\mathrm{s} \end{gathered}$$

Thus, the angular velocity of the point at$\mathrm{t}=0\mathrm{s}\mathrm{is}128\mathrm{rad}/\mathrm{s}$.

Using equation (ii), the angular acceleration is calculated as:

$$\begin{gathered} \propto \left(\mathrm{t}\right)=\frac{\mathrm{d\omega }\left(\mathrm{t}\right)}{\mathrm{dt}} \\ =\frac{\mathrm{d}(8.0\mathrm{t}+6.0{\mathrm{t}}^2)}{\mathrm{dt}} \\ =8.0+12.0\mathrm{t} \end{gathered}$$

At $\mathrm{t}=2.0\mathrm{s}$, the angular acceleration is,

$$\begin{gathered} \propto (\mathrm{t}=2.0\mathrm{s})=8.0+12.0(2.0\mathrm{s}) \\ =32\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the angular acceleration is $32\mathrm{rad}/{\mathrm{s}}^2$.

The equation of angular acceleration is a function of time. Hence, it is changing with respect to time. Therefore, the angular acceleration is not constant.