91Ó°ÊÓ

The angular displacement for 5.0s is, $\mathrm{θ}=25\text{ r²¹»å}$

Use the kinematic equation for constant angular acceleration to solve for angular acceleration, instantaneous angular velocity, and angular displacement. Calculate the average angular velocity using the total displacement and total time.

Consider the required formula:

1. $\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm ³Ù}$
2. $\mathrm{θ}={\mathrm{Ó\neg }}_0\mathrm{t}+\frac{1}{2}{\mathrm{Î\pm ³Ù}}^2$
3. ${\mathrm{Ó\neg }}_{\mathrm{avg}}=\frac{\mathrm{Δθ}}{\mathrm{Δ³Ù}}$

Consider the formula for the angular acceleration as:

$\mathrm{θ}={\mathrm{Ó\neg }}_0\mathrm{t}+\frac{1}{2}{\mathrm{Î\pm ³Ù}}^2$

Substitute the values and solve as:

$\begin{array}{c}25=0×5.0+\frac{1}{2}×\mathrm{Î\pm }×{5.0}^2\\ \mathrm{Î\pm }=2.0\text{ }\frac{\text{rad}}{{\text{s}}^2}\end{array}$

The magnitude of angular acceleration, $\mathrm{Î\pm }=2.0\text{ }\frac{\text{rad}}{{\text{s}}^2}$.

The equation for average velocity is

$\begin{array}{c}{\mathrm{Ó\neg }}_{\mathrm{avg}}=\frac{\mathrm{Δθ}}{\mathrm{Δ³Ù}}\\ =\frac{25}{5.0}\\ =5.0\text{ }\frac{\text{rad}}{\text{s}}\end{array}$

The magnitude of average angular velocity is $5.0\text{ }\frac{\text{rad}}{\text{s}}$.

The instantaneous angular velocity can be calculated using the kinematic equation for final angular velocity as$\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm ³Ù}$

Substitute the values and solve as:

The instantaneous angular velocity at 5.0 s, $\mathrm{Ó\neg }=10\text{ }\frac{\text{rad}}{\text{s}}$

The angular displacement for further 5.0 s from $\mathrm{t}=$5.0 s to $\mathrm{t}=$10 s is

$\mathrm{θ}={\mathrm{Ó\neg }}_0\mathrm{t}+\frac{1}{2}{\mathrm{Î\pm ³Ù}}^2$

Solve further as:

$\begin{array}{c}\mathrm{θ}=10×5.0+\frac{1}{2}×2.0×{5.0}^2\\ =75\text{ r²¹»å}\end{array}$

Here, ${\mathrm{Ó\neg }}_0=10\text{ }\frac{\text{rad}}{\text{s}}$ and$\mathrm{Δ}\text{t}=5.0\text{s}$.

The additional angular displacement from 5.0 s to 10.0 s, $\mathrm{θ}=75\text{ }\text{rad}$.