Question

A disk rotates at constant angular acceleration, from angular position${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{10}\mathbf{.0}\mathbf{}\mathbf{\text{\hspace{0.17em}rad}}\mathbf{}$ to angular position${\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{}\mathbf{70}\mathbf{.0}\mathbf{}\mathbf{\text{rad}}\mathbf{}$ in $\mathbf{6}\mathbf{.00}\mathbf{\text{s}}$. Its angular velocity at${\mathit{\theta }}_{\mathbf{2}}$ is$\mathbf{15}\mathbf{.0}\mathbf{}\mathbf{\text{\hspace{0.17em}rad}}\mathbf{/}\mathbf{\text{s}}$. (a) What was its angular velocity at ${\mathit{\theta }}_{\mathbf{1}}$? (b) What is the angular acceleration? (c) At what angular position was the disk initially at rest? (d) Graph $\mathit{\theta }$versus time $\mathbf{\text{t}}$and angular speed$\mathit{\omega }$ versus$\mathbf{\text{t}}$ for the disk, from the beginning of the motion (let$\mathbf{\text{t}}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}$ then).

Short answer

a) Angular velocity at${\theta }_1$ is${\omega }_1=5.00\text{\hspace{0.17em}rad}/\text{s}$

b) Angular acceleration is$\alpha =1.67\text{\hspace{0.17em}rad}/{\text{s}}^2$

c) Angular position when disk was at rest is${\theta }_0=2.50\text{\hspace{0.17em}rad}$

d) Graphs of $\theta$Vs t and$\omega$ Vs t are drawn below

Step-by-step solution

Given

i) Angular position${\theta }_1=10.0\text{rad}$

ii) Angular position${\theta }_2=70.0\text{\hspace{0.17em}rad}$

iii) Time is$\text{t}=6.00\text{\hspace{0.17em}s}$

iv) Angular velocity at ${\theta }_2$is$15.0\text{\hspace{0.17em}rad}/\text{s}$

Understanding the concept

Use the concept of angular kinematics. Use the angular displacement equation to find the angular velocity at the given theta. Using the angular kinematic equation, which relates initial and final angular velocities and acceleration, find angular acceleration. Draw the graph of angular displacement versus t and angular velocity versus t.

Formulae:

$\theta -{\theta }_0=\frac{1}{2}({\omega }_0+\omega )\text{t}\mathrm{................}\text{(i)}$

$\omega ={\omega }_0+\alpha \text{t}\mathrm{...........................}\text{(ii)}$

${\omega }^2={\omega }_0^2+2\alpha (\theta -{\theta }_0)\mathrm{.............}\left(\text{iii}\right)$

(a) Calculate the angular velocity at θ1 

Angular velocity at :${\theta }_1$

We can use the equation (i) as follows:

$\begin{array}{c}{\theta }_2-{\theta }_1=\frac{1}{2}({\omega }_1+{\omega }_2)\text{t}\\ 70-10=\frac{1}{2}({\omega }_1+15.0)\left(6.00\right)\\ \frac{60\times 2}{6.00}-15.0={\omega }_1\\ {\omega }_1=5.00\text{rad}/\text{s}\end{array}$

Hence, the angular velocity is.$5.00\text{rad}/\text{s}$

(b) Calculate the angular acceleration

Angular acceleration:

We use the equation (ii) to find angular acceleration as follows:

$\begin{array}{c}15.0=5.00+\alpha \left(6.00\right)\\ \alpha =\frac{15.0-5.00}{6.00}\\ =\frac{1.66\text{\hspace{0.17em}rad}}{{\text{s}}^2}\\ =1.67\text{\hspace{0.17em}rad}/{\text{s}}^2\end{array}$

Hence, the angular acceleration is$1.67\text{\hspace{0.17em}rad}/{\text{s}}^2$.

(c) Calculate the angular acceleration

Angular position when disk was at rest:

We have the equation (iii) as follows:

$\begin{array}{c}{\omega }_1^2={\omega }_0^2+2\alpha ({\theta }_1-{\theta }_0)\\ {5.0}^2=0^2+2\left(1.67\right)(10.0-{\theta }_0)\\ {\theta }_0=-\frac{25}{2\times 1.67}+10.0\\ =2.50\text{\hspace{0.17em}rad}\\ \end{array}$

Hence, the angular acceleration is.$2.50\text{\hspace{0.17em}rad}$

(d) Graph θversus time tand angular speedω versus t for the disk

Graph of$\theta$ Vs t and$\omega$ Vs t:

We can draw the given graphs as follows: