Question

The figure shows angular position versus time graphs for six different objects.

a) Rank these graphs on the basis of the angular velocity of each object. Rank positive angular velocities as larger than negative angular velocities. Rank from largest to smallest. To rank items as equivalent, overlap them.

b) Rank these graphs on the basis of the angular acceleration of the object. Rank positive angular accelerations as larger than negative angular accelerations.Rank from largest to smallest. To rank items as equivalent, overlap them.

Answer 1

Concepts and reason

The required concepts to solve these questions are angular velocity and angular acceleration.

Initially, find the slope of each of the line and then calculate the angular velocity for every point and rank from largest to smallest. Similarly, calculate the angular acceleration and then rank from largest to smallest.

Fundamentals

Angular velocity is defined as the rate of change of the angular position with respect to time.

The expression for angular velocity is,

$\omega = \frac{{\Delta \theta }}{{\Delta t}}$

Here, $\theta$ is angular position and $t$ is time.

Angular acceleration is defined as the rate of change of the angular velocity with respect to time.

The expression for angular acceleration is,

$\alpha = \frac{{\Delta \omega }}{{\Delta t}}$

Here, $\omega$ is angular velocity and is $t$ the time.

The expression for slope is,

${\rm{Slope}} = \;\frac{{{\rm{change in quantity along }}y{\rm{ - axis}}}}{{{\rm{change in quantity along }}x{\rm{ - axis}}}}$

(a)

The expression for slope of the line (angular velocity) is,

${\rm{Slope}} = \;\frac{{\Delta \theta }}{{\Delta t}}$

The slope of the line A and F is zero because velocity is constant.

The slope of the line B is,

${\rm{Slope}} = \;\frac{{{\theta _1} - \theta }}{{{t_1} - t}}$

Substitute $2{\rm{ rad}}$ for ${\theta _1}$ , $0{\rm{ rad}}$ for $\theta$ , $12{\rm{ s}}$ for ${t_1}$ and $0{\rm{ s}}$ for $t$ in above equation.

$\begin{array}{c}\\{\rm{Slope}} = \;\frac{{2{\rm{ rad}} - 0{\rm{ rad}}}}{{12{\rm{ s}} - 0{\rm{ s}}}}\\\\ = \frac{1}{6}{\rm{rad}} \cdot {\rm{s}}\\\end{array}$

The slope of the line C is,

${\rm{Slope}} = \;\frac{{{\theta _2} - \theta }}{{{t_2} - t}}$

Substitute $1{\rm{ rad}}$ for ${\theta _2}$ , $0{\rm{ rad}}$ for $\theta$ , $12{\rm{ s}}$ for ${t_2}$ and $0{\rm{ s}}$ for $t$ in above equation.

$\begin{array}{c}\\{\rm{Slope}} = \;\frac{{1{\rm{ rad}} - 0{\rm{ rad}}}}{{12{\rm{ s}} - 0{\rm{ s}}}}\\\\ = \frac{1}{{12}}{\rm{rad}} \cdot {\rm{s}}\\\end{array}$

Slope of the line D is,

${\rm{Slope}} = \;\frac{{{\theta _3} - \theta }}{{{t_3} - t}}$

Substitute $- 2{\rm{ rad}}$ for ${\theta _3}$ , $0{\rm{ rad}}$ for $\theta$ , $12{\rm{ s}}$ for ${t_3}$ and $0{\rm{ s}}$ for $t$ in above equation.

$\begin{array}{c}\\{\rm{Slope}} = \;\frac{{ - 2{\rm{ rad}} - 0{\rm{ rad}}}}{{12{\rm{ s}} - 0{\rm{ s}}}}\\\\ = - \frac{1}{6}{\rm{rad}} \cdot {\rm{s}}\\\end{array}$

Slope of the line E is,

${\rm{Slope}} = \;\frac{{{\theta _4} - \theta }}{{{t_4} - t}}$

Substitute $- 3{\rm{ rad}}$ for ${\theta _4}$ , $- 1\;{\rm{rad}}$ for $\theta$ , $12{\rm{ s}}$ for ${t_3}$ and $0{\rm{ s}}$ for $t$ in above equation.

$\begin{array}{c}\\{\rm{Slope}} = \;\frac{{ - 3{\rm{ rad}} - \left( { - 1} \right){\rm{ rad}}}}{{12{\rm{ s}} - 0{\rm{ s}}}}\\\\ = - \frac{1}{6}{\rm{rad}} \cdot {\rm{s}}\\\end{array}$

Rank of angular velocities from largest to smallest is,

${\rm{B}} > {\rm{C}} > {\rm{A}} = {\rm{F}} > {\rm{D = E}}$

(b)

The expression for angular acceleration is,

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

In the graph, they all have constant slopes. It means that all lines represent constant angular velocity. Thus,

$\begin{array}{c}\\\alpha = \;\frac{{d\left( {{\rm{constant}}} \right)}}{{dt}}\\\\ = 0\\\end{array}$

Rank of angular acceleration from largest to smallest is ${\rm{D = E}} = {\rm{B}} = {\rm{C}} = {\rm{A}} = {\rm{F}}$ .

Ans: Part a

Rank of angular velocities from largest to smallest is ${\rm{B}} > {\rm{C}} > {\rm{A}} = {\rm{F}} > {\rm{D = E}}$ .