Relationship Between Linear And Angular Motion

The relationship between linear and angular motion is a fundamental concept in physics, describing how objects move through space. Linear motion refers to movement along a straight path, while angular motion describes rotation around a fixed axis. These two forms of motion are intricately linked, as angular motion often results in linear displacement, and vice versa. For instance, when a car's wheels rotate (angular motion), they cause the car to move forward (linear motion). Similarly, a spinning fan blade rotates, pushing air and creating a linear breeze.

Relationship Between Linear and Angular Motion

The relationship between linear and angular motion is fundamental in physics, as it describes how objects translate and rotate in space. Linear motion refers to the movement along a straight path, characterized by parameters like velocity, displacement, and acceleration. Angular motion, on the other hand, deals with rotation around a fixed axis, described by angular velocity, angular displacement, and angular acceleration. These two forms of motion are connected through formulas that convert angular quantities into linear ones, depending on the radius of rotation. For example, the linear velocity of a point on a rotating wheel is proportional to the angular velocity and the radius of the wheel.

• Relation Between Linear and Angular Properties

1. $\vec{S}=\theta \overrightarrow{\times} \vec{r}$
2. $\vec{v}=\omega \overrightarrow{\times} \vec{r}$
3. $\vec{a}=\alpha \overrightarrow{\times} \vec{r}$

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Solved Examples Based on the Relationship Between Linear and Angular Motion

Example 1: A body starts rotating about its centre with angular acceleration $=\alpha=0.2 \mathrm{rad} / \mathrm{s}^2$. Then what is its angular displacement (in rad) in the 4^th second

1) 0.4

2) 0.5

3) 0.6

4) 0.7

Solution:
$
\theta_n=\omega_i+\frac{\alpha}{2}(2 n-1)
$
As
Here

$
\begin{aligned}
& \omega_i=0 \\
& \alpha=0.2 \mathrm{rad} / \mathrm{s}^2 \\
& n=4
\end{aligned}
$
So, $\theta_n=0+\frac{0.2}{2}(2 * 4-1)=0.7 \mathrm{rad}$

Hence, the answer is the option (4).

Example 2: Starting from rest, a fan takes five seconds to attain the maximum speed of 300 rpm Assume constant acceleration, then the time (in seconds) taken by the fan attaining half the maximum speed

1) 2

2) 2.5

3) 3

4) 3.5

Solution:

The maximum angular velocity is given by, $\omega=\frac{2 \pi N}{60}$

$
\Rightarrow \omega_{\max }=\frac{2 \pi \times 300}{60}=\frac{220}{7} \mathrm{rad} / \mathrm{sec}^2
$
Now,

$
\begin{aligned}
\alpha & =\frac{w_f-w_i}{\Delta t} \\
\alpha & =\frac{\omega_f-\omega_i}{\Delta t}=\frac{\frac{220}{7}-0}{5}=\frac{44}{7} \mathrm{rad} / \mathrm{sec}^2
\end{aligned}
$
Now, for half of $\omega_{\max }$

$
\begin{aligned}
& \frac{\omega_{\max }}{2}=0+\frac{44}{7} . t \\
& \frac{100}{2}=\frac{44}{7} . t \\
& t=2.5 \text { seconds }
\end{aligned}
$
Hence, the answer is the option (2).

Example 3: A thin uniform rod of length $l$ and mass $m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of

1) $\frac{1}{3} \frac{l^2 \omega^2}{g}$
2) $\frac{1}{6} \frac{l \omega}{g}$
3) $\frac{1}{2} \frac{l^2 \omega^2}{g}$
4) $\frac{1}{6} \frac{l^2 \omega^2}{g}$

Solution:

The kinetic energy of rotation

$\begin{aligned} & K=\frac{1}{2} I w^2 \\ & \text { wherein } \\ & I=\text { moment of inertia about the axis of rotation } \\ & w=\text { angular velocity } \\ & \text { Rotational kinetic energy = potential energy } \\ & \frac{1}{2} I \omega^2=m g h \\ & I=\frac{1}{3} m l^2 \\ & \frac{1}{2}\left(\frac{1}{3} l^2\right) \omega^2=m g h \\ & h=\frac{\omega^2 l^2}{6 g}\end{aligned}$

Hence, the answer is the option (4).

Example 4: A thin rod MN, free to rotate in the vertical plane about the fixed end N, is held horizontally. When the end M is released the speed of this end, when the rod makes an angle α with the horizontal, will be proportional to : (see figure)

1) $\sqrt{\sin } \alpha$
2) $\sin \alpha$
3) $\sqrt{\cos } \alpha$
4) $\cos \alpha$

Solution:

The kinetic energy of rotation
$
K=\frac{1}{2} I w^2
$

wherein
$I=$ moment of inertia about the axis of rotation
$w=$ angular velocity

From energy conservation,

$\begin{aligned} & \frac{m g \cdot L}{2} \sin \alpha=\frac{1}{2}\left(\frac{m L^2}{3}\right) \cdot \omega^2 \\ & \omega^2=\frac{3 g}{l} \sin \alpha \quad \omega=\sqrt{\frac{3 g}{l}} \sqrt{\sin \alpha} \\ & \text { speed of end } m=\omega l \\ & =\sqrt{3 g l} \sqrt{\sin } \alpha \\ & \therefore v \propto \sqrt{\sin } \alpha\end{aligned}$

Hence, the answer is the option (1).

Example 5: A rod of length 50 cm is provided at one end. It is raised such that if makes an angle of $30^{\circ}$ from the horizontal as shown and is released from rest. Its angular speed when it passes through the horizontal ( in rad s $\mathrm{s}^{-1}$ ) will be ( $\mathrm{g}=$ $1\left(\mathrm{~ms}^{-2}\right.$ )

1) $\frac{\sqrt{20}}{3}$
2) $\sqrt{\frac{30}{2}}$
3) $\sqrt{30}$
4) $\frac{\sqrt{30}}{2}$

Solution:

The kinetic energy of rotation

$
K=\frac{1}{2} I w^2
$

wherein
$I=$ moment of inertia about the axis of rotation
$w=$ angular velocity
Work done by gravity from initial to final point

$
\begin{aligned}
W & =m g \frac{l}{2} \sin 30^{\circ} \\
& =\frac{m g l}{4}
\end{aligned}
$
By work energy theorem

$
\begin{aligned}
W & =\frac{1}{2} I \omega^2 \\
& =\frac{1}{2} \frac{m L^3}{3} \omega^2
\end{aligned}
$

from (1) and (2)

$
\omega=\sqrt{30} \mathrm{rad} / \mathrm{sec}
$

Hence, the answer is the option (3).

Summary

The relationship between linear and angular motion is crucial in understanding how objects move. Linear motion involves movement along a straight path, while angular motion refers to rotation around an axis. These concepts are linked, as angular motion often leads to linear displacement. The provided examples illustrate how angular acceleration affects angular displacement, velocity, and kinetic energy, showing practical applications like the rotation of rods and wheels in real-world scenarios.