When an object moves in a straight line then the motion is considered as translational but when the same moves along an axis in a circular path then it is known as rotational motion. We can take an example of a bicycle wheel. When it starts spinning then the rotational axis is the line that passes through the wheel’s centre and it’s in right angle to the plane of the wheel.

As we know that in linear motion, an object is moved till it feels any external force. This force changes the motion of objects. But when an object is moving about a fixed axis in a circular or curved path then, it can also feel the force better known as**torque**. Thus, the torque is a force which is studied under rotational dynamic.

In rotational motion, the object is not treated as a particle as treated in translational motion. The rotational dynamics starts with the study of torque. Let’s discuss of the rotational dynamics and its basic terms.

As we know that in linear motion, an object is moved till it feels any external force. This force changes the motion of objects. But when an object is moving about a fixed axis in a circular or curved path then, it can also feel the force better known as

In rotational motion, the object is not treated as a particle as treated in translational motion. The rotational dynamics starts with the study of torque. Let’s discuss of the rotational dynamics and its basic terms.

To understand the Rotational dynamics, we need to know about many terms like torque, angular acceleration etc. Angular displacement is denoted by **$\theta$**. For a particle moving in a circle of radius r and assuming that it has moved an arc length of s, the angular position theta is given by,

$\theta$ = $\frac{s}{r}$

Unit: radians

1 radian = $\frac{180}{\pi}$

Angular Displacement: The angular displacement can be defined as the change in the angular position of the particle or object.

$\Delta$ $\theta$ = $\theta_2$ - $\theta_1$

$\theta_2$ = Final angular position

$\theta_1$ = Initial angular position

Angular Velocity and Angular Acceleration

The rate of change of angular displacement is called **Angular velocity**.

Unit: radian per second

$\overline{\omega}$ = $\frac{\Delta \Theta}{\Delta t}$

$\omega$ = 2$\pi$f

where,

$\omega$ = angular velocity

**Angular Acceleration:** The rate at which angular velocity changes with time is called Angular acceleration.

$\overline{A}$ = $\frac{\Delta\omega}{\Delta t}$

Here A is pronounced as alpha

Also a = r$\alpha$ = translational acceleration

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There are 3 Rotational Dynamics equations. They are,$\theta$ = $\frac{s}{r}$

1 radian = $\frac{180}{\pi}$

Angular Displacement:

$\theta_2$ = Final angular position

$\theta_1$ = Initial angular position

Angular Velocity and Angular Acceleration

Unit: radian per second

$\overline{\omega}$ = $\frac{\Delta \Theta}{\Delta t}$

$\omega$ = 2$\pi$f

where,

$\omega$ = angular velocity

$\overline{A}$ = $\frac{\Delta\omega}{\Delta t}$

Here A is pronounced as alpha

Also a = r$\alpha$ = translational acceleration

$\omega$(t) = $\omega$0+ at

q(t) = q0 + $\omega$0t + $\frac{1}{2}$at2

$\omega_0$ = $\omega_0^{2}$ + 2a (q - q0) where,

$\omega_0$ = magnitude of the initial angular velocity

$\omega_t$ = angular velocity’s magnitude after time t

q0 = Initial angular position

q(t) = Angular position after time t

The law of conservation of rotational energy states that the energy of a system remains constant. Energy can neither be destroyed nor be created. It changes from one state to another.

Hence,

For rotational energy, E = K.E + P.E

or

**m g h = $\frac{1}{2}$m v**^{2} + $\frac{1}{2}$m$\omega$^{2}

Hence,

For rotational energy, E = K.E + P.E

or

Rotational kinetic energy is the energy which is gained by an object or body on virtue of its rotation.

It is given by**, **

It is given by

As we know for translation

K E => K.E = $\frac{1}{2}$mv^{2}

Similarly, if we consider the rotational component then

K.E = $\frac{1}{2}$ I$\omega$^{2}

Thus, the mathematical formula for rotational K E is,

E_{k} = $\frac{1}{2}$ I$\omega$^{2}

Here the momentum of inertia takes the place of mass and translational velocity is replaced by angular velocity ω.

For example:

Earth time period = 23.93 hours

Angular velocity = 7.29 × 10^{-5} rad /s

Moment of inertia I = 8.04 × 10^{37} kgm^{2}

Hence rotational kinetic energy = 2.138 x 10^{29} J

The rotational kinetic energy is basically the energy associated with every part of the object. Also the kinetic energy of any object at any instance can be the sum of both rotational as well as translational kinetic energy.

The units of rotational kinetic energy = K E => K.E = $\frac{1}{2}$mv

Similarly, if we consider the rotational component then

K.E = $\frac{1}{2}$ I$\omega$

Thus, the mathematical formula for rotational K E is,

E

Here the momentum of inertia takes the place of mass and translational velocity is replaced by angular velocity ω.

For example:

Earth time period = 23.93 hours

Angular velocity = 7.29 × 10

Moment of inertia I = 8.04 × 10

Hence rotational kinetic energy = 2.138 x 10

The rotational kinetic energy is basically the energy associated with every part of the object. Also the kinetic energy of any object at any instance can be the sum of both rotational as well as translational kinetic energy.

Let us derive the units by using the formula, We know that,

E

Unit of Inertia, I = m r

Hence I can be expressed as Kg m

Also for angular momentum unit is $\frac{rad}{second}$

So, K.E (rotational) = Kg m

rad / second = watt sec = JOULES

Torque can be defined in various ways:

1. Torque is the turning effect of the force about the axis of rotation.

2. The moment of force is called torque.

3. It is rotational analogous of Force.

4. The rate at which there is a change in angular momentum is also called torque.

5. It is force which makes an object move about an axis. To measure**torque, we can define it as the product of force magnitude and perpendicular distance of the line of action of force from the axis of rotation**.

Torque = perpendicular distance from the axis of rotation x Force

**$\tau$ = r x F**

= r F sin$\theta$where,

r = a distance between the point from which torque is measured to the point where force is applied

$\tau$ is the torque and

F = force applied

Its dimensional formula is [M L^{2} T^{ -2}], which is same as that of work.

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1. Torque is the turning effect of the force about the axis of rotation.

2. The moment of force is called torque.

3. It is rotational analogous of Force.

4. The rate at which there is a change in angular momentum is also called torque.

5. It is force which makes an object move about an axis. To measure

= r F sin$\theta$

r = a distance between the point from which torque is measured to the point where force is applied

$\tau$ is the torque and

F = force applied

Its dimensional formula is [M L

1) The point can be real or imaginary, e.g., in case of a hollow or empty box the mass is physically not located at the center of mass point.

2) This mass is supposed to be located at the center of mass in order to simplify calculations.

3) The motion of the center of mass characterizes the motion of the entire object.

4) The center of mass may or may not be the same to the geometric center if a rigid body is considered.

5) It is considered as a reference point for many other calculations of mechanics.

This concept was introduced by *Leonhard Euler*.** **The resistance that is shown by the object to change its rotation is called Moment of inertia.** I **and **J **are used as symbols for denoting moment of inertia.

I = $\sum_{i=1}^{n}$ mr^{2}**where,**

**m = mass**

r = Distance from the axis of rotation

The difficulty faced to change the angular motion of any object about an axis is shown or given or measured by calculating moment of inertia about that particular axis. It includes how far each bit of mass constituting the object from the axis.

**Greater the distance of the mass would be, greater would be the rotational force required to change its rate of rotation.**

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r = Distance from the axis of rotation

a) Find the angular velocity of the disc after 4 seconds?

b) Find the angular displacement of the disc after 4 sec?

c) Find the number of turns accomplished by the disc in 4 sec?

$\omega$(4 sec) = $\omega_{0}$ + at

= 0 + (p rad/sec

= 4p rad/sec.

Angular velocity, q(4 sec) = q

= 0 + $\frac{1}{2}$ (p rad/sec

Let the number of turns be n,

then n x 2 p rad = 8 p rad and n = 4

a) Calculate the angular velocity of the disc if the time given is 8 seconds?

b) Calculate the angular displacement of the disc if the time is 8 sec?

c) Calculate the number of turns accomplished by the disc if the time is 8 sec?

$\omega$(8 sec) = $\omega_{0}$

= 0 + (p rad/sec

= 8p rad/sec.

Angular velocity, q(8 sec) = q

= 0 + $\frac{1}{2}$ (p rad/sec

= 32 p radian.

Let the number of turns be n, then n x 2 p rad = 16 p rad and n = 8

More topics in Rotational Dynamics | |

Moment of Inertia | Center of Mass |

Torque | |

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