Dynamics Rotational Motion

Introduction

There can be many types of motion exerted by a body like translational motion, rotational motion, circular motion, uniform motion, etc. In rotational motion, rigid bodies which have a definite size and definite shape are considered. The particles present in a body about a fixed axis move in a circular path. So to understand the concept of rotational motion you must know about circular motion. Here the terms are included in angular form.

What is Rotational Motion?

If a force is applied to any object at the corner side of it(not on the centre of mass), then we see that the object starts rotating with a fixed axis.

In this type of motion, a rigid body rotates in a circular path with a fixed point or line of axis. The various particles of a body move together with the same velocity at any point in time. So we can say that the motion of the body in which the particles move in a circular motion with the same angular velocity is known as rotational motion.

Rotational Motion about a fixed axis

A rotational motion occurs if all the particles present in a body move in a circular way along a single line or fixed axis. The line through which the body moves is known as the axis of rotation. So we can say that in a rotational motion, the number of particles in a body moves in a circular shape about a fixed axis of rotation.

Rotational Acceleration

As in a rotational motion, all the particles move in a circular path. The direction of particles that are moving in a circle at every point in rotation motion changes constantly. Due to the change in direction, there is a change in the velocity with time that gives rotational or angular acceleration. Like linear acceleration of translational motion, angular or rotational acceleration is defined as the ratio of change in the angular velocity to the change in time taken by the particles. It can be represented by $\mathrm{\alpha}$.

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

where $\mathrm{\omega}$ is angular velocity and t is the time taken. SI unit of angular acceleration is rad $\mathrm{s^{-2}}$. Its dimensional formula is $\mathrm{[M^0 L^0 T^{-2}]}$

Torque

It is also known as the rotational analogue of force. In linear motion, due to the force the object starts moving whereas in rotational motion, due to torque the object starts turning or rotating around a fixed point or axis. It can also be known as the moment of force or turning effect of force.

Torque is measured as the multiple of force magnitude and the distance from point of rotation.

$$\mathrm{torque = force \times perpendicular \:distance}$$

Its unit is Newton-meter(Nm). It is denoted by the Greek letter τ (tau) It is a vector quantity. Its dimensional formula is $\mathrm{[M L^{2} T^{-2}]}$. Mathematical Expression for Torque is given by

$$\mathrm{\overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F}=r Fsin \theta}$$

Where r is the position vector

F is the magnitude of the force

θ is the angle between position r and force F

Moment of Inertia

It is known as the rotational analogue of mass. The part of the mass in linear motion is the same as the part of a moment of inertia in the motion of rotation. The moment of inertia of a rigid body about a given line of rotation or axis of rotation is measured as the addition of the multiple of each particle’s mass and the square of their distances from the rotational axis. It is denoted by I.

Expression: Consider a body consisting of n particles of masses $\mathrm{m_1,m_2,m_3.........m_n}$ at distance $\mathrm{r_1,r_2,r_3,.......r_n}$ from the line of rotation or rotational axis. Then the moment of inertia of a body is:

$$\mathrm{I=m_1 r_1^{2},m_2 r_2^{2},m_3 r_3^{3},......m_n r_n^{2}}$$

$$\mathrm{I=\displaystyle\sum\limits_{i=1}^n m_ir_i^{2}}$$

SI unit of moment of inertia is $\mathrm{kilogram\:meter^{2} (kg\:m^{2})}$. It is a scalar quantity. Its dimensional formula is $\mathrm{[ML^{2} T]}$

Relation between torque(τ) and moment of inertia(I)

Let us consider a rigid consisting of n-particles having mass $\mathrm{m_1,m_2,m_3............................m_n}$ rotating about a fixed axis. Let $\mathrm{r_1,r_2,r_3...........................r_n}$ are the respective position of the particles.

Rotation of rigid body about an axis

So total torque acting on the rigid body is

$$\mathrm{\tau=\tau_1+\tau_2+\tau_3.........\tau_n}$$

$$\mathrm{\tau=F_1 r_1+F_2 r_2+F_3 r_3............F_n r_n}$$

We know that F=ma, so the above equation can be written as

$$\mathrm{\tau= m_1 a_1 r_1+m_2 a_2 r_2+m_3 a_3 r_3............m_n a_n r_n}$$

But $\mathrm{a=r \alpha}$

$$\mathrm{\tau= m_1 (r_1 \alpha)r_1+m_2 (r_2 \alpha)r_2+m_3 (r_3 \alpha)r_3 .................m_n (r_n \alpha)r_n\:\:\:\:\:\:\:\:\:\:}$$

$$\mathrm{\tau=m_1 r_1^{2} \alpha+m_2 r_2^{2} \alpha+m_3 r_3^{2} \alpha ......................m_n r_n^{2} \alpha\:\:\:\:\:\:\:\:\:\:}$$

$$\mathrm{\tau=(m_1 r_1^{2}+m_2 r_2^{2}+m_3 r_3^{2} .....................m_n r_n^{2})\alpha}$$

$$\mathrm{\tau=\displaystyle\sum\limits_{i=1}^n m_ir_i^{2}\alpha}$$

$\mathrm{\tau=I \alpha}$

$$\mathrm{(\displaystyle\sum\limits_{i=1}^n m_ir_i^{2}=I)}$$

Conclusion

In a circular motion, all the particles in a body are rotating with the continuous changing velocity over time and produce an angular acceleration of a particle in a body. When the pair of two equal and unlike parallel forces are applied to the body, a turning effect of the couple is formed. It is termed torque exerted by the couple.

FAQs

Q1. A constant torque of 1000 Nm turns a wheel of the moment of inertia of $\mathrm{100 \:kgm^{2}}$ about an axis from its center of rotation. Find the gain in angular velocity in 2s.

Ans. Given =1000 Nm , I=200 $\mathrm{kgm^{2}}$

As $\mathrm{\tau=I \alpha}$

$$\mathrm{\alpha=\frac{\tau}{I}}$$

$$\mathrm{=\frac{1000}{200}=5\:rad/s^{2}}$$

If ω be the angular velocity of the wheel after 2s,

$$\mathrm{\omega=\omega_0+\alpha t}$$

$$\mathrm{= 0 + 5 \times 2 = 10 rad/s}$$

Q2. Explain why a torque cannot be balanced by a single force?

Ans. A single force produces translational motion, whereas a torque produces rotational motion. The turning effect of the pair of equal and unlike parallel forces acting on a body along a different action line is dependent on a moment of the couple or torque exerted by the couple. Hence a torque cannot be balanced by a single force.

Q3. On what factors, the moment of inertia of a body depends?

Ans.

• The mass of the body,

• Distribution of the mass about an axis of rotation.

Q4. What is the importance of rotational motion in our daily life?

Ans. The rotational motion plays a vital role in the motion of vehicles and motors. The blades of the helicopter also possess rotational motion.

Q5. At what angle, the torque is maximum?

Ans. When the force is act perpendicularly to the lever arm. i.e. when $\mathrm{\theta=90^0}$. the torque is maximum.