IIT JEE , AIEEE Forum - Articles Category , Community shelf - Plane Rotational Motion of a Single Particle

snandkeolyar

Torque

The turning effect of a force about the axis of rotation is called moment of force or torque due to the force and is measured as the product of the magnitude of the force and the perpendicular distance between the line of action of the force and the axis of rotation.

Torque = force

perpendicular distance from the axis of rotation

..(xiii)

The SI unit of torque is N-m. Its dimensional formula is [ML²T^?2].

Expression for torque in Cartesian co-ordinates from rotation of a body in a plane( physical meaning of torque ) :
Consider a body rotating in a plane XY about origin O. Let P be the position of centre of mass of the body where entire mass ?m? of the body is supposed to be concentrated. The rotation occurs due to force ?F? applied at P along

.

is the position vector of centre of mass and

XOP =

. In small time dt.

Position co-ordinates for the centre of mass of the body of mass ?m?

Let centre of mass at P reach Q where

and

POQ = d

In vector triangle OPQ

or

Work done in rotating the body from P to Q is given as

..(xiv)

If F_x and F_y are components of force and dx and dy are components of displacement dr then

From equation (xiv) we have


		..(xv)

Let the co-ordinates of the point P be (x, y)

Position coordinates of P

	x = r cos	..(xvi)
	and y = r sin	..(xvii)

Differentiating (xvi) with respect to

, we get


	= r ( ? sin )
	= ?r (sin )	..(xviii)
		..(xix)

Again differentiating (xvii) with respect to

, we get

		..(xx)
		..(xxi)

Substituting in equation (xv), we get


	= X. F_y.d ? Y.F_x.d

		..(xxii)
		..(xxiii)

Where

denotes the torque.
Equation (xxiii) is the expression for torque in cartesian co-ordinates.
The physical meaning of torque can be given as magnitude of a quantity in rotational motion, which when multiplied by a small angular displacement gives us work done in rotational motion. The quantity corresponds to force in linear motion, which when multiplied by a small linear displacement gives the work done in linear motion.

Expression for torque in polar co-ordinates
Let the line of action

make an angle

with X-axis.

Polar co-ordinates of the body at point P

		..(xxiv)
	and	..(xxv)

If x, y are the co-ordinates of the point P where

and

then	x = r cos	..(xxvi)
	y = r sin	..(xxvii)

Therefore, substituting this value in equation (xxiii) we have

	= (r cos ) (F sin ) ? (r sin ) (F cos )
	= r F [sin cos ? cos sin ]
	= r F sin	..(xxviii)

Let

be the angle made by line of action of force F and positive vector


	= ?

Therefoe, substituting value of (

?

) in equation (xxvii) we get

= r F sin

..(xxix)

This is expression for torque in polar coordinates from O, draw ON on the line of action of

.
In

OPN, sin

=

ON = r sin

From (xxix) we have

= rF sin

= F(ON)

..(xxx)

Therefore, The torque due to a force is the product of force and perpendicular distance of line of action of force from the axis of rotation.
From the above expression we conclude that:

When ?r? is maximum, torque due to a force is maximum, e.g., we can open or close the door easily by applying the force near the edge of the door, to unscrew a nut fitted tightly we need a wrench with a long arm.

Application of torque

When length of arm (r) is long, the force (F) required to produce a given turning effect (r F) is smaller. This principle of lever arm is used in lifting heavy weights.
Torque will be maximum when sin = max = + 1 Therefore, = 90^o i.e., when force is applied in a direction perpendicular to it is easy to open/close door by applying force perpendicular to the edge of the door.
When = 0^o or 180^osin = sin 0^o or sin 180^o = 0
= r Fsin = 0, i.e. torque is zero.
door cannot be closed by applying the force parallel to the door.
The equation (xxix) in vector form is given as

..(xxxi)

Torque is vector quantity whose direction is given by right hand screw rule.
As shown in figure has two rectangular components in the direction of increasing , it is called radial component of and in the direction of increasing and perpendicular to Fr. This is called transverse component of .

Torque as vector quantity

= r F sin

= r (F ) ..(xxxii)

Torque is due to transverse component only. The radial component of force has no role to play in the torque.

Torque vector in three dimensions
For the motion taking place in x-y plane equation (xxxii) gives us an expression for torque which is z- component of torque vector. The force can rotate the body in three dimensions. We shall have three rectangular components of torque which can be obtained from equation (xxxii)







	Equating we get,
	_x = y F_z ? z F_y	..(xxxiii)
	_y = z F_x ? x F_z	..(xxxiv)
	_z = x F_y ? y F_x	..(xxxv)

From the above expression we see that the product r x F of two vectors to produce a third vector , changes sign if we interchange the vectors.

..(xxxvi)

Angular momentum
The angular momentum of a particle about an axis of rotation is the product of its linear momentum and the perpendicular distance of the line of action of linear momentum from the axis. It is denoted by L.

Therefore, angular momentum = Linear momentum x perpendicular distance from the axis of rotation.
The SI unit of angular momentum is kgm²/s and dimension of angular momentum is [ML² T^?1].

Expression for angular momentum in Cartesian Coordinates system
We know the expression for rotating a body in XY plane as

= x F_y ? y F_x

..(xxxvii)

If the x and y components of linear momentum of the body are p_x = mv_x and
p_y = mv_y. Then by Newton?s second law we have


and

Substituting in equation (xxxvii) we get


	=	..(xxxviii)

differentiating with respect to time we get




	Therefore,	..(xxxix)

Substituting (xxxix) in (xxxviii) we obtain




		..(xL)

	putting (x p_y ? y p_x) = L	..(xLi)
		..(xLii)

Thus, we obtain torque as rate of change of angular momentum (L).

Expression for Angular momentum in polar co-ordinates
Let point k (x, y) be the centre of mass of a body of mass m and linear momentum

rotating in XY plane about Z-axis.
Let

.

Polar co-ordinatesof the body have angular momentum

	x = r cos
	y = r sin	..(xLiii)

Line of action of linear momentum

makes an angle ?

? with OX and angle

with

.

..(xLiv)

Putting these values in equation (xLi) we have

	L = (rcos ) (psin ) ? (rsin ) (pcos )
	= pr [sin cos ? cos sin )
	L = pr sin ( ? )

From figure we have

+

=

?

L = pr sin

This is the expression for angular momentum of a particle in polar co-ordinates.
From O, draw ON perpendicular to the line of action

.

	In OKN

	L = p (ON)	..(xLv)

The physical meaning of angular momentum of a body about a given axis is the product of linear momentum and perpendicular distance of line of action of linear momentum vector from the axis of rotation.
The vector form of angular momentum expression is given as

and the three rectangular components of angular momentum are as
L_x = yp_z ? zp_y
L_y = zp_x ? xp_z
L_z = xp_y ? yp_x

Geometric meaning of angular momentum
Let us consider a body moving in XY plane about an axis OZ. Let the entire mass (m) of the body be concentrated at K, the centre of mass of the body at any time t. Let

. In a small time dt let the centre of mass at K reach L, where

Geometric meaning of angular momentum

Join KL which represents displacement of the centre of mass in small time dt
In vector

OKL,

From O, draw

equal and parallel to

. Join ML.
Area swept by the position vector in a small time dt is

Divide both sides by dt




	Or,	..(xLvi)

is area swept by the position vector per unit time and is called areal velocity of the position vector of the centre of mass equation (xLvi) shows angular momentum of a body about a given axis is twice the product of mass of the body and areal velocity of the position vector of the body.

Deduction of Kepler?s Second Law of Planetary Motion
Kepler?s second law states A planet revolves around the sun in an elliptical orbit in such a way that radius vector traces out equal areas in equal interval of time, i.e., areal velocity of the planet around the sun is constant.

Kepler?s Second Law of Planetary Motion

We know

A planet revolves around the sun in an elliptical orbit under the influence of the gravitational pull of the sun on the planet. This pull of force

acts along the line joining the centres of the sun and the planet and is bound towards the sun.
Therefore,

Torque

= constant
Using (xLvi)

i.e., areal velocity of the planet is constant. This proves Kepler?s Second Law.
We have shown area P₁P₂S = area P₃P₄S travelled by the planet in same time dt
As SP₁ > SP₃

area P₁P₂ < area P₃P₄or area P₃P₄ > area P₁P₂i.e., speed of the planet is going from P₃ to P₄ which is greater than its speed in going from P₁ to P₂. Thus, planet moves faster when it is closer to the sun and vice versa.
Points to remember

The turning effect of a force about the axis of rotation is called moment of force or torque.
Moment of momentum is called the angular momentum. The angular momentum of a particle about an axis of rotation is the product of its linear momentum and the perpendicular distance of the line of action of linear momentum from the axis.

rate if you find it useful

Ask & Discuss Questions with Community & Experts