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# Second law of motion in angular form

Module by: Sunil Kumar Singh. E-mail the author

Summary: Law of motion in angular form is a general law - not limited to linear or rotational motion. It is just that application of this form of law suits best to rotation that we tend to associate the law with rotational motion only.

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The heading of this module (law of motion in angular form - not law of motion in rotational form) points to the subtle difference about angular quantities as applicable to general motion vis-a-vis rotation. The title also indicates that law of motion in angular form is just yet another form of Newton's law - not specific to any particular type of motion. What it means that we can use angular concepts and angular law (including the one being developed here) to even pure translational motion along a straight line.

In earlier module, we wrote Newton's second law for rotation as :

t = I α t = I α

This relation is evidently valid for rotation, which involves circular motion in a plane perpendicular to a fixed axis of rotation. This form of Newton's second law is not applicable in general case. Reason is simple. The law as defined above involves moment of inertia, which is defined for a particle in rotation as :

I = m r 2 I = m r 2

The expression of moment of inertia involves the perpendicular linear distance "r'" of the particle from the axis of rotation. In general motion about a point, linear distance is measured as position vector of the particle about a point. Clearly, the rotational context of motion does not meet this criterion. This is illustrated in the figure below. In the case of rotation of a particle in "xz" plane, the distance "r'" is in the plane of circular path and is perpendicular to the z-axis. In the general case, the distance from a point (say origin or any other point),"r", is different.

In this module, we shall develop the form of Newton's law, which is based on the concept of angular momentum about a fixed point in the coordinate reference. The law so derived here, ofcourse, can be shown to yield the version of Newton's second law for rotation, which is a special case.

## Newton's second law for a particle in general motion

It was stated in the previous module that angular momentum is defined such that its first time derivative gives torque on the particle. This condition was specified keeping in mind about Newton's second law for translation. Following the logic, let us consider angular momentum of a moving particle moving with respect to a point and find whether first derivative actually yields torque as expected or not?

Angular momentum of the particle about the origin of the coordinate system is :

= r x p = m ( r x v ) = r x p = m ( r x v )

The "r" and "v" vectors represent position and velocity vectors respectively as shown in the figure. Taking differentiation of the terms with respect to time, we have :

t = t { m ( r x v ) } t = t { m ( r x v ) }

For constant mass,

t = m t ( r x v ) t = m t ( r x v )

t = m ( r x v t + r t x v ) t = m ( r x v t + r t x v )

By definition, the first time derivative of velocity is the acceleration and first time derivative of position vector is the velocity of the particle. Putting the appropriate terms for these quantities,

t = m ( r x a + v x v ) t = m ( r x a + v x v )

But, the vector product of a vector with itself is equal to zero as sinθ = sin° = 0. Hence,

t = m ( r x a ) t = m ( r x a )

Rearranging, we have :

t = r x m a = r x F t = r x m a = r x F

Indeed the first derivative of angular momentum equals the toque on the particle as expected. Since force on the particle is external force, we can qualify the above relation that first time derivative of angular momentum equals external torque applied on the particle. Next, we should think about a situation when more than one force acts on the particle. According to Newton's second law in translation, we have :

F net = m a F net = m a

Substituting the same in the equation above, we have :

t = r x m a = r x F net = r x F i t = r x m a = r x F net = r x F i

We note here that all forces are acting at the same point (occupied by the particle). As such, position vectors for all forces are same. Thus, we can enclose the vector product inside the summation sign,

t = r x F i = t i = t net t = r x F i = t i = t net

t net = t t net = t

In words, Newton's second law of motion for a particle in angular form is read as :

Definition 1: Newton's second law of motion in angular form
The net torque on a particle is equal to the time rate of change of its angular momentum.

### Example 1

Problem : Consider projectile motion of a particle of mass "m" and show that time derivative of angular moment about point of projection is equal to torque on the particle about that point.

Solution : For this question, we express various vector quantities in terms of unit vectors, because two directions involved in projectile motion i.e. horizontal and vertical are mutually perpendicular. This makes it easy to represent vectors in the unit vector form. We consider horizontal and vertical directions as "x" and "y" directions respectively. The torque, τ, on the particle about the point of projection "O" is :

For the problem in hand, we consider horizontal and vertical directions as "x" and "y" reference directions respectively. Now, the torque, τ, on the particle about the point of projection "O" is :

t = r x F t = r x F

The position vector of the position of the projectile at a given time "t" is :

r = x i + y j r = x i + y j

We note here that the external force on the projectile is force due to gravity. Hence,

F = - mg j F = - mg j

Putting in the equation of the torque, we have :

t = ( x i + y j ) x - mg j t = ( x i + y j ) x - mg j

t = - x mg k t = - x mg k

Negative sign shows that torque is clockwise i.e. into the plane of the projection i.e. xy plane. Next, we evaluate torque about "O" in terms of angular momentum. Here, angular momentum is given as :

= m ( r x v ) = m { ( x i + y j ) ) x v = m ( r x v ) = m { ( x i + y j ) ) x v

Differentiating with respect to time "t", we have :

t = t = m { ( x t i + x t j ) ) x v t t = t = m { ( x t i + x t j ) ) x v t

But v t v t is equal to acceleration. In the case of projectile, acceleration due to gravity is the acceleration of the projectile,

a = v t = - g j a = v t = - g j

Putting this value in the equation,

t = m { ( x t i + x t j ) ) x - g j t = m { ( x t i + x t j ) ) x - g j

t = - x mg k t = - x mg k

This example illustrates an important aspect of Newton's law in angular form. The quantities torque and angular momentum are both measured or calculated about a common point. If it is not so, then the relationship is not valid. This aspect should always be kept in mind while applying the law.

The important aspect of this form of Newton's law is that angular quantities are defined about a point - not an axis and as such not limited to rotational motion. Hence, we can employ this law even in situation where motion is strictly along a straight line. We shall, as a matter of fact, work out an example to illustrate this.

### Example 2

Problem : A bunjee jumper of mass 60 kg jumps from the middle of a bridge 200 m in length. Find the torque as time derivative of angular momentum about a point "O" as shown in the figure, before the rope attached to the jumper becomes taught.

Solution : Initially, the jumper is at a horizontal distance of 100 m from the end of the bridge. We must note here that the jumper is not tied to the bridge. The rope is loose during the fall. He or she falls through a vertical straight path. It is a linear motion for which we are required to calculate torque. It is also evident that we can treat the jumper as point mass as all body parts are moving with same motion.

Now, angular momentum of the jumper with respect to the point, O, is :

= m r v sin θ = m r v sin θ

Here, r sinθ = L/2, where "L" is the length of the brifge.

= m v L 2 = m v L 2

However, we do not know the speed of the jumper. We note here that the force on the jumper is force due to gravity. He/she falls with a constant acceleration. Thus, applying equation of motion, speed of the jumper after time "t" is :

v = u + a t = 0 + g t = g t v = u + a t = 0 + g t = g t

Substituting in the equation above :

= m g t L 2 = m g t L 2

Differnetiating with respect to time,

t = m g L 2 = 60 x 10 x 200 2 = 60000 N - m t = m g L 2 = 60 x 10 x 200 2 = 60000 N - m

(i) Applying right hand rule, we realize that torque is into the plane of the figure. If we choose the origin on the right side of the bridge, then torque is out of the plane of figure.

(ii) The torque is calculated on a particle like object even though the object is moving along straight line without any angular manifestations.

(iii) In a similar situation in the earlier module, we had noted that a particle moving in a straight line parallel to one of the axes of coordinate system has zero torque. The difference has arisen as velocity was constant in that case, whereas velocity is not constant in this case. We need force (hence torque) to cause acceleration or deceleration. The result, therefore, is consistent with the basic percept of Newton's law.

(iv) We can also calculate torque using relation t = r x F t = r x F , where F = mg. This relation should also give same result.

## Angular momentum for a particle in rotation

In rotation, a particle rotates about a fixed axis as shown in the figure. We consider haere a particle, which rotates about y-axis along a circular path in a plane parallel to "xz" plane. By the nature of the rotational motion, linear velocity,"v", and hence linear momentum, "p" are tangential to circular path and are perpendicular to the radius vector, "r".

The angle between radius vector "r" and velocity vector "v" is always 90°. This fact is not easily visualized. In order to visualize the same, we specifically consider a time instant when radius vector and moment arm, r r , are in "xy" plane. At this instant, the velocity vector, "v'", is tangential to the circle and is perpendicular to the "xz" plane. This figure clearly shows that radius vector is indeed perpendicular to velocity vector.

Thus, angular momentum of the particle in rotation is :

= m r v sin θ = m r v sin 90 0 = m r v = m r v sin θ = m r v sin 90 0 = m r v

The direction of angular momentum is perpendicular to the plane formed by radius and velocity vectors. For the specific situation as shown in the figure above, the direction of angular momentum is obtained by first shifting the velocity vector to the origin and then applying right hand rule. Importantly, the angular momentum vector makes an angle with extended z-axis as shown here.

The component of angular momentum in the direction along axis of rotation i.e. y-axis is :

y = m r v sin α y = m r v sin α

From geometry, we see that :

r sin α = r r sin α = r

y = m r v = m r x ω r = m r 2 ω y = m r v = m r x ω r = m r 2 ω

But, we know that m r 2 = I m r 2 = I . Hence,

y = I ω y = I ω

This has been the expected relation corresponding to p=mv for translational motion. The product of moment of inertia and angular velocity about a common axis is equal to component of angular momentum about the rotation axis. If we define angular momentum of the particle for rotation as the product of linear momentum and moment arm (not radius vector as in the general case), then we can say that product of moment of inertia and angular velocity about a common axis is equal to the angular momentum about the rotation axis. Thus, angular momentum of a particle in rotation is defined as :

= r p = r p

where r r is the moment arm about the axis of rotation. And,

= I ω = I ω

We must ensure, however, that all quantities in the equation above refer to the same axis of rotation. Also, we should also keep in mind that the definition of angular momentum about an axis has been equal to the component of linear momentum about the axis of rotation and is different to the one about a point.

### Angular momentum for rigid body in rotation

A rigid body is equivalent to a system of particles, having fixed mass distribution about an axis of rotation. It is not very difficult to extend the concept of angular momentum of a particle to rigid body as :

y = m r 2 ω y = m r 2 ω

Since all particles move with a constant angular velocity, we can take the same out of the summation sign :

y = ω m r 2 y = ω m r 2

But, we know that :

I = m r 2 I = m r 2

Putting this in the equation,

y = I ω y = I ω

Dropping subscript for angular momentum of rotation about y-axis,

= I ω = I ω

Thus, the angular momentum for rigid body has same form as of the particle in rotation. Only thing is that we need to use moment of inertia about the axis of rotation appropriately for the given rigid body.

## Newton's second law in rotation

Newton's law for a particle in rotation is established by taking first derivative of the angular momentum of a particle about an axis of rotation. As derived in previous section. it is given by :

= r 2 p = r 2 p

Differentiating with respect to time,

t net = t = ( ) t t net = t = ( ) t

Since particle moves around a circular path of a constant radius, I = m r 2 I = m r 2 is a constant and the same can be taken out of differentiation,

t net = ( ) t = I ( ) t = I α t net = ( ) t = I ( ) t = I α

This derivation underlines that the above relation is specific to rotation about an axis - not for general motion. On the other hand, the definition of torque as the time rate of angular momentum about a point is a general form of Newton's second law. It is easily implied here that Newton's second law for a rigid body in motion has same form with appropriate moment of inertia about the axis of rotation.

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