`star` Moment of force (Torque)

`star` Angular momentum of a particle

`star` Angular momentum of a particle

`●` As we know, If the body is fixed at a point or along a line, it has only rotational motion.

`●` We know that force is needed to change the translational state of a body, i.e. to produce linear acceleration.

`●` We may then ask, what is the analogue of force in the case of rotational motion? To look into the question in a concrete situation let us take the example of opening or closing of a door. A door is a rigid body which can rotate about a fixed vertical axis passing through the hinges. What makes the door rotate? It is clear that unless a force is applied the door does not rotate. But any force does not do the job.

`●` The rotational analogue of force is moment of force. It is also referred to as torque. (We shall use the words moment of force and torque interchangeably.) We shall first define the moment of force for the special case of a single particle.

`color{blue} ►` The magnitude of the moment of force may be written

`color{blue}{tau=(r sin theta) F = r_(bot) F...................(7.24)}`

or `color{blue} {tau = r F sin theta= r F_(bot)...............(7.24)}`

`=>` where `r_(⊥) = r sinθ` is the perpendicular distance of the line of action of `F` form the origin and `F_(⊥)(= F sinθ )` is the component of F in the direction perpendicular to r. Note that `τ = 0` if `r = 0, F = 0` or `θ = 0^o ` or `180^o` .

`\color{green} ✍️` One may note that since `r xx F` is a vector product, properties of a vector product of two vectors apply to it. If the direction of F is reversed, the direction of the moment of force is reversed. If directions of both `r` and `F` are reversed, the direction of the moment of force remains the same.

`●` We know that force is needed to change the translational state of a body, i.e. to produce linear acceleration.

`●` We may then ask, what is the analogue of force in the case of rotational motion? To look into the question in a concrete situation let us take the example of opening or closing of a door. A door is a rigid body which can rotate about a fixed vertical axis passing through the hinges. What makes the door rotate? It is clear that unless a force is applied the door does not rotate. But any force does not do the job.

`\color{green} ✍️` A force applied to the hinge line cannot produce any rotation at all, whereas a force of given magnitude applied at right angles to the door at its outer edge is most effective in producing rotation. It is not the force alone, but how and where the force is applied is important in rotational motion.

`●` The rotational analogue of force is moment of force. It is also referred to as torque. (We shall use the words moment of force and torque interchangeably.) We shall first define the moment of force for the special case of a single particle.

`●` If a force acts on a single particle at a point `P` whose position with respect to the origin `O` is given by the position vector r (Fig. 7.18), the moment of the force acting on the particle with respect to the origin `O` is defined as the vector product

`color{blue}{vectau = vecr xx vecF........................(7.23)}`

`●` The moment of force (or torque) is a vector quantity. The symbol `τ` stands for the Greek letter tau. The magnitude of `τ` is

`color{blue}{τ = rF sin theta..................(7.24 a)}`

`=>`where `r` is the magnitude of the position vector `r`, i.e. the length `OP, F` is the magnitude of force `F` and `θ` is the angle between `r` and `F` as shown.

`\color{green} ✍️` Moment of force has dimensions `M L^2 T ^(-2)`. Its dimensions are the same as those of work or energy. It is, however, a very different physical quantity than work. Moment of a force is a vector, while work is a scalar.

`\color{green} ✍️` The SI unit of moment of force is Newton-metre (Nm).

`color{blue} ►` The magnitude of the moment of force may be written

`color{blue}{tau=(r sin theta) F = r_(bot) F...................(7.24)}`

or `color{blue} {tau = r F sin theta= r F_(bot)...............(7.24)}`

`=>` where `r_(⊥) = r sinθ` is the perpendicular distance of the line of action of `F` form the origin and `F_(⊥)(= F sinθ )` is the component of F in the direction perpendicular to r. Note that `τ = 0` if `r = 0, F = 0` or `θ = 0^o ` or `180^o` .

`color{green} ✍️` Thus, the moment of a force vanishes if either the magnitude of the force is zero, or if the line of action of the force passes through the origin.

`\color{green} ✍️` One may note that since `r xx F` is a vector product, properties of a vector product of two vectors apply to it. If the direction of F is reversed, the direction of the moment of force is reversed. If directions of both `r` and `F` are reversed, the direction of the moment of force remains the same.

`●` Here we define angular momentum for the special case of a single particle and look at its usefulness in the context of single particle motion. We shall then extend the definition of angular momentum to systems of particles including rigid bodies.

`●` Like moment of a force, angular momentum is also a vector product. It could also be referred to as moment of (linear) momentum. From this term one could guess how angular momentum is defined.

`●` Consider a particle of mass `m` and linear momentum `p` at a position `r` relative to the origin `O.` The angular momentum `l` of the particle with respect to the origin `O` is defined to be

`color{blue}{I = r xx p...............(7.25)}`

`●` The magnitude of the angular momentum vector is

`color{blue}{I = r p sin theta...............(7.26)}`

`=>` where `r_(⊥) (= r sinθ)` is the perpendicular distance of the directional line of `p` from the origin and `p_(⊥)(= p sinθ )` is the component of p in a direction perpendicular to `r`.

`●` The physical quantities, moment of a force and angular momentum, have an important relation between them. It is the rotational analogue of the relation between force and linear momentum. For deriving the relation in the context of a single particle, we differentiate `I = r × p` with respect to time,

`(dI)/(dt)=d/(dt)( r xx p)`

`●` Applying the product rule for differentiation to the right hand side,

`color{green}{d/(dt) ( r xx p)= (dr)/(dt) xx p + r xx (dp)/(dt)}`

`●` Now, the velocity of the particle is `v = dr//dt` and `p = m v`

Because of this `(dr)/(dt) xx p = v xx m v=0`

as the vector product of two parallel vectors vanishes. Further, since `dp // dt = F`,

`r xx (dp)/(dt)= r xx F = tau`

`●` Thus, the time rate of change of the angular momentum of a particle is equal to the torque acting on it. This is the rotational analogue of the equation `F = dp//dt`, which expresses Newton’s second law for the translational motion of a single particle.

`●` Like moment of a force, angular momentum is also a vector product. It could also be referred to as moment of (linear) momentum. From this term one could guess how angular momentum is defined.

`●` Consider a particle of mass `m` and linear momentum `p` at a position `r` relative to the origin `O.` The angular momentum `l` of the particle with respect to the origin `O` is defined to be

`color{blue}{I = r xx p...............(7.25)}`

`●` The magnitude of the angular momentum vector is

`color{blue}{I = r p sin theta...............(7.26)}`

`=>` where `r_(⊥) (= r sinθ)` is the perpendicular distance of the directional line of `p` from the origin and `p_(⊥)(= p sinθ )` is the component of p in a direction perpendicular to `r`.

`\color{red} ✍️`We expect the angular momentum to be zero `(l = 0)`, if the linear momentum vanishes `(p = 0)`, if the particle is at the origin `(r = 0)`, or if the directional line of p passes through the origin `θ = 0^o` or `180^o`.

`●` The physical quantities, moment of a force and angular momentum, have an important relation between them. It is the rotational analogue of the relation between force and linear momentum. For deriving the relation in the context of a single particle, we differentiate `I = r × p` with respect to time,

`(dI)/(dt)=d/(dt)( r xx p)`

`●` Applying the product rule for differentiation to the right hand side,

`color{green}{d/(dt) ( r xx p)= (dr)/(dt) xx p + r xx (dp)/(dt)}`

`●` Now, the velocity of the particle is `v = dr//dt` and `p = m v`

Because of this `(dr)/(dt) xx p = v xx m v=0`

as the vector product of two parallel vectors vanishes. Further, since `dp // dt = F`,

`r xx (dp)/(dt)= r xx F = tau`

`color{red} ✍️ d/(dt) (r xx p)=tau` or `color{blue}{(dI)/(dt)= tau`

.....................(7.27)}`●` Thus, the time rate of change of the angular momentum of a particle is equal to the torque acting on it. This is the rotational analogue of the equation `F = dp//dt`, which expresses Newton’s second law for the translational motion of a single particle.

`●` To get the total angular momentum of a system of particles about a given point we need to add vectorially the angular momenta of individual particles. Thus, for a system of `n` particles,

`L=I_1+I_2+.......+I_n= sum_(i=1)^n I_i`

`●` The angular momentum of the `i^(th)` particle is given by

`I_i=r_i xx p_i`

`=>` where `r_i` is the position vector of the `i^(th)` particle with respect to a given origin and `p = (m_iv_i) ` is the linear momentum of the particle. (The particle has mass `m_i` and velocity `v_i`) We may write the total angular momentum of a system of particles as

`color{blue}{L= sum I_i = sum_i r_i xx p_i........................(7.25 b)}`

`●` This is a generalisation of the definition of angular momentum (Eq. 7.25a) for a single particle to a system of particles.

Using Eqs. (7.23) and (7.25b), we get

`color{blue}{(dL)/(dt)=d/(dt) (sum I_i)=sum_i (dI_i)/(dt)= sum_i tau_i ..................(7.28 a)}`

`=>` where `tau_i` is the torque acting on the `i^(th)` particle;

`tau_i =r_i xx F_i`

`●` The force `F_i` on the `i^(th)` particle is the vector sum of external forces `F_i^(ext)` acting on the particle and the internal forces int i F exerted on it by the other particles of the system. We may therefore separate the contribution of the external and the internal forces to the total torque

`tau= sum_i tau_i = sum_i r_i xx F_i ` as

`tau= tau_(ext) + tau_(i n t )`

where, `tau_(ext) = sum_i r_i xx F_(ext)` and

`●` We shall assume not only Newton’s third law, i.e. the forces between any two particles of the system are equal and opposite, but also that these forces are directed along the line joining the two particles.

`●` In this case the contribution of the internal forces to the total torque on the system is zero, since the torque resulting from each action-reaction pair of forces is zero. We thus have, `tau_(i n t) = 0` and therefore `tau = tau_(ext)`

since `tau= sum tau_i` it follows from Eq (7.28 a) that

`color{blue}{(dL)/(dt)= tau_(ext)...................(7.28 b)}`

`●` Thus, the time rate of the total angular momentum of a system of particles about a point (taken as the origin of our frame of reference) is equal to the sum of the external torques (i.e. the torques due to external forces) acting on the system taken about the same point.

`●` Eq. (7.28 b) is the generalisation of the single particle case of Eq. (7.23) to a system of particles. Note that when we have only one particle, there are no internal forces or torques. Eq.(7.28 b) is the rotational analogue of

`=>` Note that like Eq.(7.17), Eq.(7.28b) holds good for any system of particles, whether it is a rigid body or its individual particles have all kinds of internal motion.

`L=I_1+I_2+.......+I_n= sum_(i=1)^n I_i`

`●` The angular momentum of the `i^(th)` particle is given by

`I_i=r_i xx p_i`

`=>` where `r_i` is the position vector of the `i^(th)` particle with respect to a given origin and `p = (m_iv_i) ` is the linear momentum of the particle. (The particle has mass `m_i` and velocity `v_i`) We may write the total angular momentum of a system of particles as

`color{blue}{L= sum I_i = sum_i r_i xx p_i........................(7.25 b)}`

`●` This is a generalisation of the definition of angular momentum (Eq. 7.25a) for a single particle to a system of particles.

Using Eqs. (7.23) and (7.25b), we get

`color{blue}{(dL)/(dt)=d/(dt) (sum I_i)=sum_i (dI_i)/(dt)= sum_i tau_i ..................(7.28 a)}`

`=>` where `tau_i` is the torque acting on the `i^(th)` particle;

`tau_i =r_i xx F_i`

`●` The force `F_i` on the `i^(th)` particle is the vector sum of external forces `F_i^(ext)` acting on the particle and the internal forces int i F exerted on it by the other particles of the system. We may therefore separate the contribution of the external and the internal forces to the total torque

`tau= sum_i tau_i = sum_i r_i xx F_i ` as

`tau= tau_(ext) + tau_(i n t )`

where, `tau_(ext) = sum_i r_i xx F_(ext)` and

`\color{red} ✍️` `tau_(i n t) = sum_i r_i xx (F_i)_(i n t)`

`●` We shall assume not only Newton’s third law, i.e. the forces between any two particles of the system are equal and opposite, but also that these forces are directed along the line joining the two particles.

`●` In this case the contribution of the internal forces to the total torque on the system is zero, since the torque resulting from each action-reaction pair of forces is zero. We thus have, `tau_(i n t) = 0` and therefore `tau = tau_(ext)`

since `tau= sum tau_i` it follows from Eq (7.28 a) that

`color{blue}{(dL)/(dt)= tau_(ext)...................(7.28 b)}`

`●` Thus, the time rate of the total angular momentum of a system of particles about a point (taken as the origin of our frame of reference) is equal to the sum of the external torques (i.e. the torques due to external forces) acting on the system taken about the same point.

`●` Eq. (7.28 b) is the generalisation of the single particle case of Eq. (7.23) to a system of particles. Note that when we have only one particle, there are no internal forces or torques. Eq.(7.28 b) is the rotational analogue of

`color{blue}{(dp)/(dt)= F_(ext)`.

..................(7.17)}``=>` Note that like Eq.(7.17), Eq.(7.28b) holds good for any system of particles, whether it is a rigid body or its individual particles have all kinds of internal motion.