Torque, Angular, Momentum and Equilibriums
Torque or Moment of Force
Definition
If a body is
capable of rotating about an axis, then force applied properly on this body
will rotate it about the axis (axis of rotating). This turning effect of the
force about the axis of rotation is called torque.
Torque is the physical quantity which produces angular
acceleration in the body.
Explanation
Consider a body which can rotate about O (axis of rotation). A force F acts
Consider a body which can rotate about O (axis of rotation). A force F acts
on point P whose position vector w.r.t O is r.
Diagram
Coming Soon
F is resolved into F1 and F2. θ is the angle between F and
extended line of r.
The component of F which produces rotation in the body is F1.
The magnitude of torgue (π) is the product of the magnitudes of r and F1.
Equation (1) shows that torque is the cross-product of displacement r and force F.
Torque → positive if directed outward from paper
Torque → negative if directed inward from paper
The direction of torque can be found by using Right Hand Rule and is always perpendicular to the plane containing r & F.
Thus
Clockwise torque → negative
Counter-Clockwise torque → positive
The component of F which produces rotation in the body is F1.
The magnitude of torgue (π) is the product of the magnitudes of r and F1.
Equation (1) shows that torque is the cross-product of displacement r and force F.
Torque → positive if directed outward from paper
Torque → negative if directed inward from paper
The direction of torque can be found by using Right Hand Rule and is always perpendicular to the plane containing r & F.
Thus
Clockwise torque → negative
Counter-Clockwise torque → positive
Alternate
Definition of Torque
π = r x F
|π| = r F sin θ
|π| = F x r sin θ
But r sin θ = L (momentum arm) (from figure)
Therefore,
|π| = F L
Magnitude of Torque = Magnitude of force x Moment Arm
π = r x F
|π| = r F sin θ
|π| = F x r sin θ
But r sin θ = L (momentum arm) (from figure)
Therefore,
|π| = F L
Magnitude of Torque = Magnitude of force x Moment Arm
Note
If line of action of force passes through the axis of rotation then this force cannot produce torque.
The unit of torque is N.m.
If line of action of force passes through the axis of rotation then this force cannot produce torque.
The unit of torque is N.m.
Couple
Two forces are said to constitute a couple if they have
1. Same magnitudes
2. Opposite directions
3. Different lines of action
These forces cannot produces transiatory motion, but produce rotatory motion.
Two forces are said to constitute a couple if they have
1. Same magnitudes
2. Opposite directions
3. Different lines of action
These forces cannot produces transiatory motion, but produce rotatory motion.
Moment
(Torque) of a Couple
Consider a couple composed of two forces F and -F acting at points A and B (on a body) respectively, having position vectors r1 & r2.
If π1 is the torque due to force F, then
π1 = r1 x F
Similarly if π2 is the torque due to force – F, then
π2 = r2 x (-F)
The total torque due to the two forces is
π = π1 + π2
π = r1 x F + r2 x (-F)
π = r1 x F – r2 x (-F)
π = (r1 – r2) x F
π = r x F
where r is the displacement vector from B to A.
The magnitude of torque is
π = r F sin (180 – θ)
π = r F sin θ ……………….. {since sin (180 – θ) = sin θ}
Where θ is the angle between r and -F.
π = F (r sin θ)
But r sin θ is the perpendicular distance between the lines of action of forces F and -F is called moment arm of the couple denoted by d.
π = Fd
Thus
[Mag. of the moment of a couple] = [Mag. of any of the forces forming the couple] x [Moment arm of the couple]
Moment (torque) of a given couple is independent of the location of origin.
Consider a couple composed of two forces F and -F acting at points A and B (on a body) respectively, having position vectors r1 & r2.
If π1 is the torque due to force F, then
π1 = r1 x F
Similarly if π2 is the torque due to force – F, then
π2 = r2 x (-F)
The total torque due to the two forces is
π = π1 + π2
π = r1 x F + r2 x (-F)
π = r1 x F – r2 x (-F)
π = (r1 – r2) x F
π = r x F
where r is the displacement vector from B to A.
The magnitude of torque is
π = r F sin (180 – θ)
π = r F sin θ ……………….. {since sin (180 – θ) = sin θ}
Where θ is the angle between r and -F.
π = F (r sin θ)
But r sin θ is the perpendicular distance between the lines of action of forces F and -F is called moment arm of the couple denoted by d.
π = Fd
Thus
[Mag. of the moment of a couple] = [Mag. of any of the forces forming the couple] x [Moment arm of the couple]
Moment (torque) of a given couple is independent of the location of origin.
Centre of Mass
Definition
The centre of mass of a body, or a system of particles, is a point on the body that moves in the same way that a single particle would move under the influence of the same external forces. The whole mass of the body is supposed to be concentrated at this point.
The centre of mass of a body, or a system of particles, is a point on the body that moves in the same way that a single particle would move under the influence of the same external forces. The whole mass of the body is supposed to be concentrated at this point.
Explanation
During translational motion each point of a body moves in the same manner i.e., different particles of the body do not change their position w.r.t each other. Each point on the body undergoes the same displacement as any other point as time goes on. So the motion of one particle represents the motion of the whole body. But in rotating or vibrating bodies different particles move in different manners except one point called centre of mass. The centre of mass of a body or a system of particle is a point which represents the movement of the entire system. It moves in the same way that a single particle would move under the influence of same external forces.
During translational motion each point of a body moves in the same manner i.e., different particles of the body do not change their position w.r.t each other. Each point on the body undergoes the same displacement as any other point as time goes on. So the motion of one particle represents the motion of the whole body. But in rotating or vibrating bodies different particles move in different manners except one point called centre of mass. The centre of mass of a body or a system of particle is a point which represents the movement of the entire system. It moves in the same way that a single particle would move under the influence of same external forces.
Centre of
Mass and Centre of Gravity
In a completely uniform gravitational field, the centre of mass and centre of gravity of an extended body coincides. But if gravitational field is not uniform, these points are different.
In a completely uniform gravitational field, the centre of mass and centre of gravity of an extended body coincides. But if gravitational field is not uniform, these points are different.
Determination
of the Centre of Mass
Consider a system of particles having masses m1, m2, m3, …………….. mn. Suppose x1, and z1, z2, z3 are their distances on z-axis, all measured from origin.
Consider a system of particles having masses m1, m2, m3, …………….. mn. Suppose x1, and z1, z2, z3 are their distances on z-axis, all measured from origin.
Equilibrium
A body is said to be in equilibrium if it is
1. At rest, or
2. Moving with uniform velocity
A body in equilibrium possess no acceleration.
1. At rest, or
2. Moving with uniform velocity
A body in equilibrium possess no acceleration.
Static
Equilibrium
The equilibrium of bodies at rest is called static equilibrium. For example,
1. A book lying on a table
2. A block hung from a string
The equilibrium of bodies at rest is called static equilibrium. For example,
1. A book lying on a table
2. A block hung from a string
Dynamic
Equilibrium
The equilibrium of bodies moving with uniform velocity is called dynamic equilibrium. For example,
1. The jumping of a paratrooper by a parachute is an example of uniform motion. In this case, weight is balanced by the reaction of the air on the parachute acts in the vertically upward direction.
2. The motion of a small steel ball through a viscous liquid. Initially the ball has acceleration but after covering a certain distance, its velocity becomes uniform because weight of the ball is balanced by upward thrust and viscous force of the liquid. Therefore, ball is in dynamic equilibrium.
The equilibrium of bodies moving with uniform velocity is called dynamic equilibrium. For example,
1. The jumping of a paratrooper by a parachute is an example of uniform motion. In this case, weight is balanced by the reaction of the air on the parachute acts in the vertically upward direction.
2. The motion of a small steel ball through a viscous liquid. Initially the ball has acceleration but after covering a certain distance, its velocity becomes uniform because weight of the ball is balanced by upward thrust and viscous force of the liquid. Therefore, ball is in dynamic equilibrium.
Angular
Momentum
Definition
The quantity of rotational motion in a body is called its angular momentum. Thus angular momentum plays same role in rotational motion as played by linear momentum in translational motion.
Mathematically, angular momentum is the cross-product of position vector and the linear momentum, both measured in an inertial frame of reference.
ρ = r x P
The quantity of rotational motion in a body is called its angular momentum. Thus angular momentum plays same role in rotational motion as played by linear momentum in translational motion.
Mathematically, angular momentum is the cross-product of position vector and the linear momentum, both measured in an inertial frame of reference.
ρ = r x P
Explanation
Consider a mass ‘m’ rotating anti-clockwise in an inertial frame of reference. At any point, let P be the linear momentum and r be the position vector.
ρ = r x P
ρ = r P sinθ ……….. (magnitude)
ρ = r m V sinθ ………. {since P = m v)
where,
V is linear speed
θ is the angle between r and P
θ = 90º in circular motion (special case)
The direction of the angular momentum can be determined by the Righ-Hand Rule.
Also
ρ = r m (r ω) sin θ
ρ = m r2 ω sin θ
Consider a mass ‘m’ rotating anti-clockwise in an inertial frame of reference. At any point, let P be the linear momentum and r be the position vector.
ρ = r x P
ρ = r P sinθ ……….. (magnitude)
ρ = r m V sinθ ………. {since P = m v)
where,
V is linear speed
θ is the angle between r and P
θ = 90º in circular motion (special case)
The direction of the angular momentum can be determined by the Righ-Hand Rule.
Also
ρ = r m (r ω) sin θ
ρ = m r2 ω sin θ
Units of
Angular Momentum
The units of angular momentum in S.I system are kgm2/s or Js.
1. ρ = r m V sin θ
= m x kg x m/s
= kg.m2/s
The units of angular momentum in S.I system are kgm2/s or Js.
1. ρ = r m V sin θ
= m x kg x m/s
= kg.m2/s
2. ρ = r P sin θ
= m x Ns
= (Nm) x s
= J.s
= m x Ns
= (Nm) x s
= J.s
Dimensions of
Angular Momentum
[ρ] = [r] [P]
= [r] [m] [V]
= L . M . L/T
= L2 M T-1
[ρ] = [r] [P]
= [r] [m] [V]
= L . M . L/T
= L2 M T-1
Relation
Between Torque and Angular Momentum
OR
Prove that
the rate of change of angular momentum is equal to the external torque acting
on the body.
Proof
We know that rate of change of linear momentum is equal to the applied force.
F = dP / dt
Taking cross product with r on both sides, we get
R x F = r x dP / dt
τ = r x dP / dt ……………………….. {since r x P = τ}
Now, according to the definition of angular momentum
ρ = r x P
Taking derivative w.r.t time, we get
dρ / dt = d / dt (r x P)
=> dρ / dt = r x dP / dt + dr / dt x P
=> dρ / dt = τ + V x P ……………… {since dr / dt = V}
=> dρ / dt = τ + V x mV
=> dρ / dt = τ + m (V x V)
=> dρ / dt = τ + 0 …………….. {since V x V = 0}
=> dρ / dt = τ
Or, Rate of change of Angular Momentum = External Torque …………….. (Proved)
We know that rate of change of linear momentum is equal to the applied force.
F = dP / dt
Taking cross product with r on both sides, we get
R x F = r x dP / dt
τ = r x dP / dt ……………………….. {since r x P = τ}
Now, according to the definition of angular momentum
ρ = r x P
Taking derivative w.r.t time, we get
dρ / dt = d / dt (r x P)
=> dρ / dt = r x dP / dt + dr / dt x P
=> dρ / dt = τ + V x P ……………… {since dr / dt = V}
=> dρ / dt = τ + V x mV
=> dρ / dt = τ + m (V x V)
=> dρ / dt = τ + 0 …………….. {since V x V = 0}
=> dρ / dt = τ
Or, Rate of change of Angular Momentum = External Torque …………….. (Proved)
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