Samacheer Class 11 Physics - Motion of System of Particles and Rigid Bodies

Torque is defined as the moment of the external applied force about a point or axis of rotation. The expression for torque is,
\(\vec{\tau}\) = \(\vec{r}\) x \(\vec{F}\)

1) The torque is zero when r and P are either parallel or anti parallel
i.e θ = 0, if parallel sin θ = 0, τ = 0
θ = 180°, if anti parallel sin 180 = 0, τ = 0
2) The torque is zero if the force acts at the reference point as \(\vec { r }\) = 0, τ = 0

A pair of forces which are equal in magnitude but opposite in direction and separated by a perpendicular distance so that their lines of action do not coincide that causes a turning effect is called a couple.

Principle of moments states that the sum of the clockwise moments is equal to sum of the anti-clockwise moments about the axis rotation.
d 1 f 1 = d 2 f 2

The center of gravity of a body is the point at which the entire weight of the body acts, irrespective of the position and orientation of the body.

* For rotational motion moment of Inertia is a measure of rotational Inertia.
* The Moment of Inertia of a body is not a variable quantity.
It depends not only on the mass of the body but also on the way the mass is distributed around the axis of rotation.

The radius of gyration of an object is the perpendicular distance from the axis of rotation to an equivalent point mass, which would have the same mass as well as the same moment of inertia of the object.

In the case of pure roiling, for all points on the edge, the magnitude of V t Translation is V rotational are equal (V tr = V rot ).
As V Trans = V cm and V rot = Rω, the condition
for pure rolling without slipping is Vcm = Rω.

Sliding:
* Velocity of center of mass is greater than Rω i.e. V CM > Rω.
* Velocity of transnational motion is greater than velocity of rotational motion.
* Resultant velocity acts in the forward direction.
Slipping:
* Velocity of center of mass is lesser than Rω. i.e. V CM < Rω
* Velocity of translation motion is lesser than velocity of rotational motion.
* Resultant velocity acts in the backward direction.
III. Long Answer Questions:

The different types of equilibrium are:
i) Translational equilibrium – The body is set to be in translational equilibrium, when the linear momentum remains constant. The net force acting on body is zero
Example: Let F1, F2, F3 …… Fn be n forces acting on body in different directions, and the net force on the body is the vector sum of different forces is the resultant, which is equal to zero, then the body is in translational equilibrium.
ii) Rotational equilibrium – The body is said to be in rotational equilibrium when the angular momentum is constant. Net Torque on the body is equal to zero
Example: If a metre scale is balanced by a knife edge at its center then the body is in rotational equilibrium as clockwise moment is equal to Anticlockwise moment.
iii) Static equilibrium – When the linear momentum and angular momentum on body are zero, where no net force and no net torque, the body is in static equilibrium.
Example: A book kept in the table
iv) Dynamic equilibrium – Here linear momentum and angular momentum are kept constant. Net force and Net torque on a body is equal to zero.
v) Stable equilibrium:
Consider a china dish in which a small sphere is placed. It comes to rest at the bottom of the dish, said to be in stable equilibrium where potential energy is minimum. If displaced, the sphere tries to attain it original position after released. So the body is in stable equilibrium. In stable equilibrium.
a) Linear momentum and angular momentum are zero
b) The body tries to come back to equilibrium if slightly disturbed and released.
c) The centre of mass of the body shifts slightly higher if disturbed from equilibrium.
d) Potential energy is minimum and it is increased if disturbed.
Unstable equilibrium:
Consider an inverted china dish and place the sphere on the top of the dish. The body roll down to a new position if disturbed. It never returns to its original position stating unstable equilibrium. Incase of unstable equilibrium
a) Linear momentum and angular momentum are zero.
b) The body cannot come back to its original equilibrium if slightly disturbed and released.
c) The centre of mass of the body shifts slightly lower if disturbed from equilibrium.
d) potential energy is not minimum and it decreases if disturbed.
Neutral equilibrium:
Consider the sphere placed in a plane surface. The position of center of mass never changes, potential energy never changes if disturbed. So the equilibrium is said to be neutral equilibrium. In case of neutral equilibrium.
a) Linear and angular momentum are zero.
b) The body remains at same equilibrium if slightly disturbed and released.
c) The center of mass of the body does not shift higher or lower if disturbed from equilibrium
d) Potential energy remains same even if disturbed.

The center of gravity of an irregularly shaped lamina by pivoting it at various points by trail and error. The lamina remains horizontal when pivoted at the point where the net gravitational force acts, which is the centre of gravity shown figure.
When the body is supported at the centre for gravity, the sum of torques acting on all point masses of the rigid body becomes zero. Moreover the weight is compensated by the normal reaction force exerted by the pivot. The body in static equilibrium and hence it is horizontal.

Consider a cyclist negotiating a circular level road of radius ‘r’ with a velocity v. Cycle and cyclist are taken to be one system with mass m. The centre of gravity of the system is c and it goes in circle of radius r with centre o.
Consider OC as x – axis, vertical through 0 is z axis.
The system as a frame is rotating about z axis.
The system is at rest in the rotating frame.
centrifugal force on the system = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\)
This force will act through centre of gravity.
The forces acting on the system are
* Gravitational force (mg)
* Normal force (N)
* Frictional force (f)
* Centrifugal force \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\)
As the system is in equilibrium in the rotational frame of reference, the net external force and external torque must be zero. Consider all torques about point A for rotational equilibrium.
τ net = 0.
Tongue due to gravitational force about point A = mg AB. Taken as negative as it produces clockwise momentum.
Torque due to centripetal force = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\)(BC)
Taken as positive as it produces an anticlockwise moment.
-mg AB + \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) BC = 0
g AB = \(\frac{\mathrm{v}^{2}}{\mathrm{r}}\) BC
from ∆ ABC
AB = AC sin θ
BC = AC cos θ

Consider a uniform red of mass (m) and length \((\ell)\). Consider an axis passing through the geometric centre of the rod perpendicular to its length. Consider an infinitesimally small mass (dm) at a distance x from 0 origin of length dx. THe moment of Inertia (dl) of this mass (dm) about the axis is
dl = (dm)x²
As the mass is uniformly distributed, the mass per unit length (λ) if rod = λ = \(\frac{\mathrm{M}}{\ell}\).
The mass of infinitesimally small length dx as
dm = λ dx = \(\frac{\mathrm{M}}{\ell}\) dx
Moment of Inertria (I) of the entire rod can be obtained by integrating dl.
I = ∫ dI = ∫(dm) x² = ∫ \(\frac{\mathrm{M}}{\ell}\)x²dx
I = \(\frac{\mathrm{M}}{\ell}\) ∫x² dx
As the mass is distributed on either side of the origin the limits for integration are taken from

Consider a uniform ring of mass ‘M’ and radius R
Consider an axis passing through the center of the ring 0 and perpendicular to its plane take an infinitesimally small mass (dm) of length (dx) of the ring dm is located at a distance R from 0. The moment of Inertia (dl) of this small mass dm is
dI = (dm) R²
The length of the ring = Circumference of the ring \((\ell)\) = 2πR
As the mass is distributed uniformly
mass per unit length – λ = \(\frac {Mass}{length}\) = \(\frac { MR }{ 2πR }\)
The mass of infinitesimally small length is

Consider a disc of mass M and Radius R. This disc is made up of many infinitesimally small rings each of mass (dm) and thickness dr.
The moment of Inertia (dl) of this small ring is dl = (dm) r²
As the mass is uniformly distributed. The mass per unit area.
σ = \(\frac { Mass }{ Area }\) = \(\frac{M R}{\pi R^{2}}\)
The mass of infinitesimally small ring is
dm = σ (2π r. dr) \(\frac{\mathrm{M}}{\pi \mathrm{R}^{2}}\) 2 π r.dr
Where 2 π r.dr is the area of the elemental ring
∵ dm = \(\frac{2 \mathrm{M}}{\mathrm{R}^{2}}\) rdr
∵ dI = \(\frac{2 \mathrm{M}}{\mathrm{R}^{2}}\)r³dr
∵ The moment of Inertia of the entire disc is

When no external torque acts on a body, the net angular momentum of a rotating rigid body remains constant. This is known as the law of conservation of angular momentum.
τ = \(\frac { dL }{ dt }\)
If τ = 0
then L is a constant.
As angular momentum L = Iω, the conservation of angular momentum can be written as
I i ω i = I f ω f
Iω = constant
The above equation indicates if I increase ω will decrease and vice versa, to keep angular momentum constant.
Example:
Let us consider an ice dancer. The dancer spins slowly when the hands are stretched out and spins faster when the hands are brought close to the body.
Stretching the hands away from body increases the moment of Inertia, thus angular velocity decreases resulting in a slower spin. When the hands are brought to the body the moment of Inertia decreases and thus angular velocity in-creases resulting in faster spin.

Law:
Parallel axes theorem states that the moment of Inertia of a body about any axis is equal to the sum of its moment of Inertia about a parallel axis through its center of mass and the product of the mass of the body and the square of the perpendicular distance between two axis.
Let I c be the moment of Inertia of the body of mass m about an axis (AB) passing through the center of mass.
Let I be the moment of Inertia of a parallel axis (DE) at a distance d from AB is
I = Ic + Md²
Proof:
Consider a rigid body as in figure. Let Ic be the moment of Inertia of the body about an axis AB passing through center of mass c.
DE is another axis parallel to AB at a perpendicular distance d from AB. Let I be the moment of Inertia about the axis DE.
Consider a point P of mass ‘m’ on the body at a distance, x from c.
The moment of Inertia of the point mass about the axis DE = m(x+d)².
The moment of Inertia I of the whole body about
DE = I = ∑ m (x+d)²
I = ∑ m (x² + d² + 2xd)
I = ∑ mx² + md² + 2mxd
I = ∑ mx² + ∑ md² + 2d ∑ mx
Here ∑ mx² = moment of Inertia of the body about the center of mass
∴ I c = ∑ mx²
The term ∑mx = 0, because x can take positive and negative values w.r. to the axis AB.
I = I c + Imd²
but ∑m = M
∵ I = I c + Md²
Hence proved

Consider a round object of mass m and radius R is rolling down an inclined plane without slipping.
There are two forces acting on the object along the inclined plane.
One is the component of gravitational force mg sin θ and the other is the static frictional force ‘f’.
The other component of gravitational force mg cos θ is cancelled by the normal force N exerted by the plane.
Considering free body diagram for the inclined plane.
For translational motion, mg sin 0 is the supporting force and f is the opposing force
mg sin θ – f = ma → (1)
For rotational motion, we have to consider the torque w.r. to the centre of the object. Here mg sin 0 cannot cause torque is it passes through it, so frictional force causes the torque.
τ = I∝ = fR → (2)
W.K.T a =R∝ and moment of Inertia I = mk²
IV. Conceptual Questions:

The weight of the tree exerts a torque about the point where the cut is made. This causes the rotation of the tree about the cut. So the reason why the cut is made on the side facing the direction of fall.

When porter is carrying a sack of rice, on his back, the position of center of gravity changes. In order to bring it in the middle for stable equilibrium, porter bends forward.

It is much easier to balance a meter scale on your finger than balancing on a match stick. While balancing the meter scale in stable equilibrium. The vertical line passes through the center of gravity of scale should pass through the base of supporting material. The finger has more base over than a match stick.

Empty water bottle has large moment of Inertia than filled water bottle. So least value of radius of gyration is for bottle filled with water. So the water bottle filled with water reaches bottom first.
Suggests that for a given angle of inclination, the object with least value of radius of gyration k reach the bottom of incline first.

Angular momentum = L = Iω
Rotational K.E = K R = \(\frac { 1 }{ 2 }\)Iω²
By comparing
The shape of the graph is a straight line. The \(\frac { 1 }{ slope }\) of the curve gives the value of the moment of Inertia I. If angular momentum of two objects are same, rotational K.E one same having same moment of Inertia.

Yes, the normal force is concentrated at the center of mass. When force F is applied normal reaction of the floor shifts to right the block topples when N reaches edge.

Any two forces acting an a body can be combined to form a single force then it is called a resultant. In some cases effect of resultant is not same as the effect of the two forces acting
Example: A large body can be considered as made of a number of mass particles and all mass particles interact with each other. But the vector sum of all these internal forces is zero.
V. Numerical Problems:

m = 100 x 10 -3 kg = 100g
2r = 10 x 10 -2 m r = 5 cm
r = 5 x 10 -2 m v = 20 cm/s
V = 20 cms -1 = 20 x 10 -2 m/s
E = ?
for rolling body total K.E is
= \(\frac { 1 }{ 2 }\) mv² + \(\frac { 1 }{ 2 }\) Iw²
K.E = 3/4 x 100 x 20 x 20 x 10 -3 x 10 -4
= \(\frac { 3 }{ 4 }\) x 10 -3 x 4
= 3 x 10 -3 J

y = x + 4
Comparing it with straight line equation y = mx + c
m = tan θ = 1 θ = 45°
A line makes an angle θ = 45° with five x axis and intercepts on the y axis is + 4.
Draw a ⊥ r from 0 to the line y = x + 4
OA = d = 4sing 45 = \(\frac{4}{\sqrt{2}}\)
Angular momentum about 0 = Linear momentum x ⊥ r distance from 0

Here mass is distributed on length taking an element of small length dx at a distance x as in figure.
dm = \(\frac{\mathrm{M}}{\ell}\).dx
Moment of Inertia of small element

:
Mass of each H 2 atom = 1.7 x 10 -27 kg
Distance of each H 2 atom from the axis of rotation = 2 x 10 -10 m
Moment of Inertia about the axis = I = mr² + mr²
I = 2mr²
I = 2 x 1.7 x 10 -27 x (2 x 10 -10 )mr²
= 13.6 x 10 -47 kgm²
I = 13.6 x 10 -47 kgm²
I = 1.36 x 10 -46 kgm²

Force acting on the wheel after the plane touches the surface.
(Note: 747 has 16 wheels below wings and 2 wheels below nose)
If the wheel is not accelerating
N = W
Torque about certain wheel
τ = fR
But f = µN = µW
τ = µWR
Angular acceleration x = \(\frac { τ }{ I }\) = \(\frac{\mu \mathrm{WR}}{\frac{1}{2} \mathrm{MR}^{2}}\)
∵ τ = Iα
= \(\frac { 2 µ W }{ MR }\)
The time taken by the bowing wheel to stop skidding.
Part – II:
11th Physics Guide Motion of System of Particles and Rigid Bodies Additional Important Questions and Answers
I. Multiple choice questions: