Physics Notes: System of Particles and Rotational Motion

Centre of Mass

The center of mass of a system is the point where the entire mass of the system is concentrated, and all external forces act upon it.
For rigid bodies, the center of mass is independent of its state (rest or accelerated motion).
Centre of Mass of System of n Particles:
- If a system has n particles with masses m1, m2, m3, …, mn and position vectors r1, r2, r3, …, rn, the position vector of the center of mass is:
  r{cm} = \frac{m1r1 + m2r2 + m3r3 + … + mnrn}{m1 + m2 + m3 + … + m_n}
Centre of Mass of Two Particle System:
- Choosing O as the origin:
  r{cm} = \frac{m1r1 + m2r2}{m1 + m_2}
Position of center of mass from m2: \frac{m1d}{m1 + m2}
If position vectors of particles of masses m1 and m2 are r1 and r2 respectively, then:
If in a two-particle system, particles of masses m1 and m2 are moving with velocities v1 and v2 respectively, the velocity of the center of mass is:
v{cm} = \frac{m1v1 + m2v2}{m1 + m_2}
If the accelerations of the particles are a1 and a2 respectively, then the acceleration of the center of mass is:
a{cm} = \frac{m1a1 + m2a2}{m1 + m_2}
The center of mass of an isolated system has a constant velocity.
An isolated system remains at rest if initially at rest, or moves with the same velocity if initially in motion.
The position of the center of mass depends upon the shape, size, and mass distribution of the body.
The center of mass of an object need not lie within the object.
In symmetrical bodies with homogeneous mass distribution, the center of mass coincides with the geometrical center.
The position of the center of mass of an object changes in translatory motion but remains unchanged in rotatory motion.

Translational Motion

A rigid body performs pure translational motion if each particle of the body undergoes the same displacement in the same direction in a given interval of time.

Rotational Motion

A rigid body performs pure rotational motion if each particle of the body moves in a circle, and the centers of all circles lie on a straight line called the axis of rotation.

Rigid Body

If the relative distance between the particles of a system does not change upon applying force, it is called a rigid body.
General motion of a rigid body consists of both translational and rotational motion.

Moment of Inertia

The inertia of rotational motion is called the moment of inertia, denoted by I.
Moment of inertia is the property of an object that opposes any change in its state of rotation about an axis.
The moment of inertia of a body about a given axis is the sum of the products of the masses of its constituent particles and the square of their respective distances from the axis of rotation.
Its unit is kg \cdot m^2, and its dimensional formula is [ML^2].
The moment of inertia of a body depends upon:
- Position of the axis of rotation
- Orientation of the axis of rotation
- Shape and size of the body
- Distribution of mass of the body about the axis of rotation
The physical significance of the moment of inertia in rotational motion is the same as mass in linear motion.

Radius of Gyration

The root mean square distance of its constituent particles from the axis of rotation is called the radius of gyration, denoted by K.
The product of the mass of the body M and the square of its radius of gyration K gives the moment of inertia of the body about the rotational axis.
I = MK^2
K = \sqrt{\frac{I}{M}}

Parallel Axes Theorem

The moment of inertia of any object about any arbitrary axis is equal to the sum of the moment of inertia about a parallel axis passing through the center of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes. I = I_{CM} + Mr^2
- I is the moment of inertia about the arbitrary axis.
- I_{CM} is the moment of inertia about the parallel axis through the center of mass.
- M is the total mass of the object.
- r is the perpendicular distance between the axes.

Perpendicular Axes Theorem

The moment of inertia of any two-dimensional body about an axis perpendicular to its plane (Iz) is equal to the sum of the moments of inertia of the body about two mutually perpendicular axes lying in its own plane and intersecting each other at a point where the perpendicular axis passes through it. Iz = Ix + Iy
- Ix and Iy are the moments of inertia of the plane lamina about perpendicular axes X and Y, respectively, which lie in the plane lamina and intersect each other.
The parallel axes theorem is applicable for any type of rigid body (2D or 3D), while the perpendicular axes theorem is applicable for laminar (2D) bodies only.

Moment of Inertia of Homogeneous Rigid Bodies

Thin Circular Ring:
- Axis passing through its center and perpendicular to its plane: I = MR^2
- About a tangent perpendicular to its plane: I = \frac{3}{2}MR^2
- About a tangent in its plane: I = \frac{5}{4}MR^2
- About a diameter: I = \frac{1}{2}MR^2
Thin Rod:
- About an axis passing through its center and perpendicular to its length: I = \frac{ML^2}{12}
- About an axis passing through its one end and perpendicular to its length: I = \frac{ML^2}{3}
Solid Cylinder:
- About its geometrical axis: I = \frac{MR^2}{2}
- About an axis passing through its outer face along its length: I = MR^2
- About an axis passing through its centre and perpendicular to its length: I = \frac{ML^2}{12} + \frac{MR^2}{4}
- About an axis passing through its diameter of circular surface: I = \frac{ML^2}{3} + \frac{MR^2}{4}
Rectangular Plate:
- About an axis passing through its center and perpendicular to its plane: I = \frac{M(a^2 + b^2)}{12}
Thin Spherical Shell:
- About its any diameter: I = \frac{2}{3}MR^2
- About its any tangent: I = \frac{5}{3}MR^2
Solid Sphere:
- About its any diameter: I = \frac{2}{5}MR^2

Equations of Rotational Motion

\omega = \omega_0 + \alpha t
\theta = \omega_0 t + \frac{1}{2} \alpha t^2
\omega^2 = \omega_0^2 + 2 \alpha \theta
- \theta is the displacement in rotational motion.
- \omega_0 is the initial velocity.
- \omega is the final velocity.
- \alpha is the angular acceleration.

Torque

Torque or moment of a force about the axis of rotation: \tau = r \times F = rF \sin(\theta) \hat{n}
It is a vector quantity.
If the force tends to rotate the object clockwise, the torque is negative; if anticlockwise, it is positive.
Its SI unit is Newton-meter, and its dimension is [ML^2T^{-2}].
In rotational motion, torque: \tau = I\alpha where \alpha is angular acceleration and I is the moment of inertia.

Angular Momentum

The moment of linear momentum is called angular momentum, denoted by L.
Angular momentum: L = I \omega = mvr
In vector form: L = I \omega = r \times mv
Its unit is Joule-second, and its dimensional formula is [ML^2T^{-1}].
Torque: \tau = \frac{dL}{dt}

Conservation of Angular Momentum

If the external torque acting on a system is zero, its angular momentum remains conserved.
If \tau{ext} = 0, then L = I \omega = constant I1 \omega1 = I2 \omega_2

Angular Impulse

Total effect of a torque applied on a rotating body in a given time is called angular impulse.
Angular impulse is equal to the total change in angular momentum of the system in the given time.

MCQ I

7.1 For which of the following does the centre of mass lie outside the body ?
- (d) A bangle
7.2 Which of the following points is the likely position of the centre of mass of the system shown in Fig. 7.1?
- (b) B
7.3 A particle of mass m is moving in yz-plane with a uniform velocity v with its trajectory running parallel to +ve y-axis and intersecting z-axis atz = a (Fig. 7.2). The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is:
- (b) 2mva êx
7.4 When a disc rotates with uniform angular velocity, which of the following is not true?
- (d) The angular acceleration is non-zero and remains same.
7.5 A uniform square plate has a small piece Q of an irregular shape removed and glued to the centre of the plate leaving a hole behind (Fig. 7.3). The moment of inertia about the z-axis is then
- (c) the same
7.6 In problem 7.5, the CM of the plate is now in the following quadrant of x-y plane,
- (c) III
7.7 The density of a non-uniform rod of length 1m is given by (x) = a(1+bx2) where a and b are constants and o  x  1. The centre of mass of the rod will be at
- (a) 4(3 + b) / 3(2 + b)
7.8 A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed  . A person of mass M is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (asseen from the round). The speed of the round afterwards is
- (a) 2

MCQ II

7.9 Choose the correct alternatives:
- (a) For a general rotational motion, angular momentum L and angular velocity  need not be parallel.
- (c) For a general translational motion , momentum p and velocity v are always parallel.
7.10 Figure 7.4 shows two identical particles 1 and 2, each of mass m, moving in opposite directions with same speed v along parallel lines. At a particular instant, r1 and r2 are their respective position vectors drawn from point A which is in the plane of the parallel lines . Choose the correct options:
- (a) Angular momentum l of particle 1 about A is l = mvd 1
- (d) TotalangularmomentumofthesystemaboutAisl = mv(d2 − d1)  represents a unit vector coming out of the page.
7.11 The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it ?
- (a) The forces may be acting radially from a point on the axis.
- (b) The forces may be acting on the axis of rotation.
- (d) The torque caused by some forces may be equal and opposite to that caused by other forces.
7.12 Figure 7.5 shows a lamina in x-y plane. Two axes z and z pass perpendicular to its plane. A force F acts in the plane of lamina at point P as shown. Which of the following are true? (The point P is closer to z-axis than the z-axis.)
- (a) Torque  caused by F about z axisis along -k ˆ .
- (b) Torque  caused by F about z axis is along -k ˆ .
7.13 With reference to Fig. 7.6 of a cube of edge a and mass m,state whether the following are true or false. (O is the centre of the cube.)
- (a) The moment of inertia of cube about z-axis is I z = I x + I y
- (d) I x = I y

VSA

7.14 The centre of gravity of a body on the earth coincides with its centre of mass for a ‘small’ object whereas for an ‘extended’ object it may not. What is the qualitative meaning of ‘small’ and ‘extended’ in this regard? For which of the following the two coincides? A building, a pond, a lake, a mountain?
- The qualitative meaning of 'small' is that the object's dimensions are much smaller than the Earth's radius, so the variation in gravitational acceleration across the object is negligible. 'Extended' means the object's dimensions are comparable to or larger than the scale over which the Earth's gravitational field varies significantly.
- For a building, a pond, and a lake, the center of gravity and center of mass coincide because their dimensions are small compared to the Earth's radius. For a mountain, they may not coincide because of its extended size and the variation in gravitational acceleration with height.
7.15 Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about an axis passing through their axes of symmetry?
- A solid sphere has a smaller moment of inertia than a hollow cylinder of the same mass and radius because the mass in the solid sphere is distributed closer to the axis of rotation, while in the hollow cylinder, all the mass is located at the radius R.
7.16 The variation of angular position  , of a point on a rotating rigid body, with time t is shown in Fig. 7.7. Is the body rotating clock-wise or anti-clockwise?
- The body is rotating clockwise because the angular position is decreasing with time.

SA

7.19 The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?
- No, it does not necessarily mean that the vector sum of the torques is zero about any arbitrary point. If the vector sum of forces is non-zero, the torque will generally be different about different points.
7.20 A wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium? How would you set a half-wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?
- The wheel is in mechanical equilibrium because the net external force and net external torque on the entire wheel are zero. The centripetal acceleration experienced by individual particles is due to internal forces within the wheel, not external forces.
- To set a half-wheel into uniform motion about an axis passing through its center of mass, you would need to apply an initial torque to start the rotation. Once in motion, if there are no external torques (e.g., friction), it will continue rotating at a constant angular velocity without the need for external forces to sustain the motion.
7.21 A door is hinged at one end and is free to rotate about a vertical axis (Fig. 7.10). Does its weight cause any torque about this axis? Give reason for your answer.
- No, the weight of the door does not cause a torque about the vertical axis of the hinges. This is because the line of action of the weight (acting at the center of mass of the door) passes through the axis of rotation provided the axis is perfectly vertical. Therefore, the lever arm (the perpendicular distance from the axis of rotation to the line of action of the force) is zero, resulting in zero torque.

LA

7.24 Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed 1 and 2 are brought into contact face to face with their axes of rotation coincident.
- (a) Does the law of conservation of angular momentum apply to the situation? why?
  - Yes, the law of conservation of angular momentum applies to this situation because there are no external torques acting on the system of two discs.
- (b) Find the angularspeed of the two-disc system.
  - \omega = \frac{I1 \omega1 + I2 \omega2}{I1 + I2}
- (c) Calculate the loss in kinetic energy of the system in the process.
  - Loss = \frac{I1 I2 (\omega1 - \omega2)^2}{2(I1 + I2)}
- (d) Account for thisloss.
  - The loss in kinetic energy is due to the work done by the frictional forces between the two discs as they are brought into contact and their surfaces slip against each other until they reach a common angular speed. This work is dissipated as heat.