System of Particles and Rotational Motion Flashcards
Centre of Mass
Point which moves in the same way in which a single particle having the total mass of the system and acted upon by the same external force would move
Centre of mass of a two-particle system
R = (m1r1 + m2r2) / (m1 + m2)
Prove centre of mass of two particle system from ab-initio
Proved
Centre of mass coordinates of n-particle system
xCM = (m1x1 + m2x2 + ….. + mnxn) / (m1+m2+m3+m4+…mn)
Similarly, y and z
Binary Systems example
- Binary stars
- Diatomic molecule
- Earth-moon system
Rigid Body
Does not undergo size and shape, however large the external force may be acting on it
Axis of rotation
- Body possessing rotational motion if all particles move along circles in parallel planes
- Centres of these cricles lie on a fixed line perpendicular to parallel planes
Equation of rotational motion
- w = w0 + (alpha)(t)
- theta = w0 + 1/2(alpha)(t)^2
- w^2 - w0^2 = 2(alpha)(theta)
Torque
- Depends on magnitude of force
- Depends on Perpendicular distance
- Measurement of product of magnitude of force and perpendicular distance between line of action of force and axis of rotation
- r * F
Principal of moments
When a body is in rotational equilibrium, sum of clockwise moments about any point is equal to the sum of anticlockwise moments about that point or algebraic sum of moments about any point is zero
Relation between Power and torque
P = torque * omega
Angular momentum
Moment of linear momentum of particle about that axis
L = r * p
Relation between torque and angular momentum + derivation
d(L) / dt = torque
d (r * p) / dt = dr/dt * p +r * dp/dt
= 0 + r * F (0 as v * p = 0)
= torque
2 kinds of motion of a rigid body
- Translation - Particles move with same velocity
- Rotational motion - About an axis
Rotational Equilibrium
Resultant of torques due to all forces acting on body about any point must be zero
Translational equilibrium
Resultant of all external forces acting on the body must be zero
Moment of inertia
Moment of inertia of a rigid body about a fixed axis is defined as sum of products of masses of the particles constituting the body and the squares of their respective distances from axis of rotation
Moment of Inertia formula
I = sigma (mr^2)
Factors on which I depends upon
- Mass of body
- Size and shape of body
- Distribution of mass about axis of rotation
- Position and orientation of axis of rotation
Relation between rotational KE and I
KE = 1/2 * I * (omega)^2
Total KE of a rolling body
Translational KE + Rotational KE
= 1/2 mv^2 + 1/2 * I (omega)^2
Relation between MI and angular momentum
L = (I) * (omega)
Relation between MI and Torque
torque = I * alpha
Radius of gyration
I = MK^2 or K = root (I / M)
Moment of inertia for:
1. Thin rod of length L (passing through centre)
2. Thin rod of length L (through one end and perpendicular to length)
3. Circular ring (passing through centre and perpendicular to plane)
4. Circular ring (passing through diameter)
5. Circular ring (tangent in its plane)
6. Circular disc (passing through centre and perpendicular to plane)
7. Circular disc (passing through diameter)
8. Circular disc (tangent in its plane)
9. Right circular solid
10. Hollow cylinder
11. Solid sphere (through diameter)
12. Hollow sphere (symmetry axis)
- I = 1/12 * ML^2
- I = 1/3 * ML^2
- I = MR^2
- I = MR^2 / 2
- I = 3/2 * MR^2
- I = 1/2 * MR^2
- I = 1/4 * MR^2
- I = 5/4 * MR^2
- I = 1/2 * MR^2
- I = MR^2
- I = 2/5 * MR^2
- I = 2/3 * MR^2
