Formula sheet: System of Particles and Rotational Motion

System of Particles and Rotational Motion15 min readLesson 1 of 8

1.1 Centre of mass and the two motions A system of particles can be treated as a single body located at its centre of mass, with the total mass. The external force on the system determines the acceleration of that point, while the internal forces cancel in pairs. The internal motion can be described as motion about the centre of mass, so every rigid body problem splits into a translation of the centre of mass and a rotation about it. The centre of mass of a continuous body is found by replacing the sum over particles with an integral over the mass distribution. For a symmetric body the point lies on the axis of symmetry, and for an irregular body it may fall outside the material.

R_{cm} = \frac{1}{M} i \sum m_{i} r_{i}, F_{ext} = M a_{cm}

Centre of mass of a system and the motion of that point

τ = r \times F, τ = r F sin θ

Torque of a force about a point

1.2 Rotation, angular momentum and equilibrium Angular velocity describes the rate of rotation and is the same for every point of a rigid body. Angular acceleration is the rate of change of angular velocity. The torque of a force about an axis measures its turning effect, and the torque changes the angular momentum of a body, just as force changes linear momentum. A particle carries angular momentum equal to the cross product of its position with its momentum. Multiply the moment of inertia by the angular velocity to get the corresponding value for a rigid body about a fixed axis. When the net external torque on a system is zero, the total angular momentum stays constant. A rigid body is in equilibrium when both the net force and the net torque on it are zero.

L = r \times p, L = I ω, τ = \frac{d L}{d t}

Angular momentum of a particle and of a rigid body, and the torque relation

\sum F = 0, \sum τ = 0

The two equilibrium conditions for a rigid body

1.3 Moment of inertia The moment of inertia of a body about an axis measures how its mass is distributed about that axis, and it plays the role of mass in rotational motion. A given body has different moments of inertia about different axes, and pulling mass farther from the axis increases it. The standard results for a ring, a disc, a rod, a sphere and a cylinder should be remembered along with their axes. The parallel axis theorem takes the centre-of-mass value and adds the total mass times the square of the gap to a parallel axis. The perpendicular axis theorem applies to a flat lamina and relates the moments about two perpendicular axes in the plane to the moment about the axis perpendicular to the plane. These two theorems generate most of the results needed in numericals.

I = \sum m_{i} r_{i}^{2}, I = \int r^{2} d m

Moment of inertia of a discrete system and of a continuous body

I = I_{cm} + M d^{2}, I_{z} = I_{x} + I_{y}

Parallel axis theorem and perpendicular axis theorem

1.4 Rotational dynamics and rolling The rotational form of the second law is torque equals moment of inertia times angular acceleration, for a fixed axis. Rotational kinetic energy is half the moment of inertia times the square of the angular velocity. A rolling body carries both translational and rotational kinetic energy, and the rolling condition ties the speed of the centre of mass to the angular speed through the radius. For a body rolling down an incline without slipping, the acceleration is g sin theta divided by one plus the ratio of the moment of inertia to mass times radius squared. The factor from the moment of inertia makes different bodies reach the bottom with different accelerations: a solid sphere beats a hollow sphere, and a ring is slowest. Friction is essential for rolling, but in the ideal case it does no work because the contact point is momentarily at rest.

τ = I α, K_{rot} = \frac{1}{2} I ω^{2}, K_{rolling} = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}

Rotational second law and kinetic energy of rotation and rolling

a = \frac{g sin θ}{1 + \frac{I}{m R ^{2}}}, v = ω R

Acceleration of a body rolling without slipping down an incline

1.5 Common moments of inertia and units A ring of radius R about its central axis has moment of inertia M R squared, a disc has half of that, a solid sphere has two fifths of M R squared and a hollow sphere two thirds. A rod of length L about a perpendicular axis through its centre has M L squared over twelve, and about its end M L squared over three. Torque is in newton metre, angular velocity in radian per second, angular acceleration in radian per second squared, angular momentum in joule second and moment of inertia in kilogram metre squared.

Exam

Every rigid body problem starts with the axis. The moment of inertia, the torque and the angular momentum all depend on that choice, so write the axis and the sign convention on the diagram before touching the formulas.

in hinglish

Is chapter do hisson mein baanta jaata hai: system of particles aur rotational motion . Centre of mass system ka ek point hai jahan total mass rakhi hui maani jaati hai, aur external force usi point ko accelerate karti hai. Torque r x F hai, aur rigid body ke liye tau = I alpha . Angular momentum L = I omega hota hai, aur net torque zero ho to L conserved rehta hai. Moment of inertia mass distribution par depend karta hai, isliye ek hi body ka alag axes ke liye alag I hota hai. Parallel axis theorem I = I_cm + M d^2 aur perpendicular axis theorem I_z = I_x + I_y yaad rakho. Rolling mein v = omega R hota hai, aur incline par acceleration g sin theta /(1 + I/(mR^2)) milta hai; solid sphere sabse pehle pahunchti hai. Numerical mein pehle axis choose karo, phir I nikaalo, phir equation likho.