Questions: System of Particles and Rotational Motion

Q1. Define the centre of mass and write its formula for a system of particles.

Answer: The centre of mass of a system is the point whose position is the mass weighted average of the positions of all the particles. For a system of masses m1, m2 and so on at positions r1, r2 and so on, its position vector is the sum of each mass times its position divided by the total mass. The external force on the system equals the total mass times the acceleration of this point, which is why the internal forces can be ignored for the translational motion.

Hinglish: Centre of mass woh point hai jahan total mass concentrated maani jaati hai, aur uska position mass weighted average hota hai: R = (sum m r)/(sum m). External force total mass guna is point ka acceleration deta hai, isliye internal forces translation ke liye ignore ki ja sakti hain.

Q2. Two particles of masses 2 kg and 6 kg are at x = 1 m and x = 5 m. Find the centre of mass.

Answer: The centre of mass is the weighted average: 2 times 1 plus 6 times 5, divided by 8, which is 32 over 8, so 4 m. The point lies closer to the 6 kg mass, as it must. If the masses were interchanged, the result would be 2 m, which is a good check on the formula because the heavier mass always pulls the centre toward itself.

Hinglish: Weighted average (2x1 + 6x5)/8 = 32/8 = 4 m. Point bhaari mass ke qareeb aata hai; masses badal do to 2 m aata hai, jo formula ka sahi check hai.

Q3. Define torque and give its unit and dimensions.

Answer: Torque about a point is the cross product of the position vector of the point of application of the force with the force vector, so its magnitude is force times the perpendicular distance from the point to the line of action. Its direction is given by the right hand rule. The unit is the newton metre and the dimensional formula is mass length squared time to the power minus two. Torque is not energy, although the dimensions are the same, because torque is a vector and energy is a scalar.

Hinglish: Torque r x F hai, magnitude force guna perpendicular distance, direction right hand rule se. Unit newton metre, dimensions [M L^2 T^-2]. Torque aur energy ke dimensions same hain par torque vector hai aur energy scalar, isliye torque ko joule nahi kehte.

Q4. State the two conditions of equilibrium for a rigid body.

Answer: A rigid body in equilibrium has zero net external force and zero net external torque about any point. The first condition prevents translational acceleration and the second prevents angular acceleration. Both are needed: two equal and opposite forces along the same line satisfy the force condition and produce no torque, while two equal and opposite forces along different lines produce a couple with zero net force but a non-zero torque, which rotates the body and is not equilibrium.

Hinglish: Equilibrium ke liye net force zero aur net torque zero dono chahiye. Force zero hone se translation nahi hoti aur torque zero hone se rotation nahi hoti. Same line par equal opposite forces se dono zero hote hain, par different lines par couple ban jaata hai jisme net force zero par torque non-zero hota hai, isliye woh equilibrium nahi hai.

Q5. Define moment of inertia and explain why it depends on the axis.

Answer: The moment of inertia about an axis is the sum of each mass element multiplied by the square of its perpendicular distance from the axis. It measures the resistance of the body to a change in its rotational motion about that axis. It depends on the axis because the distances of the mass elements change when the axis changes: a rod has a smaller moment about its centre than about its end, even though the mass is the same. Moving mass farther from the axis always increases the value.

Hinglish: Moment of inertia I = sum m r^2 hai, jahan r axis se perpendicular distance hai. Yeh rotational motion ke change ka virodh batata hai. Axis badalne par distances badal jaate hain, isliye rod ka I centre ke around end se chhota hota hai. Mass ko axis se door le jaane par I hamesha badhta hai.

Q6. State the parallel axis theorem and use it for a rod about its end.

Answer: The parallel axis theorem says that the moment of inertia about an axis is the moment about a parallel axis through the centre of mass plus the total mass times the square of the distance between the two axes. For a rod of mass M and length L, the centre value is M L squared over twelve and the distance from the centre to the end is L over two. The theorem then gives M L squared over twelve plus M L squared over four, which is M L squared over three, the standard end value.

Hinglish: Parallel axis theorem I = I_cm + M d^2 hai. Rod ke liye centre par ML^2/12, distance L/2, isliye end par ML^2/12 + ML^2/4 = ML^2/3 milta hai.

Q7. A disc of mass 2 kg and radius 0.5 m rotates at 4 rad s^-1. Find its rotational kinetic energy.

Answer: The moment of inertia of a disc about its central axis is half M R squared, which is half of 2 times 0.25, so 0.25 kg metre squared. The rotational kinetic energy is half the moment times the square of the angular speed, so half of 0.25 times 16, which is 2 J. Doubling the angular speed would quadruple the kinetic energy, because the energy depends on the square of the speed.

Hinglish: Disc ka I = 1/2 M R^2 = 1/2 x 2 x 0.25 = 0.25 kg m^2. Rotational KE 1/2 I omega^2 = 1/2 x 0.25 x 16 = 2 J. Angular speed double karne par KE chaar guna ho jaati hai.

Q8. Explain the conservation of angular momentum with the example of a skater.

Answer: When the net external torque on a system about an axis is zero, the total angular momentum about that axis stays constant. A skater spinning with arms out has a large moment of inertia and a small angular speed. Pulling the arms in reduces the moment of inertia, and since the product is fixed, the angular speed rises. The kinetic energy rises too, and the extra energy comes from the work the skater does in pulling the arms inward against the outward tendency of the mass.

Hinglish: Net external torque zero ho to angular momentum constant rehta hai. Skater arms bahar rakhe to I bada aur omega chhota. Arms andar karte hi I kam ho jaata hai, aur I omega fix hone se omega badh jaata hai. Kinetic energy badhti hai kyunki muscles work karti hain.

Q9. Why does a solid sphere reach the bottom of an incline before a ring of the same radius?

Answer: Both bodies lose the same potential energy, but it is shared between translational and rotational kinetic energy. The fraction that goes into rotation is set by the ratio of the moment of inertia to mass times radius squared, which is two fifths for a solid sphere and one for a ring. The ring puts a larger share of its energy into rotation, leaving less for translation, so it moves slower and takes longer. The acceleration formula g sin theta over one plus that ratio shows the same result directly.

Hinglish: Dono ka potential energy same girta hai par woh translation aur rotation mein bant jaata hai. Rotation ka hissa I/(mR^2) se tay hota hai, jo solid sphere ke liye 2/5 aur ring ke liye 1 hai. Ring zyada energy rotation mein kharch karti hai, isliye translation slow rehti hai. Acceleration formula g sin theta/(1 + I/(mR^2)) se bhi yahi dikhta hai.

Q10. A flywheel of moment of inertia 5 kg m^2 is at rest. A constant torque of 20 N m acts for 4 s. Find the angular speed and the rotational kinetic energy.

Answer: The angular acceleration is torque over moment of inertia, 20 over 5, which is 4 radian per second squared. The angular speed after 4 s is 16 radian per second. The rotational kinetic energy is half of 5 times 256, which is 640 J. The work done by the torque is the torque times the angular displacement, and the displacement is half times alpha times t squared, which is 32 radian, giving 20 times 32, the same 640 J.

Hinglish: Angular acceleration 20/5 = 4 rad/s^2, aur 4 s baad omega 16 rad/s. Rotational KE 1/2 x 5 x 256 = 640 J. Torque ka work tau x theta, jahan theta = 1/2 x 4 x 16 = 32 rad, isliye 20 x 32 = 640 J, same answer.

Quiz below: five questions on centre of mass, torque, moment of inertia, angular momentum and rolling. Attempt them before rereading the formula sheet.