Physics · System Of Particles And Rotational Motion · NEET
During coupling, the only torque is the internal friction between the two disc faces. Internal torques come in action-reaction pairs, so they cancel and do NOT change total L. That is why L1 + L2 stays constant. But that same friction slides the surfaces against each other and turns some rotational kinetic energy into heat. So use L conservation to find ω, never energy conservation. Energy is lost unless ω1 = ω2 to begin with.
No, and NEET mixes them up on purpose. Same-axis (coaxial, face-to-face): they end at ONE common ω, use L1 + L2 = (I1+I2)ω. Rim-in-contact (edges touch, like gears): the discs keep separate axes and DO NOT reach one common ω. Instead their contact-point speeds match: R1·ω1 = R2·ω2, so ω1/ω2 = R2/R1. Read the question: 'brought face-to-face' or 'common axis' = coaxial; 'edges/rims touch' = surface-speed matching.
Step 1: write total angular momentum before = I1·ω1 + I2·ω2 (add signs if directions differ). Step 2: after coupling both move as one body with moment of inertia I1 + I2 at speed ω. Step 3: set before = after: I1·ω1 + I2·ω2 = (I1+I2)·ω. Step 4: solve ω = (I1·ω1 + I2·ω2)/(I1+I2). For two identical discs (I1 = I2 = I) this becomes the simple average ω = (ω1+ω2)/2.
Give one direction + and the other -. If disc 2 spins the opposite way, ω2 is negative in the formula: ω = (I·ω1 - I·ω2)/(2I). If they are equal and opposite (ω1 = ω, ω2 = -ω), the common speed is zero - they stop each other. All the initial kinetic energy is then lost as heat.
Yes. Energy lost = KE(before) - KE(after). For two identical discs, this reduces to a clean form: loss = (1/4)·I·(ω1 - ω2)². Notice it depends on the DIFFERENCE in speeds, so if ω1 = ω2 there is zero loss (nothing slides), and the loss is largest when they spin oppositely.
Two discs of equal moment of inertia I, rotating about their common axis with angular speeds ω1 and ω2, are brought face-to-face into contact (axes coinciding). The loss of energy in the process is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
ω = (I1·ω1 + I2·ω2)/(I1 + I2). For two identical discs it simplifies to the average ω = (ω1 + ω2)/2.
Yes. The coupling force is internal friction, which produces no net external torque, so total angular momentum before equals total after.
The two faces slide against each other while gripping. This kinetic friction converts rotational kinetic energy into heat, so final KE is always less than initial KE (unless the discs already had the same speed).
When discs touch at their rims and keep separate axes, they do NOT share one ω. Their contact points have equal linear speed: R1·ω1 = R2·ω2, giving ω1/ω2 = R2/R1 (smaller disc spins faster).
Take one direction as negative in the formula. Equal and opposite speeds give ω = 0 (they stop), and all the kinetic energy is lost as heat.
