Key relationship
mvb = (Ml²/3+mb²)ω; Kafter = ½Iω².
Physics · Class 11
During a brief collision the hinge can exert an impulse, so linear momentum of particle plus rod need not be conserved. Its moment about the hinge is zero, allowing angular momentum about that point to determine the final rotation.
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mvb = (Ml²/3+mb²)ω; Kafter = ½Iω².
During a brief collision the hinge can exert an impulse, so linear momentum of particle plus rod need not be conserved. Its moment about the hinge is zero, allowing angular momentum about that point to determine the final rotation.
A hinge can supply external linear impulse even when angular momentum about the hinge is conserved.
Particle hits perpendicular to the rod and sticks at distance b. Hinge impulse has zero moment about the hinge, so angular momentum there is conserved during impact. Gravity is ignored after impact.
Choose an appropriate conservation law for a constrained impact.
Try both scenarios, change one control, and predict the result before checking the graph and readouts.
Change one input while keeping the others fixed, as you would in a controlled experiment.
m=0.5 kg, v=4 m/s, b=1 m gives L=2 kg·m²/s. For M=2 kg and l=2 m, I=2×4/3+0.5=3.167 kg·m² and ω=0.632 rad/s.
The model connects strike a hinged rod with an observable reading. Check the equation and assumptions before applying it.
SI units · algebra · graph reading
Which quantity is conserved during this sticking impact?
The hinge impulse has no lever arm about the hinge; the impact is inelastic and loses kinetic energy.
Course connection: Class XI Chapter 6. Consult the roadmap for CBSE, NEET, JEE Main and Advanced scope.