Physics, asked by Sharma2547, 1 year ago

A ring or a cylinder falls early when thrown from same height

Answers

Answered by dharmendraamanp30nu0
0
Before answering, we need to realize that there is an underlying assumption that all the objects have the same mass. Consequently, the bodies have different densities.

Given that, lets start our analysis. For any process, energy is conserved. For this case, there are 3 different types of energy: Translational kinetic energy (energy because it is moving forward), Rotational kinetic energy (due rotation) and potential energy. More formally, we can write as:

E=mgh+mv2/2+Iω2/2E=mgh+mv2/2+Iω2/2

Another assumption made here is that the objects are rolling without slipping. Mathematically, this means:

v=rωv=rω

If this assumption is accounted for, our original equation becomes:

E=mgh+mv2/2+Iv2/2r2E=mgh+mv2/2+Iv2/2r2

Now lets analyze. For all the bodies mmhhrr and gg are the same. Difference lies in II and it is inversly proportional to the velocity (or rotation). II is given by 

Σmr2Σmr2

For the hollow cylinder, the mass is placed the farthest from the center and thus rr is large, II is large, and consequently it is slow. For the case of the sphere, the mass distribution occurs more closer to the center than in any other case, thus has lower II and highest velocity. There is an intuitive answer to this as well, but I guess the math will provide you with good understanding of the equations.
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