Prove that angular momentum of a particle is equal to the twice the product of mass and areal velocity
Answers
Consider a mass 'm' and linear momentum p=mv rotating in the X-Y plane. At any time, let this particle lie at ' P' whose position vector OP= r.
In a small intervel 'dt' let the particle reach ' Q' when OQ=(r + dr). Join PQ which represents the displacement dr of the particle is small time dt.
Area swept over by the position of the particle in a small time dt,
Area OPQ = dA = 1/2 (OP × PQ)
= 1/2(r ×dr)
Dividing both side by dt,
dA/dt = 1/2{(r) × (dr/dt)} = 1/2(r × v)
= 1/2{(r) ×(p/m)}
= 1/2(L/m) ------->
From the last equation
Where L is the angular momentum,
Therefore
L = 2m × (dA/dt)
L = 2m × areal velocity
Hence proved
Nb; Refer photo attached
HOPE YOU UNDERSTAND!
Answer:
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