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Question:

A hoop of radius r and mass m rotating with an angular velocity ω₀ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

rω₀/3

rω₀/2

rω₀/4

rω₀

Solution:

From conservation of angular momentum,
mr²ω₀ = mvr + mr²(v/r)
or m
r²ω₀ = 2mvr
or v = rω₀/2