Question

A ring of mass m is just loosely fit on a frictionless rod AB of length 2L which
rotates in horizontal plane with constant angular velocity  about a vertical axis as
shown in fig. Initially the ring is located at a distance L from end A. Now the ring
left free and allowed to slide along the rod. In the subsequent motion, choose
correct options.

Answer 1

Answer:

Explanation:

Solution :

Angular impulse =

change in angular momentum

∴(J×r⊥)=Iω

∴

(

J

×

r

⊥

)

=

I

ω

where J=

J

=

linear impulse =Iωrt

=

I

ω

r

t

=(m)(2l3).ωl

=

(

m

)

(

2

l

3

)

.

ω

l

=43m/ω