A rod AB of length 2m and mass 2kg
is lying on smooth horizontal x - y
plane with its centre at origin O as
shown figure. An impulse J of
magnitude 10 Ns is applied
perpendicular to AB at A. Then
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A rod AB of length 2m and mass 2kg
is lying on smooth horizontal x - y
plane with its centre at origin O as
shown figure. An impulse J of
magnitude 10 Ns is applied
perpendicular to AB at A. Then
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Answer:
1/3 .l think this is the answer for these question
Your question is incomplete without the options and diagram.
The correct question would be
A rod AB of length 2m and mass 2kg is lying on smooth horizontal x - y plane with its centre at origin O as shown figure. An impulse J of
magnitude 10 Ns is applied perpendicular to AB at A. Then the distance of p from the centre of the rod after the rest would be __________.
Choose the correct option
i) 2/3 m.
ii) 1/4 m.
iii) 1/3 m.
iv) 1/2 m.
(Refer the attachment for the diagram)
The correct option is iii) 1/3 m.
Then the distance of p from the centre of the rod after the rest would be 1/3m.
v = j/m = 10/2 = 5m/s.
ω = j (1/2) / 1
= j1 / 2 (m /1 ^ 2) ^ 2
= 6j / m1
= 6*10/2*2
= 15 rad/s
At a distance r from the centre v, r and ω should cancel each other at the point P.
Hence
r = v/ω
= 5/15
= 1/3 m.
Therefore
The correct option is iii) 1/3 m.