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A rod of lenght 2m and mass 2 kg is lying on smooth horizental x-y plane along Y- axis with its center at origin O as shown in the figure . An impulse J of magnitude 10 Ns is applied perpendicular to AB at A.

Q.1The distance of point P from point A which is at rest just after the impact is :

(A) 2/3M

(B)4/3M

(C)1/2M

(d) 3/4M

Q.2 Coordinate of point A of the rod after time t=pi/30

(A)(root3/2+pi/9,1/2)

(b)

(c)(1/2+pi/6,0)

(D)(1+pi/6,0)

625_46961_JUHI.GIF

juhi ranjan , 13 Years ago
Grade
anser 1 Answers
Chetan Mandayam Nayakar

Last Activity: 13 Years ago

Let I be moment of inertia about axis perpendicular to x-y plane and passing through O.

(Linear) velocity of O=10Ns/2kg=5m/s

angular velocity=10*1/I

let answer be 'l'

5=(l-1)*angular velocity

consider A={x(t),y(t)},y(t)=t*1m*angular velocity*cos(ang.vel.*time))

x(t)=t*1m*angular velocity*sin(ang.vel.*time)+t*linear velocity

 

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