Two smooth spheres "A" and "B" are at opposite sides of a diameter of a circular smooth tube of radius R. The tube is placed on horizontal floor. At t=0 sphere "A" moves with constant angular velocity "w" and "B" with "2w" in the clockwise direction.During collission the velocities are exchanged. a) Find the time taken for first second and third collission. b) Find the distance travelled by the sphere "A" before second collission

68 Points
13 years ago

Dear ajit ,

you can apply the concept of relative velocity in case of angular motion also. so here the angular velocityof b w.rt a is w in clockwise direction.

therefore

a) for first collision - time taken is the time when the body b moves a full circle w.r.t to a if we are applying relative concepts ....... so  wrelative   = vrelative/R ; wrelative   =  R/Rt1 ;    t=1 / wrelative

therefore  t1 =  1/w

for third collision : consider the scenario when the first hit takes , the velocities are exchanged i.e the angular velocities are also exchanged . so now A has 2w and B has w as angular velocity. so now in the same time A will hit B from behind. So , taking symmetry third collision shud take 3t1   as time. therefore  t3 = 3t1.  i.e  3/w

b)before first collision the angular velocity of A was w . so distance travelled  =  velocity * t1

=   wR * 1/w =    R

between first and second collision -- distance travelled  =  2WR*1/w =  2R

so total distance travelled  =  3R

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