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A THIN ROD OF LENGTH 1M IS FIXED IN A VERTICAL POSITION INSIDE A TRAIN, WHICH IS MOVING HORIZONTALLY WITH CONSTANT ACCELERATION 4M/S2. A BEAD CAN SLIDE ON THE ROAD AND FRICTION COEFFICIENT BETWEEN THEM IS 0.5. IF THE BEAD IS RELEASED FROM REST AT THE TOP OF THE ROD, IT WILL REACH THE BOTTOM IN .............................................
accleration of train = 4m/s2 bead will tend to move in horizontal direction due to seudo force = ma normal reaction bw rod & bead = ma (a is accleration of train) friction = u(N) = uma weight = mg net downward force = mg-uma m(net accleration) = m(g-ua) net accleration(A) = g - ua = 10-4*0.5 = 8m/s2 time taken by particle to reach the bottom is t s = At2/2 (s=L =1m (given) 1 = 8t2/2 t = 1/2 sec
accleration of train = 4m/s2
bead will tend to move in horizontal direction due to seudo force = ma
normal reaction bw rod & bead = ma (a is accleration of train)
friction = u(N) = uma
weight = mg
net downward force = mg-uma
m(net accleration) = m(g-ua)
net accleration(A) = g - ua = 10-4*0.5 = 8m/s2
time taken by particle to reach the bottom is t
s = At2/2 (s=L =1m (given)
1 = 8t2/2
t = 1/2 sec
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