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Question

Find the distance covered by the bar till it stops
  1. s=1atanα
  2. s=1atan2α
  3. s=2atanα
  4. s=2atan2α

A
s=2atanα
B
s=2atan2α
C
s=1atan2α
D
s=1atanα
Solution
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From Newton's second law for the bar in projection form, Fx=mwx along x direction we get
mgsinαkmgcosα=mw
or, vdvdx=gsinαaxgcosα, (as k=ax),
or vdv=(gsinαaxgcosα)dx
or v0vdv=gx0(sinαxcosα)dx
So, v22=g(sinαxx22acosα) (1)
From (1) v=0 at either
x=0, or x=2atanα
As the motion of the bar is unidirectional it stops after going through a distance of 2atanα.
From (1), for vmax,
ddx(sinαxx22acosα)=0, which yields x=1atanα
Hence, the maximum velocity will be at the distance x=tanαa
Putting this value of x in (1) the maximum velocity,
vmax=gsinαtanαa
127828_130254_ans_7df2057946174aa69a2ca17140026691.png

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