Center of MassHard
Question
In a vertical plane inside a smooth hollow thin tube a block of same mass as that of tube is released as shown in figure. When it is slightly disturbed it moves towards right. By the time the block reaches the right end of the tube, displacement of the tube will be (where ′R′ is mean radius of tube). Assume that the tube remains in vertical plane.
Options
A.
B.
C.
D.R
Solution
Since there is no ext. force on system
m (R − x) + m (− x) = 0
x = R/2.
Alternate : Let the tube displaced by x towards left, then ;
mx = m (R − x) ⇒ x = R/2
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