A meter stick is held vertically with one end on the floor and is then allowed to fall. The speed of the other end when it hits the floor, assuming that the end of the stick does not slip and g=9.8 m/s2, is:
Correct answer: B. 5.4 m/s
- A. 3.2 m/s
- B. 5.4 m/s
- C. 7.8 m/s
- D. 9.2 m/s
Explanation
This is the following solution: Let m and l be the mass and the length of the meter stick respectively then its center will be situated at l/2 the angular speed and m.l pf the stick is W and I respectively The M.I of the rod about an axis passing through its one end I=(1/3)Ml² =(1/3)m(1m)² =1/3m According to the conservation of energy, Ei= EfK. Ei+ P. Ei= K. Ef + P. EsIn this case,0+ mgl/2 = (1/2)Iw²+0w²=mgl/I W=square root(mgl/I) The speed of the stick is, V=1W V=1 square root (mgl/I) But L is =1 and 1=1/3m V=square root(mgl/(1/3)m) V=square root (3gl) Substitute all the values in the above equation; V=square root (3(9.81m/s²)×1m) V=5.42m/s Therefore, the speed of the other end of the stick just before it hits the floor is 5.42m/s.
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