meterstick (L = 1 m) has a mass of m = 0.132 kg. Initially it hangs from two sho
ID: 1446889 • Letter: M
Question
meterstick (L = 1 m) has a mass of m = 0.132 kg. Initially it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark. 1) What is the tension in the left string? In N 2) Now the right string is cut! What is the magnitude of the initial angular acceleration of the meterstick about its pivot point? (You may assume the rod pivots about the left string, and the string remains vertical) In rad/s2 3) What is the tension in the left string right after the right string is cut? In N
Explanation / Answer
1. As the strings are vertical, and placed syummetric, the tension in both of them is the same, hence
By force balance, 2T = 0.132*9.8
T = 0.6468 N
2. Initial angular acc = a
Initial torque = mg*0.25 = 0.132*9.8*0.25 = 0.3234 Nm
Moment of inertia of the bar about the pivot = ml^2/12 + m(0.25)^2 = 0.132/12 + 0.132(0.25)^2 = 0.01925 kgm^2
Ia =T
so a = T/I = 16.8 rad/s/s
3. Final tension in the other string = T
T = mg = 1.2936 N
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