Exercicios Resolvidos do Halliday sobre Rotação- Cap11- Exercicio 79

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79. We choose positive coordinate directions (different choices for each item) so that each is accelerating
positively, which will allow us to set a2 = a1 = Rα (for simplicity, we denote this as a). Thus, we choose
rightward positive for m2 = M (the block on the table), downward positive for m1 = M (the block
at the end of the string) and (somewhat unconventionally) clockwise for positive sense of disk rotation.
This means that we interpret θ given in the problem as a positive-valued quantity. Applying Newton’s
second law to m1, m2 and (in the form of Eq. 11-37) to M , respectively, we arrive at the following three
equations (where we allow for the possibility of friction f2 acting on m2).
m1g − T1 = m1a1
T2 − f2 = m2a2
T1R− T2R = Iα
(a) From Eq. 11-13 (with ω0 = 0) we find
θ = ω0t+
1
2
αt2 =⇒ α = 2θ
t2
.
(b) From the fact that a = Rα (noted above), we obtain a = 2Rθ/t2.
(c) From the first of the above equations, we find
T1 = m1 (g − a1) = M
(
g − 2Rθ
t2
)
.
(d) From the last of the above equations, we obtain the second tension:
T2 = T1 − Iα
R
= M
(
g − 2Rθ
t2
)
− 2Iθ
Rt2
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