Question:
Suppose the bob of the previous problem has a speed of $1.4 \mathrm{~m} / \mathrm{s}$ when the string makes an angle of $0.20$ radian with the vertical. Find the tension at this instant. You can use $\cos \theta=1-\theta^{2} / 2$ and $\sin \theta=$ 0 for small $\theta$.
Solution:
At angle $\theta$
$T=m g \cos \theta+\frac{m v^{2}}{R}$
$\mathrm{T}=\operatorname{mg}\left(1-\frac{\theta^{2}}{2}\right)+\frac{m v^{2}}{R}$
$T=(0.1)(10)\left(1-\frac{(0.2)^{2}}{2}\right)+\frac{(0.1)(1.4)^{2}}{(1)}$
$\mathrm{T} \simeq 1.16$
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