Question:
The bob of a simple pendulum of length $1 \mathrm{~m}$ has mass $100 \mathrm{~g}$ and a speed of $1.4 \mathrm{~m} / \mathrm{s}$ at the lowest point in its path. Find the tension in the string at this instant.
Solution:
At lowest point
$T=m g+\frac{m v^{2}}{R}$
$=(0.1)(10)+\frac{(0.1)(1.4)^{2}}{(1)}$
$\mathrm{T}=1.2 \mathrm{~N}$
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