Physic Labs

Problem 1

A simple pendulum consists of a small bob of mass m=0.50m=0.50 kg on a light inextensible string of length L=1.25L=1.25 m attached to a fixed support. The string is displaced to one side until it makes 60∘60^\circ with the vertical and is released from rest. Neglect resistance and take g=10g=10 m/s2^2. (a) Find the bob's height above the lowest point just before release. (b) Find its speed through the lowest point. (c) Find the string tension there. (d) Find the angle to the vertical where kinetic energy equals potential energy above the lowest point.
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12mv2=mgh02⇒v=gh0=10×0.625=2.50 m/s\frac12mv^2=\frac{mgh_0}{2}\Rightarrow v=\sqrt{gh_0}=\sqrt{10\times0.625}=2.50\ \mathrm{m/s}
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Since K=UK=U and the total is mgh0mgh_0, each energy form equals mgh0/2mgh_0/2. Solving 12mv2=mgh0/2\frac12mv^2=mgh_0/2 gives the speed there, independent of mass.