Question

A uniform cord of length $25cm$ and mass $15g$ is initially stuck to a ceiling. Later, it hangs vertically from the ceiling with only one end still stuck. What is the change in the gravitational potential energy of the cord with this change in orientation? (Hint: Consider a differential slice of the cord and then use integral calculus.)

Answer 1

Consider a differential element of length $dx$ at a distance x from one end (the end that remains stuck) of the cord. As the cord turns vertical, its change in potential energy is given by

$dU=-\left(\lambda dx\right)gx$

where $\lambda =m/h$ is the mass/unit length and the negative sign indicates that the potential energy decreases. Integrating over the entire length, we obtain the total change in the potential energy:

$\Delta U=\int dU=-{\int }_0^h\lambda gxdx=-\frac{1}{2}\lambda g{h}^2=-\frac{1}{2}mgh.$

With $m=15g$ and $h=25cm,$ we have $\Delta U=-0.018J.$