Question

A uniform cord of length25 cmand mass15 gis 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)

Short answer

The change in the gravitational potential energy of the cord, $∆U=-0.018\mathrm{J}$

Step-by-step solution

Step 1: Given

1. The mass of cord is, $m=15\mathrm{g}=0.015\mathrm{kg}$
2. The length of a cord is, $L=25\mathrm{cm}=0.25\mathrm{m}$

Determining the concept

Here, mass of the cord is distributed all over its length. So, consider the linear mass density that is the ration of mass per unit length. Using this density in the equation of potential energy, find the change in potential energy for the small length element of the cord. Then by integrating it from 0 to L (length of the cord), the total change in potential energy can be found.

Formula is as follow:

$∆dU=\left(-kdy\right)gy$

Determining thechange in the gravitational potential energy of the cord

As per the given hint, let us consider the differential slice of the cord with length ‘dy’.

Let ‘k’ be the ration of mass per unit length of the cord.

Now, consider one slice at distance ‘y’ from the ceiling.

The change in potential energy of this slice will be,

$∆dU=\left(-kdy\right)gy$

where, k is themass/unit length that is k = m/L.

So, to get the total change in potential energy, integrate the equation from 0 to L.

$$\begin{gathered} {\int }_0^L∆dU={\int }_0^L-kdy\times gy \\ ∆U=-kg{\int }_0^{L}ydy \\ ∆U=-kg{\left[\frac{y^2}{2}\right]}_0^L=-kg\times \frac{L^2}{2} \\ ∆U=-\frac{m}{L}\times g\times \frac{L^2}{2}=-\frac{1}{2}mgL \end{gathered}$$

By substituting the given values,

$$\begin{gathered} ∆U=-\frac{1}{2}\times 0.015\times 9.81\times 0.25 \\ ∆U=-0.018\mathrm{J} \end{gathered}$$

Hence, the change in the gravitational potential energy of the cord is, $∆U=-0.018\mathrm{J}$

Therefore, the change in potential energy of the cord by considering its linear mass density and integrating the equation for limit 0 to L.