Chapter 6, Problem 92GP

To determine

a. The angle $\theta$ when the person’s releases the rope.

b. The tension in the rope just before the person’s releases it.

c. Maximum tension in the rope during the swing.

Answer to Problem 92GP

Solution:

a. The angle $\theta$ is $35°$ .

b. The tension of the rope is 450 N.

c. The maximum tension is 651 N.

Explanation of Solution

Given

The mass of the student is $m=56\text{kg}$ .

The speed of the student is $v=6.0\text{m/s}$ .

The length of the rope is $l=10.0\text{m}$ .

Formula used

The running kinetic energy of the student is converted to potential energy when he makes a swing. Therefore, the formula to be used is $\frac{1}{2}m{v}^2=mgl\mathrm{.....}(1)$

Here, l is the height from the lowest point.

To calculate the tension of the rope, the necessary formula is $T=mg\cos \theta \mathrm{....}(2)$ .

The tension is maximum at the beginning of the swing. Therefore, its value at that position is $T_m=\frac{m{v}^2}{r}+mg\cos \theta .\mathrm{....}(3)$

Calculations

a.

The height reached by the student is:

  $\begin{array}{l}l=\frac{v^2}{2g}\\ l=\frac{{(6.0)}^2}{2\times 9.8}\text{m}\\ l=1.84\text{m}\end{array}$

Therefore, from the figure,

  $\begin{array}{l}\cos \theta =\frac{10.0-l}{10.0}\\ \cos \theta =\frac{10.0-l}{10.0}\\ \theta ={\cos }^{-1}\left(\frac{10.0-1.84}{10}\right)\\ \theta =35°\end{array}$

Part b

From the figure above:

  $\begin{array}{l}T=mg\cos \theta \\ T=56\times 9.8\times \cos 35°\\ T=450\text{N}\end{array}$

Part c

From equation (3),

  $\begin{array}{l}{T}_m=\frac{m{v}^2}{r}+mg\cos \theta \\ T_m=\frac{56\times {6.0}^2}{10.0}+56\times 9.8\times \cos 35°\\ T_m=651\text{N}\end{array}$

Conclusion:

a. The angle $\theta$ is $35°$ .

b. The tension of the rope is 450 N.

c. The maximum tension is 651 N.