# A stunt man whose mass is 70 kg swings from the end of a 4.4 m long rope along the arc of a...

## Question:

A stunt man whose mass is 70 kg swings from the end of a 4.4 m long rope along the arc of a vertical circle. Assuming that he starts from rest when the rope is horizontal, find the tensions in the rope that are required to make him follow his circular path at each of the following points.

(a) at the beginning of his motion

(b) at a height of 1.5 m above the bottom of the circular arc

(c) at the bottom of the arC

## Energy conservation principle:

The energy conservation principle consists of a physical law that states that if no external forces act on the system, the total energy of the system should conserve. Energy is not created nor destroyed, it only transforms. This means that a system may present energy transformations, but the overall total energy of the system will remain the same. The total energy of a system can be determined by the sum of the total kinetic and potential energy of the system.

$$E_t=K_e+U_e\\ E_t=0.5mv^2+mgh $$

## Answer and Explanation:

#### Part a):

We gather the information when the tension of the rope forms an angle of 90°. We have the following information.

**Information:**

- Mass of the stuntman: {eq}m=70\ kg {/eq}

- Length of the rope, and initial height: {eq}L=4.4\ m {/eq}

- Speed at initial position: {eq}v_i=0\ m/s {/eq}

- Height at an angle {eq}\theta {/eq}: {eq}H=L(1-\cos\theta) {/eq}

**Energy conservation principle at any angle {eq}\theta {/eq}:**

{eq}E_i=E_f\\ U_{gi}=K_{ef}+U_{gf}\\ mgL=\dfrac{1}{2}mv^2+mgh\\ mgL=\dfrac{1}{2}mv^2+mgL(1-\cos\theta)\\ 0=0.5mv^2-mgL\cos\theta\\ v=\sqrt{2gL\cos\theta} {/eq}

We proceed to analyze the centripetal force working on the stuntman.

{eq}F_c=m\dfrac{v^2}{r}\\ T-mg\cos\theta=m\dfrac{v^2}{L}\\ T=mg\cos\theta+m\dfrac{v^2}{L}\\ T=m(g\cos\theta+\dfrac{v^2}{L})\\ T=m(g\cos\theta+\dfrac{(\sqrt{2gL\cos\theta})^2}{L})\\ T=m(g\cos\theta+2g\cos\theta)\\ T=3mg\cos\theta\\ {/eq}

As the initial angle forms 90°, the tension will be of:

{eq}T=3mg\cos90\\ T=3mg(0)\\ T=0 {/eq}

#### Part b):

We determine the angle when the stuntman is 1.5 meters above. We use trigonometric ratios:

{eq}\cos\theta=\dfrac{L-h}{L}\\ \cos\theta=\dfrac{4.4-1.5}{4.4}\\ \theta=\cos^{-1}(0.66)\\ \theta=48.7^\circ {/eq}

We calculate the tension using the formula we deduced in the previous part.

{eq}T=3mg\cos\theta\\ T=3\times 70\ kg\times9.8\ m/s^2\cos48.7\\ T=1358.28\ N {/eq}

#### Part c):

At the bottom of the arc, the angle formed is of 0°

{eq}T=3mg\cos\theta\\ T=3\times 70\ kg\times9.8\ m/s^2\cos0\\ T=2058\ N {/eq}

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