Question

A person of mass 55 kg swings on a rope length 4 m from rest (when the rope makes an angle of 30 degrees with the vertical) and collides perfectly inelastically with a 70 kg person on the ground when the rope is exactly vertical. Determine the maximum angle the rope makes with the vertical after the 55 kg person picks up the 70 kg person.

Answer 1

Answer:

θ = 19.66°

Explanation:

To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:

$U=mgh$

h: distance to the ground

g: gravitational acceleration = 9.8m/s^2

m: mass of the first person = 55 kg

In the image attache below you can notice that the height h is:

$h=4-4cos(30\°)=0.53m$

Then, the potential energy is:

$U=(55kg)(9.8m/s^2)(0.53m)=285.67J$

When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:

$U'=(m_1+m_2)gh'=285.67\ J$

From the previous equation you can get h':

$h'=\frac{285.67J}{(55kg+70kg)(9.8m/s^2)}=0.2332m$

Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:

$h'=4-4cos(\theta)\\\\\theta=cos^{-1}(\frac{4-h'}{4})=cos^{-1}(\frac{4-0.2332}{4})=19.66\°$

hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°