Answer to Question #126252 in Physics for Sammy W-S

Question

starting from rest a 50 kg person swings along a circular arc from a rope over a lake. the final point is the same height as the initial point. the distance of the rope is 6.4m. Ignore air resistance and the mass and elasticity of the rope.
a) The centre of mass of the person standing on the platform is 4 m above the surface of the water. calculate gravitational potential energy when the person is at the initial point relative to the waters surface.
b) The centre of mass of the person when they are in the middle/ lowest point along the arc and closest to the water at 2.4 m above the surface of the water. calculate the person Speed at this point.
c) Compare his impact speed with the water when he lets go of the rope at the lowest point of the arc, and at the final point of the ark. is the velocity at the lowest point; greater than the velocity at the final point, lesser than, or equal to.

Answer 1

The period of oscillations of the person in the very first problem is

$T=2\pi\sqrt{\frac{l}{g}}=2\pi\sqrt{\frac{6.4}{9.8}}=5.1\text{ s}.$

a) The gravitational potential energy when the person is 4 meters above the waters surface is

$E_{Pi}=mgh=50\cdot9.8\cdot4=1960\text{ J}.$

b) According to conservation of energy principle:

$E_{Pi}=E_{Pf}+E_K,\\ E_{Pi}=E_{Pf}+\frac{mv_x^2}{2},\\ v_x=\sqrt{\frac{2(E_{Pi}-E_{Pf})}{m}}=\\\space\\ =\sqrt{\frac{2(1960-50\cdot9.8\cdot2.4)}{50}}=5.7\text{ m/s}.$

c) At the final point of the ark, the velocity is 0 because at this point the direction of motion changes to opposite.

The vertical component of velocity when the person hits the water is

$v_v=\sqrt{2gh_f}=\sqrt{2\cdot9.8\cdot2.4}=6.9\text{ m/s}.$

Since the person released the rope at the lowest point of the ark, the tangential velocity ads to this vertical component as a vector, and the resulting speed of impact is

$v=\sqrt{v_y^2+v_x^2}=\sqrt{6.9^2+5.7^2}=8.9\text{ m/s}.$