A father pulls his son, whos mass is 25 kg and who is sitting on a swing with ropes of length 3 meters, backward until the ropes make an angle of 33.6 degrees with respect to the vertical. He then releases his son from rest. What is the speed of the son at the bottom of the swinging motion? For this question i don't understand the logic behind L - Lcos33.6 degrees.
draw the figure. The length of the "swing" is from the piviot point to the bottome, or L. Now raise the swing to some angle theta. The hypotenuse is still L, but the downward length from the piviot to a point level with the swing is L*cosTheta.
Which with a little math and examining your diagram, the swing was raised L-LcosTheta
To understand the logic behind the equation L - Lcos(33.6 degrees) in this question, let's break down the situation and the forces acting on the son.
In the given scenario, the son is initially pulled backward by the father until the ropes make an angle of 33.6 degrees with respect to the vertical. At this point, the son is released from rest and starts swinging.
When the son is released, he undergoes circular motion due to the tension in the ropes. This tension provides the centripetal force required to keep the son moving in a circular path.
Now, let's consider the forces acting on the son at the bottom of the swinging motion, where his speed is maximum. At this point, two forces are acting on the son:
1. Tension (T): The tension in the swing ropes acts upward and provides the centripetal force required to keep the son moving in a circular path.
2. Weight (mg): The weight of the son acts downward due to gravity.
To find the speed of the son at the bottom, we need to equate the magnitude of the centripetal force (provided by tension) to the magnitude of the gravitational force (weight).
The equation for the centripetal force (Fc) is given by:
Fc = m * v^2 / r
Where:
m = mass of the son
v = velocity (speed) of the son
r = radius of the circular path (length of ropes)
In this case, the radius of the circular path is given as L - Lcos(33.6 degrees).
Why is it L - Lcos(33.6 degrees)?
Imagine a vertical line starting from the top of the swing ropes when they are at rest. This vertical line bisects the angle that the ropes make with the vertical when the son is pulled backward. The length of the ropes (L) can be divided into two components: the vertical component (Lcos(33.6 degrees)) and the horizontal component (Lsin(33.6 degrees)).
When the son is at the bottom of the swinging motion, the vertical component of the ropes length (Lcos(33.6 degrees)) is subtracted from the total length of the ropes (L) since the son is not at the highest point of the swing.
By substituting the values in the equation for the centripetal force and equating it to the weight (mg), we can solve for the velocity (v):
m * v^2 / (L - Lcos(33.6 degrees)) = m * g
Canceling out the mass (m) from both sides, we get:
v^2 / (L - Lcos(33.6 degrees)) = g
Rearranging the equation, we obtain:
v^2 = g * (L - Lcos(33.6 degrees))
Taking the square root of both sides, we get:
v = √(g * (L - Lcos(33.6 degrees)))
Thus, this equation allows us to calculate the speed of the son at the bottom of the swinging motion.