Calculating angles?

Question

The rope of a swing is 3.30 m long. Calculate the angle from the vertical at which a 81.0 kg man must begin to swing in order to have the same KE at the bottom as a 1450 kg car moving at 1.53 m/s (3.42 mph).

Answer 1

A 1450 kg car moving at a speed of 1.53 m/s has a Kinetic energy of,
KE = ½ mv^2
KE = ½ (1450 kg) * (1.53 m/s)^2
KE = 1697 Joules

In order for the 81.0 kg man to have a KE equal to this he must have originally been at some height (h) above the ground such that his gravitational Potential Energy equaled the car’s Kinetic Energy.
PE = mgh
PE = (1697 Joules) = (81.0 kg) * (9.81 m/s^2) * h,
Solving for h,
h = (1697 Joules) / ((81.0 kg) * (9.81 m/s^2))
h = 2.14 meters above the ground.

Since the man was on a swing, when he is a certain distance, h, off the ground, the rope on the swing will make some angle (theta) with the vertical.
The swing’s rope is 3.30 meters long, so if the man is 3.30 meters off the ground, the angle with respect to the vertical will be 90° since he is at the very top of the swing. If the man is at the very bottom of the swing (we will assume he is scraping the ground at this point), the angle with the vertical will be 0
If you draw a diagram of the situation (which I highly recommend doing), we can see that the height off the ground will be a function of the cosine of the angle. The length of the rope multiplied by the cosine of the angle will give the distance of the man down the pole of the swing. But we want the distance up from the ground. Not to worry, we need to do is subtract the distance down from the top of the swing from the total height of the swing (which is also the length of the rope, we assume).

Rope length * cos (theta)
Will give us the distance from the vertex of rotation of the swing (down the pole), but we want the distance up from the ground.
So,
Height = Length of Rope – Length of Rope * cos (theta)
Height = Length (1 – cos (theta))
We know the height and we know the rope length, but we want to find theta.
Solving for theta we see that,
1 - (h / l) = cos (theta),
Now taking the arccos of both sides gives,
Theta = arccos (1 - (h / l)) = cos^-1 1 - (h / l))

Plugging in the values we know,
Theta = arccos (1 – (2.14 m / 3.30 m))
Theta = 69.4° to the vertical

Answer 2

Negelcting swing mass, assuming man begins to swing from a position at rest. Calling the bottom of swing trajectory as height = 0.

car Kinetic Energy = .5m(v^2)
=.5(1450)(1.53^2)
=1697 Joules

Energy initial of man = mgh
Energy final of man = .5m(v^2) = 1697

Energy initial = Energy final
mgh=1697
(81)(9.8)(h)=1697
h=2.14 meters

the height of the starting point of the swing is directly proportional to the angle:

for example if the height off the ground were the length of the swing, the angle would be:
(3.3/3.3) = 1
1(90) = 90 degrees...

since our height is 2.14...
(2.14)/(3.3)=.648
.648(90)= 58.3 degrees (like, you would pull the man 58.3 degrees back from the at rest vertical)

Answer 3

Calculate the KE needed and then the height so that mgh equals this KE. Then the angle is
arccos(Θ) = (3.3-h)/3.3

HTH ☺


Doug