Will Rogers spun a lasso in a vertical cycle. The diameter of the loop was 6 feet, and the loop spun 50 times each minute. If the lowest point on the rope was 6 inches above the ground, write an equation to describe the height of this point above the ground after t seconds.
Can someone please explain to me how to solve this problem?
If the point in question is at the lowest height at t=0 minutes, then
y = -cos(kt)
diameter is 6, so the radius is 3
y = -3cos(kt)
the lowest point is 6" = 0.5 ft, so the axis of the loop is 3.5 ft off the ground:
y = 3.5 - 3cos(kt)
Since cos(kt) has period 2π/k = 1/50,
k = 100π
y = 3.5 - 3cos(100πt)
y=feet, t=minutes
No, it is negative because I assumed that at t=0 the point is at its lowest elevation. (As my first sentence stated.) If we had started when the point was level with the circle's center, we'd have sin(kt). If we started with the point at its maximum height, then we'd have cos(kt).
I had to make some kind of assumption, in the absence of any specified by the problem.
So because the question states "lowest point", the cos is negative?
To solve this problem, we need to understand the motion of the loop as it spins.
First, let's find the height of the lowest point on the rope above the ground at any given time.
We know that the diameter of the loop is 6 feet, so the radius (half the diameter) is 3 feet. The lowest point on the rope will be at a distance of 3 feet above the ground. However, we need to convert this to inches, as the given height is in inches.
Since 1 foot is equal to 12 inches, the lowest point on the rope is 3 * 12 = 36 inches above the ground.
Now, let's consider the vertical motion of the loop. We are told that the loop spins 50 times per minute. This means that for each complete cycle, the time taken is 1/50th of a minute, or 1/50th of 60 seconds. So, the time taken for one cycle is 60/50 = 6/5 seconds.
Let's assume that at t = 0 seconds, the lowest point on the rope is at its maximum height (6 inches above the ground).
As the loop spins, this lowest point will go up and down. The motion is periodic, and we can model it with a sinusoidal function. Specifically, we can use the sinusoidal function for vertical motion:
h(t) = A * sin(ωt + φ) + D
Where:
- h(t) is the height of the lowest point on the rope above the ground at time t,
- A is the amplitude of the motion,
- ω is the angular frequency of the motion,
- φ is the phase shift (horizontal displacement), and
- D is the vertical displacement.
In our case, the amplitude A is half the diameter of the loop, which is 3 feet or 36 inches. The vertical displacement D is the initial height of the lowest point on the rope, which is 6 inches.
The angular frequency ω is related to the number of cycles per second (50) as follows:
ω = 2π * cycles per second
Since we have 50 cycles per minute, we can convert this to cycles per second by dividing by 60:
ω = (50 cycles/minute) * (1 minute/60 seconds) = 5/6 cycles/second
Therefore, the equation to describe the height of the lowest point on the rope above the ground after t seconds is:
h(t) = 36 * sin((5/6)t + φ) + 6
The phase shift φ depends on the initial position of the rope at t = 0. If the rope starts at its maximum height (6 inches above the ground), then φ = 0. If it starts at its minimum height (30 inches below the highest point), then φ = π.
Depending on the given information, you can adjust φ accordingly to represent the correct starting position of the rope.