Asked by Andrew
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?
Can someone please explain to me how to solve this problem?
Answers
Answered by
Steve
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
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
Answered by
Andrew
So because the question states "lowest point", the cos is negative?
Answered by
Steve
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.
I had to make some kind of assumption, in the absence of any specified by the problem.
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