A tractor tire has a diameter of 6 feet and is revolving at a rate of 45 rpm. At t=0, a certain point is at height 0. Write an equation to describe the height of the point above the ground after t seconds.
The answer is apparently d=-3cos(3pi/2)t +3
But I thought it's supposed to be sine since the height is at 0 when t=0, and why is w (3pi/2) instead of 90pi?
since cos(kt) has a maximum at t=0, -3cos(kt) has a minimum of -3 at t=0.
So, since the bottom of the tire touches the ground, the axle is 3 feet up, and if the point starts at the bottom of the circle, it is indeed given by 3-3cos(3pi/2 t). The height is from the ground, not from the axle.
Naturally, since sin(x) = cos(pi/2 - x), they are really the same thing, just starting at different times (phase shift).
If you use sin(kt), then 3+3sin(kt) starts at axle height, since sin(kt)=0 at t=0.
The reason for the 3pi/2 is that 45 rpm is 3/4 rev/second. That is the frequency, not the period. The period is 4/3 seconds. Since the period of cos(kt) is 2pi/k, we have
2pi/k = 4/3, so k = 3pi/2
To understand why the equation involves cosine instead of sine and why the angular frequency (ω) is (3π/2) instead of 90π, let's break down the problem step by step.
The height of the point above the ground can be described using a sinusoidal function. In this case, the tire is revolving, which creates periodic motion. The angular velocity (ω) of the tire is given by the rate of revolution, which is 45 rpm.
The angular velocity (ω) can be calculated in radians per second by converting from rpm. One revolution is equal to 2π radians, and there are 60 seconds in a minute, so:
ω = (45 rpm) * (2π radians/1 revolution) * (1 minute/60 seconds)
= (45 * 2π) / 60
= 3π/2 radians/second
Now let's consider the starting position of the point (when t = 0). It is given that the height is 0 at t = 0, which means the sinusoidal function needs to be shifted vertically.
Since the height is 0 when t = 0, the formula that describes the height of the point above the ground is of the form:
h(t) = A * sin(ωt + φ) + k
Where:
- A is the amplitude of the function (which determines the vertical range of the motion)
- ω is the angular frequency (which determines the speed at which the motion repeats)
- φ is the phase shift (which determines the starting position of the function)
- k is the vertical shift (which determines the equilibrium position)
In this case, the starting height is 0, so k = 0. The amplitude (A) is not given, but we can calculate it using the diameter of the tire.
The diameter of the tire is 6 feet, which means the radius (r) is 3 feet. The amplitude (A) of the motion is half the diameter, so A = 3 feet.
Now let's substitute the given values into the equation:
h(t) = 3 * sin((3π/2)t + φ) + 0
To determine the phase shift (φ), we need to consider the starting position of the point at t = 0. If the point is at its highest position, the phase shift is 0. However, if the point is at its lowest position, the phase shift is π. Since the equation given in the answer involves a cosine function instead of a sine function, it suggests that the point is at its lowest position at t = 0, and the phase shift is π.
Using the identity sin(θ + π) = -sin(θ), we can rewrite the equation:
h(t) = -3 * sin((3π/2)t) + 0
Finally, we can simplify the equation by removing the unnecessary terms:
h(t) = -3 * sin((3π/2)t)
Therefore, the correct equation to describe the height of the point above the ground after t seconds is:
h(t) = -3 * sin((3π/2)t)
To understand why the equation for the height of the point is d=-3cos(3π/2)t + 3, let's break it down step by step.
First, it's important to note that the equation uses cosine because the point's height is oscillating up and down periodically with time. When an object moves in a circular motion, like the point on the tractor tire, the height of the point can be represented by a cosine function.
The coefficient in front of the cosine function, -3, indicates the maximum height of the point above the ground. In this case, the maximum height is 3 feet, as given in the problem.
The angular frequency (ω) in the equation corresponds to the rate of revolution of the tire. In this case, the tire is revolving at a rate of 45 revolutions per minute. To convert this to radians per second, we need to multiply by 2π, because there are 2π radians in one revolution. So, the angular frequency (ω) is (45 rev/min) * (2π rad/rev) * (1 min/60 s) = π/2 rad/s. This is why the value of ω is 3π/2 in the equation.
The variable t represents time in seconds, indicating how long the tire has been revolving. By plugging in different values of t, we can find the height of the point at different points in time.
Finally, the "+3" term in the equation represents the initial height of the point above the ground. At t=0, the height of the point is 0 because it starts at ground level. However, in the equation, we want the height to be denoted as a positive value, so we shift the entire function up by adding 3.
Overall, the equation d=-3cos(3π/2)t + 3 describes the height of the point above the ground after t seconds, with -3cos(3π/2)t representing the oscillation of the point with time and +3 denoting the initial height.